Application of Algebra

This section is a continuation of what was discussed in year 1, sets and binomial expansion. De Morgan’s Laws is an important theorem in sets as its application transcends beyond union, intersection and complements. These laws are central to Boolean Algebra where it is used to simplify expressions. In Digital Circuit Design, it is used to transform logic gates with the intention of not only optimising digital circuitry but also reducing cost and saving space. In Computer programming, it is used to simplify logical expressions in code, which can make code cleaner, simpler, and more readable. Database systems use these laws to optimise queries in SQL (Structured Query Language), which enhances data retrieval more efficiently.

The Binomial Theorem is a strong tool that makes the process of expanding binomial expressions much simpler. In developing this theorem, you further your algebra skills and prepare yourselves for higher classes of mathematics.

The Binomial Theorem is applied in finance, investment, probability, statistics, computer science, genetics, engineering, game theory, marketing, and physics. It helps in calculating compound interest, risk assessment, modeling success/failure outcomes, understanding data structure complexities, predicting inheritance probabilities and enhancing decision-making in real-world scenarios.

KEY IDEAS

• Binomial Coefficients: The coefficients ⁿCᵣ represent the number of ways to choose r elements from n elements and can be found using Pascal’s Triangle. ⁿCᵣ = !_______ (n − r)!r !

• Terms in the Expansion: Each term in the expansion is of the form: ⁿCᵣ aⁿ−k bᵏ• The Binomial Theorem provides a formula for expanding expressions of the form (a + b)ⁿwhere n is a non-negative integer.

• The complement of an empty set (ϕ) is equal to the universal set (μ).

• The intersection of a set A and its complement (A′) is equal to the null set (∅ )

• The union of a set A and its complement (A′) is equal to the universal set ( μ )

De Morgan’s Laws are rules which are used to relate intersections and unions through complements. Generally, according to De Morgan’s Laws:

1. The complement of the union of sets is equal to the intersection of the individual complements.

This means that if A, B and C are sets, (A ∪ B)′ = A′∩ B′ (A ∪ B ∪ C)′ = A′ ∩ B′ ∩ C′ Complements of the unions Equals The intersection of the individual complements

2. The complement of the intersection of sets is equal to the union of the individual complements.

This means that if A, B and C are sets, (A ∩ B)′ = A′ ∪ B′ (A ∩ B ∩ C)′ = A′ ∪ B′ ∪ C′ Complements of the intersection Equals The union of the individual complements Let us use the activity below to validate these rules.

Activity 1.1: De Morgan’s Laws

1. Make a copy of the diagram below:

2. List the elements in α, β and θ

3. List the elements in α′, β′ and θ′

4. Find (a) (α ∪ β ∪ θ)′ (b) α′ ∩ β′ ∩ θ′ (c) (α ∩ β ∩ θ)′ (d) α′ ∪ β′ ∪ θ′

5. Compare your result in 4(a) to 4(b). What conclusion can you draw?

6. Compare your result in 4(c) to 4(d). What conclusion can you draw?

Compare your answer to the suggested solution below.

2. α = {1, 2, 3, 5,7, 9,11, 15,17,18,19}, β = {2, 5, 11, 13, 15, 17, 18, 21, 29}, θ = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}

3. α′= {4, 6, 8, 10, 13, 21, 23, 29, 31}, β′= {1, 3, 4, 6, 7, w8, 9, 10, 19, 23, 31}, θ′= {1, 4, 6, 8, 9, 10,15, 18, 21 }

4. (a) (α ∪ β ∪ θ)′ = {4, 6, 8, 10}

(b) α′ ∩ β′ ∩ θ′ = {4, 6, 8, 10}

(c) (α ∩ β ∩ θ)′ = {1, 3, 4, 6, 8, 9, 10, 13, 19, 21, 23, 29, 31}

(d) α′ ∪ β′ ∪ θ′ = {1, 3, 4, 6, 8, 9, 10, 13, 19, 21, 23, 29, 31}

5. Results in 4a and 4b are equal. This confirms De Morgan’s Law

6. Results in 4c and 4d are equal. This confirms De Morgan’s Law.

Example 1.1

Study the diagram carefully and use it to answer the questions.

1. List the elements in R, M, Q and ε

2. Show that R′ ∩ M′ = (R ∪ M)′

3. Show that (R ∩ M ∩ Q)′ = R′ ∪ M′ ∪ Q′

Solution

1. R = {s, u, b, d, e, r, m, a, t, o, g, l, y, p, h, i, c}, M = {u, n, c, o, p, y, r, i, g, h, t, a, b, l, e} and Q = {a, e, u, c, o, n, x, q, v, w} are subsets of ε = {s, u, b, d, e, r, m, a, t, o, g, l, y, p, h, i, c, n, x, q, v, w, j, k}

2. R′ ∩ M′ = (R ∪ M)′ Let us first solve for R′ ∩ M′ {b, x, v, n, q, w, j, k} ∩ {m, h, s, d, b, x, v, q, w, j, k} R′ ∩ M′= {b, x, v, q, w, j, k}……………………..(1) Next, let us solve for (R ∪ M)′ (R ∪ M)′ = (s, u, b, d, e, r, m, a, t, o, g, l, y, p, h, i, c, n)′ (R ∪ M)′ = {b, x, v, q, w, j, k}…… … … … … … … ..(2) Since equation (1) is equal to equation (2), it shows that R′ ∩ M′ = (R ∪ M)′

3. Let’s first solve for (R ∩ M ∩ Q)′ (R ∩ M ∩ Q)′ = (a, e, u, c, o)′ = {s, b, d, r, m, t, g, l, y, p, h, i, n, x, q, v, w, j, k} ……………..(1) Next, let’s find R′ ∪ M′ ∪ Q′ R′ ∪ M′ ∪ R′ = {b, x, v, n, q, w, j, k} ∪ {m, h, s, d, b, x, v, q, w, j, k} ∪ {m, h, p, c, y, r, i, g, b, t, j, k} = {s, b, d, r, m, t, g, l, y, p, h, i, n, x, q, v, w, j, k} ……………..(2) Since equation (1) is equal to equation (2), it shows (R ∩ M ∩ Q)′ = R′ ∪ M′ ∪ Q′

We have learned about De Morgan’s Laws of Set Theory. Now, we will look at its application. Before we consider this application, let’s familiarise ourselves with a few properties of complements as it will be central to our understanding.

Properties of Complement

1. Union of a set and its complement.

The union of a set A and its complement (A′) is equal to the universal set (μ ) i.e A ∪ A′ = μ

2. Intersection of a set and its complement The intersection of a set A and its complement (A′) is equal to the null set ( ∅) i.e. A ∩ A′ = { } = ϕ

3. Double complement.

The complement of the complement of set A will give the same set, A. i.e.

(A′)′ = A

4. Complement of an empty set.

The complement of an empty set (ϕ) is equal to the universal set (μ). i.e.

ϕ′ = μ

5. Complement of a universal set.

The complement of the universal set is equal to the null set. i.e. μ′ = ϕ

Example 1.2

Rewrite: a. (X′ ∪ Y)′ b. (X ∩ Y′)′

Solution

a. (X′ ∪ Y)′ Recall that from De Morgan’s laws, we have (A ∪ B)′ = A′ ∩ B′ So, for (X′ ∪ Y)′, we will replace A with X′ and B with Y This implies (X′ ∪ Y)′ = (X′)′ ∩ Y′ From the double complement property, (X′)′ = X Therefore, (X′ ∪ Y)′ = X ∩ Y′

b. (X ∩ Y′)′ Recall that from De Morgan’s laws, we have (A ∩ B)′ = A′ ∪ B′ So, for (X ∩ Y′)′, we will replace A with X and B with Y′ This implies (X ∩ Y′)′ = X′ ∪ Y″ From the double complement property, (Y′)′ = Y Therefore, (X ∩ Y′)′ = X′ ∪ Y

Example 1.3

Show that:

a. (A ∪ B′ ∪ C′)′= A′∩ B ∩ C

b. (A′ ∩ B ∩ C′)′ = A ∪ B′ ∪ C

Solution

a. Recall that (A ∪ B ∪ C)′ = A′ ∩ B′ ∩ C′ This means, (A ∪ B′ ∪ C′)′ = A′ ∩ B″ ∩ C″ = A′∩ B ∩ C

b. Recall that (A ∩ B ∩ C)′ = A′ ∪ B′ ∪ C′ So, (A′ ∩ B ∩ C′)′ = A″ ∪ B′ ∪ C″ = A ∪ B′∪ C

Example 1.4

Rewrite [(P ∪ Q′)′ ∩ P]′

Solution

Method 1 Let us start with the outer bracket.

Using [A ∩ B]′ = A′∪ B′ [(P ∪ Q′)′ ∩ P]′= (P ∪ Q′)″∪ P′ [(P ∪ Q′)′ ∩ P]′ = (P ∪ Q′) ∪ P′ Since the union is commutative, P ∪ Q′= Q′∪ P [(P ∪ Q′)′ ∩ P]′ = (Q′∪ P) ∪ P′ Since the Union of sets is associative, we have (Q′ ∪ P) ∪ P′ = Q′∪(P ∪ ′) [(P ∪ Q′)′ ∩ P]′= Q′∪(P ∪ P′) Recall that (P ∪ P′) = μ (universal set) [(P ∪ Q′)′ ∩ P]′ = Q′ ∪ μ = μ Method 2 We will start with the inner bracket and work our way out.

[(P ∪ Q′)′ ∩ P]′ (P ∪ Q′)′ = P′ ∩ Q″ = P′∩ Q [(P ∪ Q′)′ ∩ P]′= [(P′ ∩ Q) ∩ P]′ Since the intersection of sets is commutative P′∩ Q = Q ∩ P′ [(P ∪ Q′)′ ∩ P]′ = [(Q ∩ P′) ∩ P]′ Since intersection is associative, (Q ∩ P′) ∩ P = Q ∩ (P′ ∩ P) [(P ∪ Q′)′ ∩ P]′ = [Q ∩ (P′ ∩ P)]′ Recall that P′∩ P = ∅ [(P ∪ Q′)′ ∩ P]′ = [Q ∩ ∅]′ [(P ∪ Q′)′ ∩ P]′ = [∅]′ [(P ∪ Q′)′ ∩ P]′ = μ

Example 1.5

Using the Venn diagram, shade the region represented by:

(X ∪ Y)′ ∩ Z X′ ∩ (Y ∪ Z)

Solution

(X ∪ Y)′ ∩ Z Since the expression involves three sets, X, Y and Z, we will construct a three-set Venn diagram. We will represent the universal set with μ An example is shown below:

(X ∪ Y)′ represent elements which are not in X ∪ Y . We will use vertical lines to shade this region.

On the same diagram let us shade set Z. We will use horizontal lines.

The intersection of (X ∪ Y)′ and Z is the double-shaded area.

Therefore, (X ∪ Y)′ ∩ Z = In year 1, we learned about set theory laws. Examples of these laws include commutative, associative and distributive. We also learned to identify the regions in a three-set Venn diagram. In this lesson, we will apply these concepts to answer real-life problems.

How to complete a three-set Venn diagram

1. Start with the intersection of the three sets and proceed to the regions involving two sets.

2. Then work on the regions involving one set and finally the regions in the universal set that do not intersect with any of the three sets.

To summarise, start from the centre (the region with the most overlaps) and work your way out.

Example 1.6

Set A = {odd numbers less than 15} Set B = {4 < W ≤ 11} Set D = {Prime numbers less than 14} All of which are subsets of:

Set R = {0, 1, 2, 3, 4, … , 16}.

a. List the members of the following sets:

b. i. A

ii. B

iii. D

iv. A ∪ B

v. (A ∪ B) ∩ D′

vi. (A′∩ B) ∪ D

c. Represent A, B, D and R using a Venn diagram

Solution

a. i. A={1, 3, 5, 7, 9, 11, 13}

ii. B={5, 6, 7, 8, 9, 10, 11}

iii. D={2, 3, 5, 7, 11, 13}

iv. A ∪ B = {1, 3, 5, 7, 9, 11, 13} ∪ {5, 6, 7, 8, 9, 10, 11}

v. = {1, 3, 5, 6, 7, 8, 9, 10, 11, 13}

vi. (A ∪ B) ∩ D′ = {1, 3, 5, 6, 7, 8, 9, 10, 11, 13} ∩ {1, 4, 6, 8, 9, 10, 12, 14, 15, 16}

vii. = {1, 6, 8, 9, 10}

viii. (A′ ∩ B) = {2, 4, 6, 8, 10, 12, 14, 15, 16} ∩ {5, 6, 7, 8, 9, 10, 11}

ix. = {6, 8, 10} (A′ ∩ B) ∪ D = {6, 8, 10} ∪ {2, 3, 5, 7, 11, 13} = {2, 3, 5, 6, 7, 8, 10, 11, 13}

b. We first find the three intersections. i.e. A ∩ B ∩ D = {5, 7, 11} c.

Then proceed to the two intersections. There are three of these.

They are A ∩ B only = A ∩ B ∩ D′ , A ∩ D only = A ∩ B′∩ D and B ∩ D only = A′∩ B ∩ D A ∩ B only = A ∩ B ∩ D′ = {9} A ∩ D only = A ∩ B′∩ D = {3, 13} B ∩ D only = A′∩ B ∩ D = { } Let’s update our table with these values ...

Next, find the single intersections. There are three of them.

A only = { 1 } B only = { 6, 8, 10} D only = { 2 } Finally, we find (A ∪ B ∪ D)′ = {0, 4, 12, 14, 15, 16} Find below the completed diagram

Example 1.7

A survey of 70 investors reveals the following.

30 invest in bonds, 40 in stocks and 21 in Annuities.

10 had investments in bonds and annuities, 15 in Bonds and Stocks, 12 in stocks and annuities, and 7 had investments in all three asserts. The remaining investors had investments in other assets.

Create a Venn diagram for this information.

Solution

First, draw a Venn diagram with three intersecting sets representing the three main sets: Bonds, Stocks, and Annuities.

We will let B represent Bonds, A represent Annuities and S represent stocks.

This means n(B) = 30 , n(A) = 21 and n(S) = 40 Next, we fill in the region where all the three sets intersect.

From the question, this figure is 7.

Next, we fill in the region where the two sets intersect. These are:

a. Bonds and Annuities. A total of 10 had investments in bonds and annuities but 7 of them has investments in all three asserts. This leaves 10 − 7 = 3 investing in only Bonds and Annuities.

b. Bonds and Stocks. A total of 15 had investments in bonds and Stocks but 7 of them had investments in all three assets. This leaves 15 − 7 = 8 investing in only Bonds and Stocks.

c. Stocks and Annuities. A total of 12 had investments in stocks and annuities but 7 of them has investments in all three asserts. This leaves 12 − 7 = 5 investing in only stocks and Annuities.

Let us add these values to the Venn diagram.

You will realise there is only one region remaining in each set. These are the regions that represent only one set.

a. Only Bonds. A total of 30 had investments in bonds but we have already accounted for 3 + 7 + 8 = 18 of them, leaving 30 − 18 = 12 having investments in only Bonds.

b. Only Annuities. A total of 21 had investments in annuities but we have already accounted for 3 + 7 + 5 = 15 of them, leaving 21 − 15 = 6 having investments in only Annuities.

c. Only Stocks. A total of 40 had investments in stocks but we have already accounted for 5 + 7 + 8 = 20 of them, leaving 40 − 20 = 20 investing in only stocks.

We will add these values to the Venn diagram.

Finally, we calculate the number of investors who have no investments in the three assets. To do this, find the sum of all values we have already calculated for the intersecting sets and deduct the result from the total number of investors surveyed.

The sum of all values already calculated in the intersecting sets:

= 12 + 3 + 6 + 8 + 5 + 7 + 20 = 61 This means 70 − 61 = 9 investors did not invest in any of the three asserts.

Record this value outside the circles but inside the universal set. This completes the Venn diagram.

Example 1.8

A survey of 97 countries that participated in the Olympics reveals that 80 won bronze, 57 won silver and 47 won gold. It also reveals that 30 won gold and silver, 50 won silver and bronze, 38 won gold and bronze and 3 countries did not win any medal.

a. Illustrate the information on a Venn diagram.

b. Calculate the number of countries who won all three medals.

c. Calculate the number of countries who won exactly two medals.

Solution

a. Let B=Bronze medal, G=Gold medal, and S=Silver medal.

b. This implies the number of countries that won bronze medals is n(B) = 80 ,

c. The number of countries that won silver medals is n(S) = 57 and

d. The number of countries that won Gold medals is n(G) = 47 We will first draw a Venn diagram and fill the region where all three sets intersect.

Since we were not provided with this information, Let x be the number of countries who won all three medals.

Next, we fill in the region where the two sets intersect (countries that won exactly two of the three medals). These are A total of 30 countries won Gold and Silver. Out of these, x won all three medals.

This shows that (30 − x) countries won only Gold and Silver medals.

A total of 50 countries won Silver and Bronze. Out of these, x won all three medals. This shows that (50 − x ) countries won only Silver and Bronze medals.

A total of 38 countries won Gold and Bronze. Out of these, x won all three medals.

This shows that (38 − x) countries won only Gold and Bronze.

At this point, we have only one region remaining in each set. These are the regions that represent only one set.

Only Bronze. A total of 80 countries won bronze but we have already accounted for x + 50 − x + 38 − x = (88 − x) of them, leaving 80 − (88 − x) = ( − 8 + x) winning only bronze.

Only Silver. A total of 57 countries won silver but we have already accounted for x+ 50 − x + 30 − x = (80 − x) of them, leaving 57 − (80 − x) = ( − 23 + x) winning only silver.

Only Gold. A total of 47 countries won bronze but we have already accounted for x + 30 − x + 38 − x = (68 − x) of them, leaving 47 − (68 − x) = ( − 21 + x) winning only bronze.

Update the diagram with these values.

According to the question, 3 countries did not win any medals. This shows that the set (G ∪ S ∪ B)′ = 3 The next step is to solve for x To do this, add all entries and equate the result to the total number of countries under consideration (97).

x + (30 − x) + (50 − x) + (38 − x) + (− 8 + x) + (− 23 + x) + (− 21 + x) + 3 = 97 x + 69 = 97 x = 97 − 69 x = 28

e. Therefore, 28 countries won all three medals.

We can choose to update the diagram with this value.

f. Number of countries that won exactly two medals = 2 + 22 + 10 = 34 Expanding binomial expressions Hopefully you recall the work we did in year one on the binomial expansion. We used the Pascal triangle and the Combination method to expand and simplify certain expressions. Let’s remind you of this in the activity below.

Activity 1. 2: Revision

1. In groups or pairs, write down or generate the Pascal triangle up to (x + y)⁵2. Use your results in step 1 to simplify the expressions,

a. (2x + y)³,

b. (a − 3b)⁴c. iii (2 + √x)⁵.

3. Apply the combination method to also simplify the above task (i – ii)

4. Compare your results in (2) and (3) and write down any observations.

5. Show your solutions, or write up, to classmates in other groups or to your mathematics teacher.

The Binomial Expansion helps us to expand and simplify complex expressions, which would have been difficult using the idea of algebraic expressions. Under algebraic expressions we could only expand with a positive power, but the binomial theorem will help us to expand any expression with a rational number as the power.

Activity 1.3 – Derivation of Binomial Theorem

1. Look for the ⁿCᵣ button on your calculator.

2. Task: use the ⁿCᵣ on your calculator to evaluate

i) ³C₀

ii) ³C₁

iii) ³C₂

iv) ³C₃

3. Hopefully the answers you have are: 1 3 3 1.

This was used in year one to expand certain expressions.

The combination of n items taking r at a time is:

(n/r ) = ⁿCᵣ = n!_______ (n − r)r ! where n! is read as n factorial and is explained as:

n! = n(n − 1)(n − 2)(n − 3)…2 × 1 For example, 5 != 5(5 − 1)(5 − 2)(5 − 3)(5 − 4) = 5 × 4 × 3 × 2 × 1 = 120 You could also find 5 ! , using a calculator.

Look for ! on your calculator to find 5 ! , to confirm the answer obtained above.

Example 1.9

In groups or in pairs, solve the following without using calculator

a. ⁶C₃

b. ⁴C₂ Confirm your answers with a calculator.

Solution

a. ⁶C₃= 6 !________ (6 − 3)!3 ! = 6 × 5 × 4 × 3 × 2 × 1/3 × 2 × 1 × 3 × 2 × 1 = 6 × 5 × 4/3 × 2 × 1 = 120/6 = 20

b. ⁴C₂= 4 !________ (4 − 2)!2 ! = 4 × 3 × 2 × 1/2 × 1 × 2 × 1 = 4 × 3/2 × 1 = 12/2 = 6

Activity 1.4: General Form of Binomial Theorem

1. Using the definition (n/r ) = ⁿCᵣ = n !_______ (n − r)r !, work in pairs or groups and discuss how the following evaluations were done:

a. ⁿC₁ = n!_________ (n − 1)!× 1 ! = n(n − 1)!_________ (n − 1)!× 1 = n

b. ⁿC₂ = n!_________ (n − 2)!× 2 ! =n × (n − 1)(n − 2)!______________ (n − 2)!× 2 × 1 =n(n − 1)______ 2

c. Iii ⁿC₃ = n!_________ (n − 3)!× 3 ! = n(n − 1)(n − 2)(n − 3)!_________________ (n − 3)!× 3 × 2 × 1 = n(n − 1)(n − 2)___________ 6

2. Write down your challenges or observations for a class discussion with your classmates and teacher.

It must be noted that n_(C)₀ = 1 and 0 != 1 In general, the Binomial Theorem is written as;

(a + x)ⁿ= n_(C)₀ ₐₙ ₓ₀ + n_(C)₁ ₐₙ₋₁ ₓ₁ + n_(C)₂ ₐₙ₋₂ ₓ₂ + n_(C)₃ ₐₙ₋₃ ₓ₃ + … + n_(C)ᵣ ₐₙ−r xʳ+n_(C)ₙ ₐ₀ ₓₙ = aⁿ+ naⁿ⁻¹x + n(n − 1)_______ 2 ! aⁿ⁻²x²+ n(n − 1)(n − 2)___________ 3 ! aⁿ⁻³x³+ … + xⁿThere is a special case when a = 1 Substituting a = 1 into:

(a + x)ⁿ= aⁿ+ naⁿ⁻¹x + n(n − 1)______ 2 ! aⁿ⁻²x²+ n(n − 1)(n − 2)___________ 3 ! aⁿ⁻³x³+ …+ xⁿ= 1ⁿ+ n.1ⁿ⁻¹x + n(n − 1)______ 2 ! 1ⁿ⁻². x²+ n(n − 1)(n − 2)___________ 3 ! 1ⁿ⁻³. x³+ …+ xⁿTherefore, we have:

(1 + x)ⁿ= 1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿLet us go through the following worked examples to assist us in the expansion of the form (1 − x)ⁿor (1 + x)ⁿExample 1.10 By using the Binomial Theorem, fully expand the expression (1 + x)⁵.

Solution

Using the binomial theorem, (1 + x)ⁿ= 1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ In this case n = 5 so we substitute this in:

(1 + x)⁵= 1 + 5x + 5 × 4/2 ! × x²+ 5 × 4 × 3/3 ! 1 × x³+ 5 × 4 × 3 × 2/4 ! x⁴+ x⁵Simplify the coefficients to obtain:

= 1 + 5x + 5 × 4/2 × 1 × x²+ 5 × 4 × 3/3 × 2 × 1 × x³+ 5 × 4 × 3 × 2/4 × 3 × 2 × 1 × x⁴+ x⁵= 1 + 5 x + 10x²+ 10x³+ 5x⁴+ x⁵Example 1.11 By using the Binomial Theorem, fully expand the expression(1 + 2x)⁴.

Solution

Using the binomial theorem:

(1 + x)ⁿ= 1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿIn this case n = 4 and ‘x′ = 2x, so we substitute these in:

(1 + 2x)⁴= 1 + 4(2x) + 4 × 3/2 ! × (2 x)²+ 4 × 3 × 2/3 ! 1 × (2x)³+ (2 x)⁴Simplify the coefficients to obtain:

= 1 + 4(2x) + 4 × 3/2 × 1 × 4x²+ 4 × 3 × 2/3 × 2 × 1 × 8x³+ 16 x⁴= 1 + 8 x + 24 x²+ 32x³+ 16x⁴Example 1.12 Find the first four terms of the binomial √1 + 2x using the binomial theorem.

Solution

√1 + 2x = (1 + 2x)¹__ 2 In this case n = 1/2 and ‘x′ = 2x, so we substitute these in:

(1 + 2x)¹__ 2 = 1+1/2(2x) + 1/2(1/2 − 1)_______ 2 ! (2x)²+ 1/2(1/2 − 1)(1/2 − 2)_____________ 3 ! (2x)³= 1+1/2(2x) + 1/2(−1/2)______ 2 × 1 (4x²) + 1/2(−1/2)(−3/2)__________ 3 × 2 × 1 (8x³) = 1+ x − 1/2 x²+ 1/2 x³Example 1.13 Find the first four terms of the binomial ³√1 − 3x using the binomial theorem.

Solution

3 √1 − 3x = (1 − 3x)¹__ 3 In this case n = 1/3 and ‘x′ = − 3x , so we substitute these in:

(1 − 3x)¹__ 3 = 1+ 1/3(− 3x) + 1/3(1/3 − 1)_______ 2 ! ( − 3x)²+ 1/3(1/3 − 1)(1/3 − 2)_____________ 3 ! ( − 3x)³= 1+1/3(− 3x) + 1/3(−2/3)______ 2 × 1 ( − 3x)²+ 1/3(−2/3)(−5/3)__________ 3 × 2 × 1 ( − 3x)³= 1+1/3(− 3x) + 1/3(−2/3)______ 2 × 1 9x²+ 1/3(−2/3)(−5/3)__________ 3 × 2 × 1 ( − 27x³) = 1+1/3(− 3x) + −2__ 9/2 9x²+ 10__ 27/6 ( − 27x³) = 1 − x − x²− 5/3 x³Example 1.14 Expand up to the fifth term the expression 1______ (1 − x)²Solution Rewriting, 1______ (1 − x)², using the rules of indices 1 __ (1 − x)²= (1 − x)⁻²(1 − x)⁻²= 1 − 2(− x) + −2(− 2 − 1)_ 2 ! (− x)²+ −2(− 2 − 1)(− 2 − 2)_____________ 3 ! (− x)³+ −2(− 2 − 1)(− 2 − 2)( − 2 − 3)___________________ 4 ! ( − x)⁴(1 − x)⁻²= 1 − 2(− x) + −2(− 3)_ 2 × 1 (− x)²+ −2(− 3)(− 4)_ 3 × 2 × 1 (− x)³+ −2(− 3)(− 4)( − 5)____________ 4 × 3 × 2 × 1 ( − x)⁴(1 − x)⁻²= 1 + 2x + 3 x²+ 4 x³+ 5 x⁴Example 1.15 Find the first four terms of (1 − 16x)¹__ 4

Solution

(1 − 16x)¹__ 4 = 1+1/4(− 16x) + 1/4(1/4 − 1)_______ 2 ! ( − 16x)²+ 1/4(1/4 − 1)(1/4 − 2)_____________ 3 ! ( − 16x)³= 1+1/4(− 16x) + 1/4(−3/4)______ 2 × 1 ( − 16x)²+ 1/4(−3/4)(−7/4)__________ 3 × 2 × 1 ( − 16x)³= 1+1/4(− 16x) + − 3 _ 16/2 (256x²) + 21__ 64/6 ( − 4096x³) = 1 − 4x − 24x²− 224 x³Activity 1.5: Deciding whether binomial theorem can be used in all cases Discuss in groups whether the binomial theorem can be used in all cases.

(1 + x)ⁿ=1 + n x + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ1. Consider (2 + 4x)⁴, can the binomial theorem be used?

2. How will you write the terms in the bracket to have 1 as one of the terms?

3. Let us go through the following tasks to help us write (2 + 4x)⁴in the form (1 + x)ⁿand use the binomial theorem to expand and simplify.

Task 1: Take the given binomial expression (2 + 4x) Task 2: Factorise the constant out from the binomial to obtain 2(1 + 2x) Task 3: Rewrite your expression obtained with the power as given i.e.

2⁴(1 + 2x)⁴Task 4: Expand the binomial (1 + 2x)⁴after which you multiply your result by 2⁴Now follow the task to obtain all the terms for the given expression.

Expected Answer: 16 + 128x + 348x²+ 512x³+ 256x⁴Let’s solve the following example using the binomial theorem

Example 1.16

1. Use binomial theorem to expand fully (3 + 12x)⁵2. Find the first three terms of (4/9 + 6x) 1/2 using the binomial theorem.

3. Use the binomial theorem to expand fully and simplify (10 + 4x)³4. Use the binomial theorem to find the first four terms of the expression ₄ √81 + 162x

Solution

1. (3 + 9x)⁵= 3⁵(1 + 3x)⁵3⁵(1 + 3x)⁵= 3⁵[1 + 5(3x) + 5(5 − 1)______ 2 ! (3x)²+ 5(5 − 1)(5 − 2)___________ 3 ! (3x)³+ 5(5 − 1)(5 − 2)(5 − 3)________________ 4 ! (3x)⁴+ 5(5 − 1)(5 − 2)(5 − 3)(5 − 4)_____________________ 5 ! ( 3x)⁵= 3⁵[1 + 53x + 5(4)____ 2 × 1(3x)²+ 5(4)(3)_______ 3 × 2 × 1 (3x)³+ 5(4)(3)(2)__________ 4 × 3 × 2 × 1 (3x)⁴+ 5(4)(3)(2)(1)_____________ 5 × 4 × 3 × 2 × 1 (3x)⁵] = 3⁵[1 + 53x + 5(4)____ 2 × 1 (9x²) + 5(4)(3)_______ 3 × 2 × 1 (27x³) + 5(4)(3)(2)__________ 4 × 3 × 2 × 1 (81x⁴) + 5(4)(3)(2)(1)_____________ 5 × 4 × 3 × 2 × 1( 243x⁵)] = 243[1 + 5(3x) + 10 (9x²) + 10 (27x³) + 5 (81x⁴) + ( 243x⁵)] = (3 + 9x)⁵= 243(1 + 15x + 90x²+ 270x³+ 405x⁴+ 243x⁵).

Compute the final coefficients

• 243⋅1=243

• 243⋅15=3645

• 243⋅90=21870

• 243⋅270=65610

• 243⋅405=98415

• 243⋅243=59049 Therefore the correct expansion is (3+9x)⁵= 243 + 3645x + 21870x²+ 65610x³+ 98415x⁴+ 59049x⁵.

2. (4/9 + 6x) 1/2 = (4/9) 1/2 (1 + 27/2 x) 1/2 = (4/9) 1/2 [1 + 1/2(27/2 x) + 1/2( 1/2 − 1)_______ 2 ! (27/2 x) 2 ] = (4/9) 1/2 [1 + 1/2(27/2 x) + −1__ 4/2 (27/2 x) 2 ] = (4/9) 1/2 [1 + 1/2(27/2 x) + −1__ 4/2 ( 729/4 x²)] = 2/3 [1 + (27/4 x) + −1__ 4/2 (729/4 x²)] = 2/3[1 + 27/4 x − 729/32 x²] = 2/3 + 9__ 2x − 243/16 x²3. (10 + 4x)³= 10³(1 + 2/5 x)³= 10³[1 + 3(2/5 x) + 3(3 − 1)______ 2 ! (2/5 x)²+ 3(3 − 1)(3 − 2)___________ 3 ! ( 2/5 x)³] = 10³[1 + 3(2/5 x) + 3(2)____ 2 × 1 (2/5 x)²+ 3(2)(1)_______ 3 × 2 × 1 ( 2/5 x)³] = 10³[1 + 3(2__ 5x) + 3 (2/5 x)²+ (2/5 x)³] = 10³[1 + 3(2_ 5 x) + 3(4x²_ 25 ) + ( 8_ 125 x³)] = 10³[1 + 6/5 x + 12x²____ 25 + 8x³___ 125] = 1000 [1 + 6/5 x + 12x²____ 25 + 8x³___ 125] = 1000 + 1200x + 480 x²+ 64x³4. ⁴√81 + x 4 √81 + 162x = (81 + 162x)¹__ 4 (81 + 162x)¹__ 4 = 81¹__ 4 (1 + 2x)¹__ 4 = 81¹__ 4 [1 + 1/4(2x) + 1/4(1/4 − 1) _______ 2 ! (2x)²+ 1/4 (1/4 − 1)(1/4 − 2) ____________ 3 ! ] = 81¹__ 4 1 + 1/4(2x) + 1/4(1/4 − 1) _______ 2 ! (2x)²+ 1/4 (1/4 − 1)(1/4 − 2) ____________ 3 !

= 81¹__ 4 [1 + 1/4(2x) + 1/4(−3/4)______ 2 × 1 (2x)²+ 1/4(−3/4)(−7/4)__________ 3 × 2 × 1 (2x)³] = 81¹__ 4 [1 + 1/4(2x) + − 3__ 16/2 × 1( 4x²) + 21__ 64/3 × 2 × 1( 8x³)] = 81¹__ 4[1 + 1/2 x − 3/8 x²+ 7/16 x³] = 3[1 + 1/2 x − 3/8 x²+ 7/16 x³] = 3 + 3/2 x − 9/8 x ²− 21/16 x³Example 1.17 The first three terms in the expansion of (1 + x/p)ⁿin ascending powers of x are 1 + x + 9/20 x².

Find the values of n and p.

Solution

Expand (1 + x/p)ⁿusing the binomial theorem;

(1 + x)ⁿ= 1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ= (1 + x/p)ⁿ= 1 + n x/p + n(n − 1)_______ 2 ! (ˣ_ p)²= 1 + nx + n(n − 1)_______ 2 x²_ p²Equating it to 1 + x + 9/20 x²and comparing the coefficient of x and x²n/p = 1 , n(n − 1)______ 2 p²= 9/20 n = p ……..(1) 20n(n − 1) = 18 p²……..(2) Substituting equation [n = p] from into equation (2) 20n(n − 1) = 18 p²20p(p − 1) = 18 p²20 p²− 20p = 18 p²20 p²− 18 p²= 20p 2 p²= 20p 2 p²_ 2p = 20p____ 2p p = 10 n = 10

Example 1.18

Find the value of t and n in the equation (1 + tx)ⁿ= 1 − 6x + 33/2 x²Solution Expand (1 + tx)ⁿusing the binomial theorem;

(1 + x)ⁿ= 1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ(1 + tx)ⁿ= 1 + ntx + n(n − 1)_______ 2 ! (tx)²1 + ntx + n(n − 1)_______ 2 t²x²Equating it to 1 − 6x + 33/2 x²and comparing the coefficient of x and x²nt = − 6 , n(n − 1)______ 2 t²= 33/2 nt = − 6 ……..(1) 2n(n − 1) t²= 66……..(2) (2 n²− 2n) t²= 66……..(2) 2n²t²− 2n t²= 66 2 (nt)²− 2(nt)t = 66 But nt = − 6 2 ( − 6)²− 2(− 6)t = 66 2(36) + 12t = 66 72 + 12t = 66 12t = 66 − 72 12t = − 6 12t_ 12 = − 6_ 12 t = − 1/2 nt = − 6 n(− 1/2) = − 6 − n = − 12 − n_ − 1 = −12_ − 1 n = 12

Example 1.19

Expand (1 + 2x − x²)⁶as far as the x²term.

Solution

How will you use binomial theorem to solve a question of this nature?

Discuss your views with a classmate or in your groups.

You need to write the trinomial 1 + 2x − x²to a binomial in the form (1 + x)ⁿHow we do that?

We represent 2x − x²by a variable say y, i.e. y = 2x − x²(1 + 2x − x²)⁶= (1 + y)⁶Expanding (1 + y)⁶, using the binomial theorem, (1 + x)ⁿ=1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ(1 + y)⁶= 1 + 6y + 6(6 − 1)_ 2 ! y²+ … = 1 + 6y + 6⁽⁵⁾_ 2 y²= 1 + 6y + 15 y²But y = 2x − x², so we substitute this in:

= 1 + 6(2x − x²) + 15 (2x − x²)²+ … = 1 + 6(2x − x²) + 15(4x²− 4 x³+ x⁴) + … = 1 + 12x − 6 x²+ 60 x²− 60 x³+ 15x⁴+ … = 1 + 12x + 54x²+ …

Example 1.20

Expand (1 − 3x + x²)⁵as far as the x³term.

Solution

We represent − 3x + x²by a variable say y, i.e. y = − 3x + x²(1 − 3x + x²)⁵= (1 + y)⁵Expanding (1 + y)⁵using the binomial theorem, (1 + x)ⁿ=1 + nx + n(n − 1)_______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ(1 + y)⁵= 1 + 5y + (5 − 1)______ 2 ! y²+ 5(5 − 1)(5 − 2)____________ 3 ! y³= 1 + 5y + 5(4)____ 2 y²+ 5(4)(3)______ 6 y³= 1 + 5y + 10 y²+ 10 y³But y = − 3x + x²= 1 + 5(− 3x + x²) + 10 (− 3x + x²)²+ 10 ( − 3x + x²)³= 1 + 5(− 3x + x²) + 10(9x²− 6x³+ x⁴) + 10( − 27x³+ 27x⁴− 9x⁵+ x⁶) = 1 − 15x + 5 x²+ 90 x²− 60 x³+ 10x⁴− 270 x³+ 270 x⁴− 90x⁵+ 10x⁶= 1 − 15x + 5x²+ 90x²− 60x³− 270 x³= 1 − 15x + 95x²− 330x³+ … Did you know we can use binomial expansion to approximate exponential numbers? This powerful technique allows us to simplify complex expressions and make calculations easier. We will explore how to apply binomial expansion to approximate exponential functions.

Activity 1.6: Approximating Exponential Numbers Using Binomial

Expansion (a + b)ⁿ.

Ensure you have your notebook, calculator and pen ready for the activity.

1. Choose a number to work with.

2. Examples include:1.5, 3.5, 10 etc.

3. Express your chosen number in the form of (a + b). For example: If you chose 1.5, you could write it as (1 + 0.5).

4. Select any integer to represent n . Examples include: 2, 4, −4,

5. Write your decimal and n in the form (a + b)ⁿ. For example: Using 1.5 and n = 4, you would write (1 + 0.5)⁴.

6. Apply the binomial theorem to expand and evaluate your expression from step 4

7. Choose different sets of numbers and repeat steps 1 to 5. Experiment with various decimal values and n to see how the approximations change.

8. Use your calculator to find the actual value of your chosen decimal raised to the power of n. For example, calculate 1.5⁴9. Look at the results from your binomial expansion and the calculator.

10. Discuss:

a. How close were your approximations to the actual values?

b. What did you observe about the accuracy of your approximations with different n values?

Great work! You have effectively used binomial expansion to approximate exponential numbers. Now, let us solve more examples together to deepen your understanding!

Example 1.21

Expand (1 + 2y)⁵using binomial theorem.

Hence use your results to evaluate (1.04)⁵, correct to 4 decimal places

Solution

The binomial theorem is given by (1 + x)ⁿ= 1 + nx + n(n − 1)______ 2 ! x²+ n(n −1)(n− 2)__________ 3 ! x³+ …+ xⁿ(1 + 2y)⁵= 1 + 5(2y) + 5(5 − 1)______ 2 ! (2y)²+ 5(5 − 1)(5 − 2)___________ 3 ! (2y)³+ 5(5 − 1)(5 − 2)(5 − 3)________________ 4 ! (2y)⁴+ 5(5 − 1)(5 − 2)(5 − 3)(5 − 4)_____________________ 5 ! (2y)⁵= 1 + 5(2y) + 5(4)____ 2 × 1 × 4y²+ 5(4)(3)_______ 3 × 2 × 1 × 8y³+ 5(4)(3)(2)_____________ 4 × 3 × 2 × 1 × 1 × 16y⁴+ 5(4)(3)(2)(1)_____________ 5 × 4 × 3 × 2 × 1 × 32y⁵= 1 + 5(2y) + 10 × 4y²+ 5 × 2 × 8y³+ 5 × 16y⁴+ 1 × 32y⁵= 1 + 10y + 40y²+ 80y³+ 80y⁴+ 32y⁵The (1.04) can be written as a binomial, i.e. (1 + 0.04) The expression (1.04)⁵= (1 + 0.04)⁵Comparing (1 + 0.04)⁵to (1 + 2y)⁵2y = 0.04 2y_ 2 = 0.04/2 y = 0.02 Now substitute y = 0.02 into your results; 1 + 10y + 40y²+ 80y³+ 80y⁴+ 32y⁵1 + 10(0.02) + 40(0.02)²+ 80(0.02)³+ 80(0.02)⁴+ 32(0.02)⁵1 + 10(0.02) + 40(0.0004) + 80(0.000008) + 80(0.00000016) + 32(0.0000000032) = 1 + 0.2 + 0.016 + 0.00064 + 0.0000128 + 0.0000001024 = 1.2166529024 = 1.2167 (to 4dp) Thus (1.04)⁵≈ 1.2167

Example 1.22

Find the first five terms of the expression √1 + 3x.

Hence find an estimate of the value of √1.12 to 5 decimal places

Solution

The binomial theorem is given by (1 + x)ⁿ=1 + n x + n(n − 1)______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ√1 + 3x = (1 + 3x)¹__ 2 (1 + 3x)¹__ 2 = 1 + 1/2(3x) + 1/2(1/2 − 1) ______ 2 ! (3x)²+ 1/2(1/2 − 1)(1/2 − 2) ___________ 3 ! (3x)³+ 1/2(1/2 − 1)(1/2 − 2)( 1/2 − 3) ________________ 4 ! (3x)⁴ = 1 + 1/2(3x) + 1/2(−1/2)______ 2 (3x)²+ 1/2(−1/2)−(3/2)__________ 6 (3x)³+ 1/2(−1/2)(−3/2)(−5/2)______________ 24 (3x)⁴= 1 + 3x/2 + –1__ 4/2 (3x)²+ 3__ 8/6 (3x)³+ − 15____ 16/24 (3x)⁴= 1 + 3x/2 + −1__ 4/2 (9x²) + 3__ 8/6 (27x³) + − 15____ 16/24 (81x⁴) = 1 + 3x/2 − 1/8( 9x²) + 1/48 (27x³) − 5/128 (81x⁴) = 1 + 3x/2 − 9x²_ 8 + 27x³_ 16 − 405x⁴_ 128 The (1.12) can be written as binomial, i.e. (1 + 0.12) The expression (1.12)¹__ 2 = (1 + 0.12)¹__ 2 Comparing (1 + 0.12)¹__ 2 to (1 + 3x)¹__ 2 3x = 0.12 3x_ 3 = 0.12/3 y = 0.04 Now substitute y = 0.04 into your results; 1 + 3x/2 − 9x²___ 8 + 27x³____ 16 −405x⁴_____ 128 = 1 + 3(0.04)______ 2 − 9(0.04)²_ 8 + 27(0.04)³_ 16 − 405(0.04)⁴_ 128 = 1 + 0.12/2 − 9(0.0016)________ 8 + 27(0.000064)__________ 16 − 405(0.00000256)_____________ 128 = 1 + 0.12/2 − 0.0144/8 + 0.001728/16 − 0.00020736/128 = 1 + 0.06 − 0.0018 + 0.000108 − 0.0000081 = 1.0582999 = 1.05830 (to 5dp) Thus √1.12 ≈ 1.05830

Example 1.23

Using the binomial theorem, find an estimate for the value of √101

Solution

The binomial theorem is given by (1 + x)ⁿ=1 + n x + n(n − 1)______ 2 ! x²+ n(n − 1)(n − 2)___________ 3 ! x³+ …+ xⁿ√101 = √100 + 1 = (100 + 1)¹__ 2 (100 + 1)¹__ 2 = 100¹__ 2 (1 + 1/100)¹__ 2 = 100¹__ 2[ (1 + 1/100)¹__ 2] = 10 [1 + 1/2(0.01) + 1/2(1/2 − 1) ______ 2 ! (0.01)²+ 1/2(1/2 − 1)(1/2 − 2) ___________ 3 ! (0.01)³+ …] = 10 [1 + 1/2(0.01) + 1/2(−1/2)______ 2 ! (0.01)²+ 1/2(−1/2)(−3/2)__________ 3 ! (0.01)³+ …] = 10 [1 + 1/2(0.01) + −1__ 4/2 (0.01)²+ 3__ 8/6 (0.01)³] = 10 [1 + 1/2(0.01)− 1/8 (0.0001) + 1/16 (0.000001)] = 10[1 + 0.005 − 0.0000125 + 0.0000000625 ] = 10[1.004987563 ] = 10.04987563 Thus √101 ≈ 10.0499

Example 1.24

Use binomial theorem to find approximations for:

a. (3.97)¹__ 2

b. (0.994)¹__ 4

c. √25.05

Solution

a. (3.97)¹__ 2 = (4 − 0.03)¹__ 2 = 4¹__ 2 (1 − 0.03)¹__ 2 = 4¹__ 2[(1 − 0.0075)¹__ 2] = 2[(1 − 0.0075)¹__ 2] = 2[1 + 1/2(− 0.0075) + 1/2(1/2 − 1)_______ 2 ! ( − 0.0075)²+ …] = 2[1 + 1/2(− 0.0075) + −1__ 4/2 ( − 0.0075)²+ …] = 2[1 + 1/2(− 0.0075) − 1/8 (− 0.0075)²+ … ] = 2[1 − 0.00375 − 1/8 (0.0056625) + …] = 2[1 − 0.00375 − 0.00003125 + …] = 2[0.99621875] = 1.9924375 ∴ (3.97)¹__ 2 ≈ 1.992 (to 4sf)

b. (0.994)¹__ 4 = (1 − 0.006)¹__ 4 =[1 + 1/4(− 0.006) + 1/4(1/4 − 1)_______ 2 ! ( − 0.006)²+ …] = [1 + 1/4(− 0.006) + − 1__ 16/2 ( − 0.006)²+ …] = [1 + 1/4 (− 0.006) − 1/32 (− 0.006)²+ …] = [1 − 0.0015 − 1/32 (0.000036) + …] = [1 − 0.0015 − 0.000001125 + …] = [0.998498875] ∴ (0.994)¹__ 4 ≈ 0.9985 (to 4sf)

c. √25.05 = (25 + 0.05)¹__ 2 = 25¹__ 2 (1 + 0.002)¹__ 2 = 5[(1 + 0.02)¹__ 2] = 5 [1 + 1/2(0.002) + 1/2(1/2 − 1) _______ 2 ! (0.002)²+ …] = 5 [1 + 1/2(0.002) + 1/2(−1/2)______ 2 ! (0.002)²+ …] = 5 [1 + 1/2(0.002) + −1__ 4/2 (0.002)²+ …] = 5 [1 + 1/2(0.002) − 1/8 (0.002)²+ …] = 5 [1 + 0.001 − 1/8 (0.000004) + …] = 5 [1 + 0.001 − 0.0000005] = 5[1.0009995] = 5.0049975 ∴ √25.05 ≈ 5.005 (to 4sf)

Part A

1. Use the Venn diagram to answer the questions.

a. List the elements in A, B and C

b. List the elements in A′, B′ and C′

c. Find (i) (A ∪ B ∪ C)′ (ii) A′ ∩ B′ ∩ C′ (iii) (A ∩ B ∩ C)′ (iv) A′ ∪ B′ ∪ C′

d. Compare your result in 4(a) to 4(b). What conclusion can you draw?

e. Compare your result in 4(c) to 4(d). What conclusion can you draw?

f. Describe set B

g. Describe set C

2. Study the diagram carefully and use it to answer the questions.

a. List the elements in G, E, T and ε

b. Show that G′ ∩ E′ = (G ∪ E)′

c. Show that (G ∩ E ∩ T)′ = G′ ∪ E′ ∪ T′

d. Show that (G ∪ E ∪ T)′ = G′ ∩ E′ ∩ T′ For questions 3 – 7, sets X, Y and Z are subsets of the universal set μ .

Use the information to choose the correct answer to the questions.

3. Find X ∪ X′ A. ∅ B. μ C. X D. Y

4. Find Z ∩ Z′ A. ∅ B. μ C. Z D. X

5. Find (Y′)′ A. ∅ B. μ C. X D. Y

6. Find (X ∪ X′)′ A. ∅ B. μ C. X D. X’∪ X

7. (Z ∩ Z′)′ A. ∅ B. μ C. X ∪ Y D. Z′ ∩ Z

8. Rewrite

a. (R ∪ S′)′ b. (M′∩ N)′

c. [(A′∪ B)′ ∩ C]′ d. [(P ∩ Q′)′ ∪ P]′

e. [A ∩ (A′∪ B)′]′

9. Show that

(a) [(R ∪ M)′ ∩ P]′ = R ∪ M ∪ P′

(b) [(R ∩ M) ∪ M′]′ = R′ ∩ M

10. On a Venn diagram, shade the region represented td by:

a. (A ∩ B)′ ∩ C

b. A′ ∪ (Y ∩ Z′)

11. Sets X = {prime numbers}, Y = {0 < W ≤ 11} and Z = {1, 2, 3, 4, 5, 6, 7, 13, 14, 15, 16, 18} are subsets of A = {0, 1, 2, 3, 4, … , 21}.

a. List the members of the following sets

b. i. X ii. Y

iii. X ∩ Y ∩ Z iv. X ∪ Y′ (v) (X ∪ Y) ∩ Z′

vi. (X′∩ Y) ∪ Z

c. Use a Venn diagram to represent X, Y, Z and A

12. A survey of 90 teachers reveals the following.

45 teach Management, 57 Accounting and 50 Costing.

25 teach Management and Accounting, 35 teach Accounting and Costing and 22 teach Management and Costing. 5 teachers cannot teach any of the three subjects.

a. Represent the information on a Venn diagram

b. calculate the number of teachers who can teach only one of the three subjects.

13. A survey of 41 employees shows that 25 are skilled in Carpentry, 15 in Brick laying and 24 in Tiling. 12 are skilled in carpentry and brick laying, 9 in brick laying and tilling, 17 in carpentry and tilling, and 8 are not skilled in any of the three occupations.

Find the number of employees who are skilled in

a. all three occupations

b. exactly two of the occupations

14. A class of 60 students were each required to have textbooks in Accounting, Management and Costing. However, when the class teacher checked their books, she found out that:

16 had Accounting and Management books 14 had Accounting and Costing books 18 had only two of the three books and 20 had accounting plus at least one other book Every student had at least one of the three books.

Find the number of students who had

a. All three books.

b. Only Costing and Accounting books

c. Exactly one of the three books

d. If 16 students had neither Accounting nor Costing books, find the probability that a student selected at random from the class had Management books.

15. The analysis of results of a sample of students who sat for the WASSCE in Biology, Physics and Mathematics revealed that:

80% passed in Biology 65% passed in Mathematics 70% passed in Physics 52% passed in Biology and Mathematics 58% passed in Biology and Physics 55% passed in Physics and Mathematics.

45% passed in all three subjects.

What percentage of the students:

a. passed only Mathematics

b. passed only physics

c. passed only Biology

d. passed Biology and Physics but not mathematics

e. did not pass any of the subjects.

16. A shop owner kept records of purchases made by customers over the weekend. The shop sells food, clothing and hardware. She found that, out of the customers who visited her shop, 210 purchased food products, 100 hardware and 147 clothing. 30 purchased food and hardware. 128 purchased two of the three products. She also found that of those who purchased two of the three products, 14 did not buy clothing and 20 did not buy food. 13 of the customers who visited the store did not buy any of the products.

a. How many customers visited the store over the weekend?

b. How many of the customers who visited the store:

i. purchased less than two products.

ii. all three products.

iii. Food and other products.

17. A group of 210 tourists were asked to indicate which of the languages, German, French and English they can speak. It was found that 55 can speak German, 76 can speak French and 200 can speak English. 103 can speak two of the three languages and 6 tourists cannot speak any of the three languages.

Find the number of tourists who can speak:

a. all three languages

b. only one of the three languages.

18. 87 schools in Tamale have subscriptions to at least one of the following newspapers: Graphic, Mirror and Times. 69 subscribe to Graphic, 62 to Mirror and 55 to Times. 32 have subscriptions to all three newspapers.

Find the number of schools who have subscriptions to:

a. exactly two of the newspapers

b. only one of the newspapers.

19. A publishing company surveyed its employees about their proficiency in three software: Photoshop, CorelDraw and InDesign. The results show that all the employees were proficient in at least one of the three software and 10 were proficient in all three. The number of employees proficient in CorelDraw and other software were twice those proficient in only CorelDraw. Of the employees who were proficient in two of the software, 14 were not proficient in InDesign, 8 were not proficient in CorelDraw and 6 were not proficient in Photoshop. The same number of employees proficient in only InDesign are proficient in only Photoshop. There were more employees proficient in only Photoshop and InDesign than employees proficient in only Photoshop and there were fewer employees proficient in only InDesign and CorelDraw than employees proficient in only InDesign.

a. Use a Venn diagram to represent the above data.

b. How many employees were involved in the survey?

c. How many employees were proficient in only one of the software?

Part B

1. Write down the first five terms of the binomial expansion of:

a. (1 + 4x)¹²b. (1 − 3x)⁹c. (3 + 2x)⁷d. (1 − x _ 2)⁸e. (2 − 5/2 x) 6

2. Write down the term indicated in the binomial expansion of the following functions:

a. (1 + 3x)⁶: 3ʳᵈterm

b. (1 + x _ 3)²⁰: 7ᵗʰterm

c. (a + 4b)¹²term containing a⁴3. Expand the following binomial expansions up to and including the terms in x²:

a. (1 − 2x) (1 + x)⁸b. (2 + 3x) (1 + 3x)¹⁰c. (5 + 2x) (1 − 2/3 x) 12

d. (1 + 4x) (1 − 3x)²⁰4. Expand (1 − 2x)⁷and hence find the value of (0.98)⁷correct to 4 decimal places.

5. Find the first four terms of the following using the binomial theorem.

a. (1 − 1/2 x) 1/2

b. (1 − 2x)⁻³c. (3 − x)⁻²6. Use appropriate binomial expansions to find the approximate values of:

a. ³√8.7 correct to 5 significant figures.

b. (81.162)¹__ 4 to 4 decimal places

c. √4.02 to 3 decimal places

d. 1___ √10 to 4 decimal places

e. √101 to 4 decimal places

f. (0.995)⁵to 4 decimal places

7. Show that in ascending powers of x of (1 − x/1 + x) 1/3 is 1 − 2/3 x + 2/9 x²− 4/21 x³….

8. Expand the following as far as the x³term

a. (1 + x + 2 x²)³b. (1 + 3x + x²)⁴9. a. find the 1st 3 terms in ascending powers of x of the binomial expansion(1 + py)⁹, where p is a non zero constant.

b. If the first 3 terms are 1, 36x and qx²where q is a constant, find the value of p and q.

10. a. Find the first 3 terms in ascending powers of x of the binomial expansion(2 + kx)⁷, where k is a constant.

b. Given that the coefficient of x²is 6 times the coefficient of x, find the value of k.

Studying sequences and inequalities is important because arithmetic and geometric sequences provide statistical understanding into central tendencies, growth rates, etc. Also, quadratic inequalities help in decision-making processes where certain conditions or constraints must be met. For example, in business and economics, they are used to model profit maximisation or loss minimisation under specific constraints. In physics and engineering, quadratic inequalities arise when determining safe operating conditions, or analysing forces. In architecture and design, they help define areas, volumes or structural integrity within specific boundaries. This section introduces you to the concept of the sum of sequences, convergent and divergent series, recursive sequences, arithmetic and geometric sequences. You will also learn how to solve quadratic inequalities, systems of quadratic inequalities and real-life problems involving quadratic inequalities.

KEY IDEAS

• A sequence is a set of numbers in a given order.

• Each of the numbers in a sequence is called a term.

• Maximum values are the largest possible values of variables within a system that satisfy a set of constraints, each represented by an inequality.

• Minimum values are the least possible values of variables within a system that satisfy a set of constraints, each represented by an inequality.

• When finding the solution to a quadratic inequality, it is advisable to start by sketching the associated quadratic graph.

In year one we identified number patterns and classified them into arithmetic and geometric sequences. While sequences are a list of numbers with defined patterns, a series is the sum of the terms of a sequence. This year we will be finding the sum of sequences and analysing the convergence or divergence of series.

If the general term of a series with n term is known, then the complete series can be written down in short notation as indicated by the following:

1. a + (a + d) + (a + 2d) + … +(a + [n − 1]d) = ∑ ᵣ₌₁ ⁿ(a + [r − 1]d)

2. a + ar + ar²+ …arⁿ⁻¹= ∑ ₖ₌₁ ⁿarᵏ⁻¹3. 1²+ 2²+ 3²+ …n²= ∑ ᵣ₌₁ ⁿr²4. −1³+ 2³− 3³+ 4³+ …( − 1)ⁿn³= ∑ ᵣ₌₁ ⁿ( − 1)ʳr³Notes:

a. It is sometimes more convenient to count the term of a series from zero rather than 1.

For example:

a + (a + d) + (a + 2d) + … +(a + [n − 1]d) = ∑ ᵣ₌₀ ⁿ⁻¹(a + rd) and a + ar + ar²+ …arⁿ⁻¹= ∑ ₖ₌₀ ⁿ⁻¹arᵏIn general, for a series with n terms starting at u₀ :

u₀ + u₁ + u₂ + u₃ + …uₙ₋₁ = ∑ ᵣ₌₀ ⁿ⁻¹uᵣ

b. We may also use the sigma notation for “infinite series” such as those already encountered in the sum to infinity of a geometric series.

For example: 1+1/3 + 1/3²+ 1/3³+ … = ∑ ᵣ₌₁ ∞ 1/3ʳ⁻¹or ∑ ᵣ₌₀ ∞ 1/3ʳActivity 2.1 – Sums of Sequences Perform this activity in pairs or groups.

In year one we learnt about Arithmetic and Geometric Progression. Let us revise and build on that knowledge by creating the sequence terms and find the sum of the terms of the expressions below.

1. Evaluate ∑ ₙ₌₀ ⁶2(n − 3)

2. Find the sum ∑ ₖ₌₁ ⁵⁰(3k + 2)

3. Evaluate ∑ ᵣ₌₁ ³(1/2) r

Solution

1. ∑ ₙ₌₀ ⁶2(n − 3) = 2(0 − 3) + 2(1 − 3) + 2(2 − 3) + 2(3 − 3) + 2(4 − 3) + 2(5 − 3) + 2(6 − 3) = 2(− 3) + 2(−2) + 2(− 1) + 2(0) + 2(1) + 2(2) + 2(3) = ( − 6) +(− 4) + (− 2) + 0 + 2 + 4 + 6 = 0

2. ∑ ₖ₌₁ ⁵⁰(3k + 2) u₁ = 3(1) + 2 = 5 u₅₀ = 3(50) + 2 = 152 sₖ = k (u₁ + u₅₀)_________ 2 sₖ = 50(5 + 152)_________ 2 = 3925

3. ∑ ᵣ₌₁ ³(1/2) r when r = 1, (1/2) 1 = 1/2 when r = 2, (1/2) 2 = 1/4 when r = 3, (1/2) 3 = 1/8 ⟹ 1/2 + 1/4 + 1/8 = 7/8 Method of undetermined coefficients The method of undetermined coefficients is a powerful technique for finding the general solution of linear recurrence relations with constant coefficients If the nth term of a linear sequence is denoted by xₙ, then the recurrence relation is of the form: xₙ + 1 = f( xₙ) for some of the function f.

For example, in xₙ + 1 = 4 − xₙ__ 2 , when x₂ = 2, then x₃ is expressed as x² + 1 = 4 − x₂__ 2 x₃ = 4 − 2/2 x₃ = 4 − 1 x₃ = 3

Example 2.1

The nth term of a sequence is Uₙ . If U₁ = 3 and Uₙ + 1 = Uₙ + 5 , find U₃.

Solution

Uₙ + 1 = Uₙ + 5 U₁ = 3 U₂ = U1 + 1. = U₁ + 5 , but U₁ = 3 U₂ = 3 + 5 = 8 U₃ = U2 + 1. = U₂ + 5 U₃ = 8 + 5 = 13 Sum of an Arithmetic Progression (AP)

Activity 2.2

1. Working in pairs or small groups, work through the following steps to discover how to find the general sum of an arithmetic sequence, ie a sequence with a common difference.

2. Let Sn denote the sum of the first n terms of a linear sequence, where “a” is the first term and “d” the common difference.

3. Sn = a + (a + d) + (a + 2d) + … +[a + (n − 1)d] …………………(1)

4. Writing the terms in the reverse order, we get :

5. Sn = [a + (n − 1)d] + [a + 2d(n − 2)] + … +(a + 2d) + (a + d) + a ………………(2) Adding (1) and (2), we get 2Sn = [2a + (n − 1)d] + [2a + (n − 1)d] + [2a + (n − 1)] +…= n[2a + (n − 1)d] (Since we are adding n terms we have n lots of this expression.)

∴ Sn = n/2[2a + (n − 1)d]

6. Let U₁ = a,

7. Then, Sn = n/2[2 U₁ + (n − 1)d]

8. Similarly, Sn = n/2[ U₁ + { U₁ + (n − 1)d}]

9. But Uₙ = U₁ + (n − 1)d

10. By substitution, we get

11. Sₙ = n/2[ U₁ + Uₙ] (Remember that U₁ = the first term and Uₙ = the last term)

Example 2.2

a. Find the sum of the first 35 terms of the linear sequence 2, 5, 8…

Solution

a = 2, d = 3 and n = 35 Sn = n/2[2a + (n − 1)d] S₃₅ = 35/2 [2(2) + (35 − 1)(3) = 1855

b. A sequence is defined by Uₙ + 1 = 3+ Uₙ and U₁ = 1

c. Find the sum of the first ten term of the sequence.

Solution

Uₙ + 1 = 3+ Uₙ U₁ = 1 ( = a) U₂ = 4 Common difference d = u₂ – u₁ = 4 − 1 = 3 Sₙ = n/2[2a + (n − 1)d], where a = first term, d the common difference and n the number of terms S₁₀ = 10/2 [2(1) + (10 − 1)3] S₁₀ = 5[2 + 9(3)] S₁₀ = 5[29] S₁₀ = 145

d. Find the the sum of the first 10 positive even numbers:

Solution

S₁₀= 2 + 4 + 6 + 8 +10⋯ + 18 + 20 Sₙ = n/2[U₁+ Uₙ] = n/2[ first term + last term] S₁₀ = 10/2 [2 + 20] = 5 × 22 = 110

Example 2.3

A conference hall has 189 seats, arranged in 9 rows. The organiser wants to arrange the seats such that the difference in the number of seats between any two consecutive rows is 5 seats. If there are 9 rows in total, how many seats should be in the third row?

Solution

Number of rows n = 9 Capacity of the room S₉ = 189 Difference d = 5 Let the number of seats in the first row = a Sₙ = n/2[2a + (n − 1)d] S₉ = 9/2[2a + (9 − 1)5] 189 = 9/2[2a + (8)5] 378 = 9[2a + (8)5] 378 = 18a + 360 18a = 378 − 360 18a = 18 a = 1 Number of seats in the third row a + 2d ⇒ 1 + 2(5) = 11 Sum of a Geometric Progression (GP) Remember that a geometric progression has the first term, a and a common ratio, r.

The position of terms represented by n can be written as:

Sₙ = a + ar + ar²+ ar³+ … +arⁿ⁻²+ arⁿ⁻¹…….….(1) rSₙ = ar + ar²+ ar³+ … … … +arⁿ⁻¹+ arⁿ……...…(2) eqn (2) − eqn (1) rSn − Sn = −a + arⁿSₙ(r − 1) = arⁿ− a Sₙ = a( rⁿ− 1)_______ r − 1 = − a(1 − rⁿ)________ −(1 − r) Sₙ = a( rⁿ− 1)_______ r − 1 , r ≠ 1 Sₙ = a(1 − rⁿ)_______ (1 − r) , r ≠ 1

Example 2.4

Find the sum of the first seven terms of the sequence 2, 8, 32…

Solution

Sₙ = a( rⁿ− 1)_______ r − 1 , r ≠ 1 a = 2, r = 4 and n = 7 S₇ = 2( 4⁷− 1)_______ 4 − 1 S₇ = 2(16384 − 1)__________ 3 S₇ = 2 × 16 382/3 S₇ = 10922

Example 2.5

Find the sum of the first 10 terms of 1/2, 1/4, 1/8, …

Solution

Sₙ = a(1 − rⁿ)_______ (1 − r) , r ≠ 1 a = 1/2, r = 1/2 and n = 10 S₁₀ = 1/2 (1 − ( 1/2)¹⁰) __________ ( 1 − 1/2 ) S₁₀ = 1/2(1− 1/1024) S₁₀= 1024 − 1/1024 S₁₀ = 1023/1024

A series is a sum of a sequence of numbers. For example, 1 + 2 + 3 + 4 + … Convergence of series A series is convergent if its sum approaches a finite value. Think of it like a target.

1. The series get closer and closer to the target (the finite value).

2. The sum of the series stabilises.

3. Example, 1 + 1/2 + 1/4 + 1/8 + …

4. As n increases the nth term is converging to 0 and the sum of the series is converging to 2.

5. If the limit of the nth term is 0 , the series might converge.

6. If the ratio between the terms is constant, then the series converges when |r| < 1 .

Divergence A series is divergent if its sum grows without bounds or approaches infinity.

Think of it like a runaway train:

1. The series gets farther and farther away from a finite value.

2. The sum of the series increases indefinitely.

3. Example; 1 + 2 + 4 + 8 + … this series diverges.

If the limit of the nth term is not 0 , the series diverges.

If the ratio between term is greater than 1 , the series diverges, ie |r| > 1 Sum to infinity of a convergent Geometric Progression For the general exponential sequence (GP):

Sₙ = a (1 − r n)________ (1 − r) , if − 1 < r < 1 , then rⁿbecomes smaller and smaller as n becomes larger and larger, so as n ⟶ ∞ then rⁿ→ 0 and (1 − rⁿ) → 1.

We say the limiting value is denoted S∞ is a/1 − r.

Hence, the sum to infinity of a, ar ar², … is given by S∞ = a/1 − r , provided − 1 < r < 1.

Example 2.6

Determine the sum to infinity of the geometric series 5, − 1 + 1/5, …

Solution

5, − 1 + 1/5, … a = 5, r = −1/5 The sum to infinity is S ∞ = a/1− r = 5/1−(−1/5 ) = 5____ (6/5) = 25/6 = 5 1/6

Note: S1 , S₂, S3 ,… are the partial sums of the series. If there is a number L such that: Sₙ = ∑ ᵣ₌₁ ∞ Sn = L . The number L is called the sum of the infinity series. If there is no such number, the series is said to diverge.

A recursive sequence is a sequence where each term is defined using previous terms.

For example, for an Arithmetic Sequence:

First term a₁ = 2 Recursive formula: aₙ = aₙ₋₁ + 3

1. Start with a₁ = 2

2. To find a₂, add 3 to a₁ : a₂ = 2 + 3 = 5

3. To find a₃, add 3 to a₂ : a₃ = 5 + 3 = 8

4. To find a₄, add 3 to a₃ : a₄= 8 + 3 = 11 Sequence: 2, 5, 8, 11, 14, 17… For example, for a Geometric Sequence:

First Term: a₁ = 2 Recursive formula: aₙ = 2 × aₙ₋₁

1. Start with a₁ = 2

2. To find a₂, multiply a₁ by 2 → a₂ = 2 × 2 = 4

3. To find a₃, multiply a₂ by 2 → a₃ = 2 × 4 = 8

4. To find a₄, multiply a₃ by 2 → a₄ = 2 × 8 = 16 Sequence: 2, 4, 8, 16, 32, 64, … In both examples;

• The first term (a₁) is given

• The recursive formula defines how to find the next term (aₙ)using the previous term ( aₙ₋₁) Recurrent decimals Recurrent decimals are repeated decimals. It can be seen that every recurrent decimal represents a rational number. Every rational number can be represented by a terminal decimal or a recurrent decimal. If the denominator of a rational number (written in its simplest form) has prime factors, other than 2 and 5, then the rational number is recognised as a recurrent decimal.

For example; 5/9 = 0.5555… or 0 . 5˙ , 2/11 = 0.181818…or 0 . 18¨ , 8/15 = 0.53333… or 0.5 3˙

Example 2.7

Write the recurring decimals as a series. Identify the common ratio, the first term and an explicit formula for the nth term:

a. 0.222222…

b. 0.323232…

c. 5.414141…

Solution

a. 0.2222... = 0.2 + 0.02 + 0.002 + 0.0002 + ⋯ = 0.2 + 0.2(0.1) + 0.2(0.01) + 0.2(0.001) + ⋯ = 0.2 + 0.2(0.1) + 0.2(0.1)²+ 0.2(0.1)³+ ⋯ The first term is 0.2, and the common ratio between each successive term is 0.1. The formula for the n th term will be Uₙ = 0.2( 1/10)ⁿ⁻¹b. 0.323232… = 0.32 + 0.0032 + 0.000032 + 0.00000032 + ⋯ = 0.32 + 0.32(0.01) + 0.32(0.0001) + 0.32(0.000001) + ⋯ = 0.32 + 0.32(0.01) + 0.32(0.01)²+ 0.32(0.01)³+ ⋯ The first term is 0.32, and the common ratio between each successive term is 0.01. The formula for the n th term will be Uₙ = 0.32( 1/100) n−1

c. 5.414141… = 5 + 0.41 + 0.0041 + 0.000041 + 0.00000041 + ⋯ = 5 + 0.41 + 0.41(0.01) + 0.41(0.1)²+ 0.41(0.1)³+ ⋯ From the second term (0.41), a geometric sequence with first term 0.41, and the common ratio 0.01 can be observed. The formula for the n th term will be For 0.414141…, Uₙ = 0.41( 1/100)ⁿ⁻¹Thus, 5.414141… = 5 + 0.41( 1/100)ⁿ⁻¹for n = 1, 2, 3, …

Example 2.8

Write 0.9999… as an exponential series and show that in the limit, the sum of series equals one.

Solution

0.9999…= 0.9 + 0.09 + 0.009 + 0.0009 + … This is a G.P. with first term, a = 0.9 and common ratio, r = 0.09/0.9 = 0.1 The series is an infinite series with sum; S∞ = a/1− r = 0.9/1 − 0.1 = 0.9/0.1 = 1

Example 2.9

Express 0.1 6˙ as an infinite geometric series and find the sum of the geometric series.

Solution

0.1 6˙ = 0.166666… = 0.1 + 0.06 + 0.006 + 0.0006 + … = 1/10 + 6/100 + 6/1000 + 6/10000 + … From the series, the term after 1/10 from geometric series with = 6/100, r = 6/100 ÷ 6/1000 = 1/10 Sum of the geometric series:

0.0 6˙ = S∞ = 6___ 100/1 − 1/10 = 6___ 100___ 9/10 = 1/15 ∴ 0.1 6˙ = 1/10 + 1/15 = 1/6

Example 2.10

If a sequence U₁, U₂, U₃… is define by the relation Uₙ + 1 = 3 + Uₙ for n ≥ 1 and U₁ = 1, find;

a. U₂, U₃and U₄

b. a formula for Uₙ in terms of n .

c. the sum of the 1st n terms

Solution

a. Using the relation Uₙ + 1 = 3 + Uₙ If n = 1 U₂ = 3 + U₁ = 3 + 1 = 4 U₃ = 3 + U₂ = 3 + 4 = 7 U₄ = 3 + U₃ = 3 + 7 = 10 The sequence is 1, 4, 7, 10, … , so it is a linear sequence

b. Uₙ = a + (n − 1)d and a = 1, d = 3 Uₙ = 1 + (n − 1)3 Uₙ = 1 + 3n − 3 Uₙ = 3n − 2

c. The sum of n terms of 1, 4, 7, 10, … Sₙ = n/2[2a + (n − 1)d] Sₙ = n/2[2(1) + (n − 1)3] Sₙ = n/2[2 + 3n − 3] Sₙ = n/2[3n − 1]

The mean of any two numbers, for example, a and b, is obtained by finding a half of the sum of the numbers thus: Mea n , (μ) = 1/2(a + b). The difference between a and μ is equal to the difference, or distance, between μ and b i.e., μ − a = b − μ .

Now, a, μ , b form an arithmetic sequence since there is a common difference and that for any arithmetic sequence with a and b as terms, some means namely, μ1 , μ₂ , μ3 , . .. μₖ can be found between a and b such that a, μ1 , μ₂ , μ3 , . .. μₖ, b, remains an arithmetic sequence as μ1 , − a = μ₂ − μ1 , = .. μₖ − b and the means are equally spaced.

Example 2.11

1. Find the arithmetic mean sequence of students’ scores in mathematics test:

80, 75,90,85,95.

2. A baker wants to increase the amount of sugar in a recipe from 200g to 500g in four equal steps. What arithmetic sequence represents the amount of sugar (in grammes) added and what are the intermediate values?

Solution

1. Arithmetic mean = 80 + 75 + 90 + 85 + 95/5 = 425/5 = 85

2. The baker wants to increase the sugar from 200g to 500g 500 − 200 = 300g d = 300/4 − 1 = 300/3 = 100g Arithmetic mean 200 + 100 = 300g 300 + 100 = 400g 400 + 100 = 500g The arithmetic sequence: 200, 300, 400, 500 .

The intermediate sequence 300 and 400 .

Geometric mean The geometric mean of a sequence is the measure of the central tendency of the sequence, similar to the arithmetic mean. However, it is calculated differently.

For a sequence of n numbers (a₁ × a₂ × a₃… aₙ)¹__ n

Example 2.12

1. Find the geometric mean of the sequence 2, 4, 8, 16

2. Find the geometric mean of the sequence 10, 20, 40.

Solution

1. 2, 4, 8, 16 (a₁ × a₂ × a₃… aₙ)¹__ n (2 × 4 × 8 × 16)¹__ 4 = ⁴√1024 = 5.65685…≈ 5.7

2. (10 × 20 × 40)¹__ 3 = ³√8000 = 20

In year one, we learnt how to solve linear equations and inequalities using graphing and algebraic reasoning. With this knowledge, we will be able to graph polygons when given specific linear equations or inequalities which will help us determine the maximum/minimum value for a function.

Working in pairs or groups, go through this activity to review graphing of linear inequalities.

Activity 2.3 – Graphing Linear Inequalities

1. Pick a graph book/sheet

2. Represent the following inequalities graphically

a. y ≤ 0.67x + 2

b. y ≤ − x + 2

c. y ≥ − 2

3. Write out the coordinates for the points at which the inequalities intersect.

4. What polygon did the intersecting lines create?

5. Does your graph look like the one in Figure. 2.1: Graphing Linear Inequalities?

Figure. 2.1: Graphing Linear Inequalities

Congratulations! You have adequately remembered how to graph linear inequalities.

Now, we will learn how to determine the maximum/minimum value(s) of a given function.

We can determine the maximum or minimum value of a given function based on a specific objective. Maximum values are the highest points of a function while minimum values are the lowest points of a function. In finding maximum or minimum values of a function algebraically, you have to substitute all the coordinates of the vertices of the region of interest that forms the solution to the function. For example, if you are interested in getting the maximum/minimum area of a given rectangle, you would evaluate the function at each corner or edge by substituting the coordinates. If the polygon of interest (triangle, square, etc.) is represented graphically, you need to identify the pair of coordinates and substitute in the given function. On the other hand, if equations of lines are given, you solve these equations simultaneously and determine the values for the pair of coordinates.

Example 2.13

Evaluate the expression V = 2x − 3y for the given feasible region to determine the point at which ‘V’ has a maximum value and the point at which ‘V’ has a minimum value.–9–9 –8–8 –7–7 –6–6 –5–5 –4–4 –3–3 –2–2 –1–1 11 22 33 44 55 66

–4–4

–3–3

–2–2

–1–1 11 22 33 44 55 66 00 cc aa bb x y A(-5, -2) B(0, 3) C(4, -1)

Solution

Step 1: Identify all the vertices in Figure. 2.2 A (− 5, − 2), B(0, 3) and C (4, − 1)

Step 2: Substitute the coordinates of each of the vertices in the function V = 2x − 3y and compute.

For (− 5, − 2), x = − 5, y = − 2.

V = 2(− 5) − 3( − 2) = − 4 For (0 , 3 ), x = 0 , y = 3.

V = 2(0 ) − 3( 3 ) = − 9 For (4 , − 1), x = 4 , y = − 1.

V = 2(4 ) − 3( − 1) = 11

Step 3: Choose the highest result as the maximum value and the least as the minimum value Maximum value is 11 at vertex (4, − 1).

Minimum value is − 9 at vertex ( 0 , 3 ).

Example 2.14

On the same graph, illustrate the inequalities, y ≥ 7/3 x - 5, y ≤ x – 1 and x ≥ 0.

Using the polygon created, determine the minimum and maximum values of the function m = 7x − y .

Solution

Step 1: Draw the three linear inequalities on a graph sheet.x

–4–4 –3–3 –2–2 –1–1 11 22 33 44 55 66 77 88 99 y

–6–6

–5–5

–4–4

–3–3

–2–2

–1–1 11 22 33 00 MM TT NN

Figure 2.3 : Graphical representation of inequalities in Example 2.14

Step 2: Identify the coordinates of the vertices polygon created as a result of the intersection of the lines.

M(0, –1), T(3, 2) and N(0, –5)

Step 3: Substitute the values of the coordinates in the function m = 7x − y.

For (0 , − 1), x = 0 , y = − 1.

m = 7(0 ) −( − 1) = 1 For (3 , 2 ), x = 3 , y = 2.

m = 7(3 ) − 2 = 19 For (0 , − 5), x = 0 , y = − 5.

m = 7(0 ) −( − 5) = 5

Step 4: Select the highest result as the maximum value and the least as the minimum value Maximum value is 19 at vertex (3, 2 ).

Minimum value is 1 at vertex ( 0 , − 1 ).

Example 2.15

On the same graph, illustrate the inequalities, x + y ≤ 8, y ≥ 2, x − y ≤ 3, − x + 2y ≤ 10 and x ≥ 1. Using the boundaries of the polygon created, determine the minimum and maximum values of the function w = x²+ y².

Solution

Step 1: Draw the five linear inequalities on a graph sheet.x

–2 2

–2 –1 1

–1 11 22 33 44 55 66 77 88 y 11 22 33 44 55 66 00 PP QQ RR SS TT Figure2.4:Graphical representation of inequalities in Example 2.15

Step 2: Identify the coordinates of the vertices polygon created as a result of the intersection of the lines.

P(1, 2), S(1, 5.5), Q(2, 6), T(5.5, 2.5) and R(5, 2)

Step 3: Substitute the values of the coordinates in the function w = x²+ y².

For (1, 2), x = 1 , y = 2 .

w = (1)²+ (2)²= 5 For (1, 5.5), x = 1 , y = 5.5.

w = (1)²+ (5.5)²= 31.25 For (2, 6), x = 2 , y = 6.

m = (2)²+ (6)²= 40 For (5.5, 2,5), x = 5.5 , y = 2.5.

m = (5.5)²+ (2.5)²= 36.5 For (5, 2), x = 5 , y = 2.

m = (5)²+ (2)²= 29

Step 4: Select the highest result as the maximum value and the least as the minimum value Maximum value is 40 at vertex (2, 6 ).

Minimum value is 5 at vertex ( 1 , 2 ).

Now, let’s see how our knowledge on solving linear inequalities can be used to solve real life problems.

Knowledge for knowledge’s sake is commendable. However, utility of the knowledge is even more satisfying. Let us go through this activity to explore the usefulness of knowledge on solving systems of linear equations.

Activity 2.4 – Optimising the Production of Furniture Working in pairs, or small groups, work on this activity.

Scenario Kpogas company manufactures two types of products (chairs and tables).

Each chair requires 2 hours of carpentry and 1 hour of painting, while each

table requires 3 hours of carpentry and 2 hours of painting. The company has a maximum of 60 hours of carpentry and 40 hours of painting available per week.

They make a profit of GH¢50.00 per chair and GH¢80.00 per table.

How many chairs (c) and tables (t) should be produced each week to maximise profit, while staying within the carpentry and painting constraints?

Solution

1. Write out algebraic expressions for the hours of carpentry and painting.

Hours for carpentry: 2c + 3t ≤ 60 Hours for painting: c + 2t ≤ 40

2. Write out algebraic expressions for the number of chairs and tables.

Number of chairs cannot be negative: c ≥ 0 Number of tables cannot be negative: t ≥ 0

3. Write out an equation for the expected profit.

Let P represent profit P = 50c + 80t

4. Express the system of linear inequalities as the constraints of production.

⎧ ⎪ ⎨ ⎪ ⎩ c ≥ 0 t ≥ 0 2c + 3t ≤ 60 c + 2t ≤ 40

5. Note that the equation of profit is the objective function.

6. Graph the system of linear inequalities

–5–5 55 1010 1515 2020 2525 3030 00

Figure 2.5 : Graphical representation of Kpogas company production constraints

7. Identify the coordinates of the vertices to the solution region.

(20, 0), (0, 0), (15, 10) and (0, 20)

8. Substitute the coordinates of the vertices into the profit equation.

At (20, 0), P = 50(20) + 80(0) = 1000 At (0, 0), P = 50(0) + 80(0) = 0 At (15, 10), P = 50(15) + 80(10) = 1550 At (0, 20), P = 50(0) + 80(20) = 1600

9. Write out your conclusion The maximum profit is GH¢1600.00, and it occurs when Kpogas produces no chairs and 20 tables.

Example 2.16

A bakery produces cakes (x ) and cookies (y ) daily. A cake takes 2 hours to make while a batch of cookies takes 1 hour to make. The bakery can spend at most 12 hours per day baking. Each cake requires 3 units of flour and a batch of cookies requires 1 unit of flour. The bakery has at most 15 units of flour available each day. The bakery earns GH¢5.00 profit per cake and GH¢3.00 profit per batch of cookies. Without an increase in the units of flour and hours of baking, how many of each product should be baked daily to make the most profit?

Solution

Step 1: Identify the constraints communicated by the bakery.

⎧ ⎪ ⎨ ⎪ ⎩ 2x + y ≤ 12 3x+ y ≤ 15 x ≥ 0 y ≥ 0

Step 2: Write out the equation of the profit (objective function).

P = 5x + 3y

Step 3: Graph the inequalities.x

–8–8 –6–6 –4–4 –2–2 22 44 66 88 1010 1212 y 22 44 66 88 1010 1212 00 (0, 12)(0, 12) (4, 6) (5, 0)(0, 0)

Figure 2.6:Graphical representation of constraints for baking

Step 4: Identify the coordinates of the vertices to the solution region.

(0, 12), (5, 0), (4, 6) and (0,0)

Step 5: Substitute the coordinates of the vertices into the profit equation.

At (0, 12), P = 5(0) + 3(12) = 36 At (5, 0), P = 5(5) + 3(0) = 25 At (4, 6), P = 5(4) + 3(6) = 38 At (0, 0), P = 5(0) + 3(0) = 0

Step 6: Write out your conclusion.

The bakery should bake 4 cakes and 6 batches of cookies daily to make the most profit.

Example 2.17

A farmer has 300 metres of fencing material to enclose a rectangular field for planting crops. The farmer wants to maximise the area of the field to get the best crop yield, but there is a restriction. The length of the field must be 10 metres longer than the width to accommodate the farm machinery.

What should be the dimensions of the field that will maximise the enclosed area, keeping in mind the restriction on the total fencing available?

Solution

Step 1: Represent the statement algebraically.

Let w be the width of the field Length of the field = w + 10

Step 2: Write the equation for perimeter of the field.

Since Perimeter(P)= 2(width) + 2(length), P = 2w + 2(w +10) P = 2w + 2w + 20 P = 4w + 20.

Step 3: Set up the linear inequality Since the material available for fencing is 300m, the total distance around the field can be less or equal to 300.

4w + 20 ≤ 300

Step 4: Simplify the inequality.

4w ≤ 300 − 20 4w ≤ 280 w ≤ 280/4 w ≤ 70 So, the width of the field must be less than or equal to 70 metres.

Step 5: Test values for width.

Since the highest value for the width will be 70, the length will be 70 + 10 = 80 Testing for perimeter, 2(70) + 2(80) = 300, satisfying perimeter requirement.

Testing for maximum area of the rectangular field: length × width Area = 70 × 80 = 5600m²Step 6: State your conclusion The area can be maximised when the rectangle has dimensions of 70 by 80, satisfying all constraints.

Activity 2.5 – Real Life Project Involving Linear Inequalities

1. Identify an industry or company that produces/manufactures goods in your locality.

2. Obtain the restrictions/constraints guiding the production of their goods.

3. Identify the profit made on the items of production.

4. Write a two-page report comprising:

a. where the company is located.

b. goods produced and constraints communicated by the company officials.

c. expected profit on each of the goods.

d. an algebraic representation of constraints and objective function.

e. a graphical representation of constraints (manually or by software).

f. a suggestion on the decision the company should make to minimise profit.

g. a suggestion on the decision the company should make to maximise profit.

h. conclusion based on your observations and findings.

In year one we learnt about quadratic functions, how to solve them algebraically and how to illustrate them graphically. Remember that quadratic functions are of the form M(x ) = a x²+ bx + c, where a, b, c ∈ R , a ≠ 0.

Combining this knowledge and what we know about linear inequalities, we can represent general quadratic inequalities in the forms;

1. a x²+ bx + c < w

2. a x²+ bx + c > w

3. a x²+ bx + c ≤ w

4. a x²+ bx + c ≥ w , where w is a constant.

To find the solution to a quadratic inequality of any of the forms above:

1. First determine the value of w

2. Simplify the inequality so that one side is zero

3. Factorise to determine the range of values for x

4. Represent it graphically to know the solution region

Activity 2.6 – Individual/Group Exploration of Quadratic Inequalities

1. Brainstorm how you would solve the inequalities:

2. x²− 3x − 10 > 0 and -2x²+ 5x ≤ 3.

3. Discuss multiple approaches (factorising, completing the square or using the quadratic formula).

4. Consider when the expression is equal to zero and how this can help determine the solution.

5. Draw separately the graph of each of the quadratic inequalities to understand the regions where the inequality is true or false.

6. Share your observations from the process with a peer.

Example 2.18

Find the solution to the inequality x²− 2x − 15 < 0 .

Solution

Step 1: Factorise the left-hand side of the inequality x²− 2x − 15 = (x + 3)(x − 5) Hence (x + 3)(x − 5) < 0

Step 2: Find the critical values x = 5 and x = − 3

Step 3:

Method 1: Represent the extreme values of the expected solution on a number line and test them.–4–4 –3–3 –2–2 –1–1 11 22 33 44 55 6600 Figure2.7: Representing possible solutions to quadratic inequalities on number lines Possible Solution Regions Test Values x + 3 x − 5 (x + 3)(x − 5) Sign x < − 3 − 4 − 1 − 9 9 Positive − 3 < x < 5 3 6 − 2 − 12 Negative x > 5 6 9 1 9 Positive

Step 4: Select the solution region that corresponds to the sign satisfying the inequality.

Since x²− 2x − 15 < 0 holds for only values of x for which the product of the factors is negative (less than zero), − 3 < x < 5 is the solution to the inequality.

OR

Step 3:

Method 2: Graph the quadratic inequality–15–15 –10–10 –5–5 55 1010 1515 2020 2525

–25–25

–20–20

–15–15

–10–10

–5–5 55 1010 1515 2020 2525 00 BB CCAA Figure2. 8: Quadratic graph of x²− 2x − 15

Step 4: Identify the portion of the graph that satisfies the inequality. In this case, the portion below the x-axis.–15–15 –10–10 –5–5 55 1010 1515 2020 2525

–25–25

–20–20

–15–15

–10–10

–5–5 55 1010 1515 2020 2525 00 BB CCAA Figure2.9: Quadratic graph of x²− 2x − 15 with solution Thus − 3 < x < 5

Example 2.19

Find the solution to the inequality − 5x²+ 12x − 7 > 0 .

Solution

Step 1: Factorise the left-hand side of the inequality − 5x²+ 12x − 7 = ( − 5x + 7)(x − 1) Hence (− 5x + 7)(x − 1) > 0

Step 2: Find the critical values x = 7/5 and x = 1.

Step 3: Represent the extreme values of the expected solution on a number line and test them.

Figure 2.10: Representing quadratic solution on number line 2 Possible Solution Regions Test Values − 5x + 7 x − 1 ( − 5x + 7)(x − 1) Sign x < 1 0 7 − 1 − 7 Negative 1 < x < 1.4 1.2 1 0.2 0.2 Positive x > 1.4 3 − 8 2 − 16 Negative

Step 4: Select the solution region that corresponds to the sign satisfying the inequality.

Since − 5x²+ 12x − 7 > 0 holds for only values of x for which the product of the factors is positive (greater than zero), 1 < x < 1.4 is the solution to the inequality.

Activity 2.7- Graphical Solution to − 5x²+ 12x − 7 > 0 Represent the solution to − 5x²+ 12x − 7 > 0 graphically and share the results with a classmate.

Activity 2.8 – Research on Quadratic Inequalities

1. Conduct research on the behaviour of solutions to maximum and minimum curves of quadratic functions.

Instructions:

a. Explore the nature of the solution when a quadratic function with a minimum curve is:

i. greater than zero

ii. less than zero

b. Explore the nature of the solution when a quadratic function with a maximum curve is:

i. less than zero

ii. greater than zero

c. Include graphical representations to support your argument.

d. Consult open education resources, YouTube, textbooks, etc.

e. Provide references to the sources you consult and use.

We have explored how to solve single quadratic inequalities algebraically and graphically. Now, we will learn how to solve systems of quadratic inequalities (two or more quadratic inequalities) through algebra and graphing.

In finding solutions to quadratic inequalities algebraically, you solve them simultaneously.

Example 2.20

Graph the system of quadratic inequalities y ≥ 4 x²+ 5x, y < − 3 x²− 7x .

Solution

Solving the systems of inequalities simultaneously

Step 1: Equate the two expressions 4 x²+ 5x = − 3 x²− 7x

Step 2: Simplify and make x the subject.

4 x²+ 3 x²+ 7x+ 5x = 0 7 x²+ 12x = 0 x(7x + 12) = 0 x = 0, x = −12_ 7 .

Step 3: Identify whether the point of intersection is inclusive in the solution to the systems.

Since the y < − 3 x²− 7x has the strictly < symbol, the points of intersection will not satisfy this inequality.

For y ≥ 4 x²+ 5x, At x = 0, y ≥ 4 (0)²+ 5(0), y ≥ 0.

For y < − 3 x²− 7x, At x = 0, y < − 3 (0)²− 7(0), y < 0 At x = 0 , y must be 0 in both inequalities, but since the second inequality requires y < 0 , there is no solution at this point.

At x = − 12/7 , y ≥ 4 (− 12/7 ) 2 + 5( − 12/7 ), y ≥ 3 3/49 .

At x = 0, y < − 3 (− 12/7 ) 2 − 7(− 12/7 ), y < 156/49 .

y < 3 9/49 .

At x = − 12/7 , y must be 3 9/49 in both inequalities, but since the second inequality requires y < 3 9/49 , there is no solution at this point.

Step 4: State the solution Since the points of intersection will not satisfy one of the inequalities, the solution becomes − 1 5/7 < x < 0. That is, the region between the points of intersection.

Solving the systems of inequalities graphically

Step 1: Draw the graphs of the two inequalities.x

–5

–5

–5 –4

–4

–4 –3

–3

–3 –2

–2

–2 –1

–1

–1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 y

–1

–1

–1 1 1 1 2 2 2 3 3 3 4 4 4 0 0 0

Figure 2.11: Graphical Solution to y≥4x²+ 5x, y<–3x²–7x

Step 2: Identify the region that is common to both inequalities.

R₃

Step 3: State the solution to the system of inequalities.

− 1 5/7 < x < 0

Step 3: State the conclusion.

All points in the region R₃ including but not exclusive to (− 1 , 0), (− 1 , 3) (− 0.5 , 2), (− 1 , 1) and ( − 0.5 , − 1 ) are in the solution set for the system.

Self-assessment task Find the solution to y < − 7 x²− 8x + 3 and y > 3 x²− 7x.

You should find − 0.6 < x < 0.5

Quadratic inequalities are a powerful tool used to model and solve various real- world problems that involve limits, boundaries, or optimisation of quantities.

These inequalities describe relationships where one side of the equation is a quadratic expression and are used in situations where the possible outcomes form a range or region rather than a single point.

By learning how to solve quadratic inequalities, we can:

1. Predict outcomes that must fall within thresholds like safety limits, financial losses, etc.

2. Identify ranges of values that satisfy conditions in optimisation problems such as finding the most efficient design.

3. Understand the critical points where changes in conditions occur, allowing for better decision-making and problem-solving.

Let’s go through these examples to see how the concept of quadratic inequalities is useful in real-life.

Example 2.21

An artist is planning an art exhibition and needs to create rectangular canvases with a specific aesthetic balance.

The total area available for displaying the artworks is limited to 80 square feet.

The artist wants the length of each canvas to be 2 feet less than its width.

To ensure that the artworks are visually appealing and fit within the assigned space, determine the maximum width for each canvas such that the total area used does not exceed 80 square feet.

Solution

Step 1: Represent the statement algebraically Let w be the width of each canvas Length of each canvas = w− 2

Step 2: Write the equation for area of a rectangle Since Area(A) = width × length, A = w(w − 2) A = w²− 2w

Step 3: Set up the linear inequality Since the space available for display is maximum 80 sq. feet, the total space covered by the art can be less or equal to 80 square feet.

width × length ≤ 80

Step 4: Set up the quadratic inequality and simplify.

w²− 2w ≤ 80 w²− 2w − 80 ≤ 0 (w − 10)(w + 8) ≤ 0 (Sketch the inequality to confirm the required values) ∴ − 8 ≤ w ≤ 10 Since width can only be positive, the width of each canvas must be less than or equal to 10 feet.

Step 6: State your conclusion The artist can create canvases with a width up to 10 feet, ensuring the canvases fit within the available display area at the exhibition.

Example 2.22

Millicent would like to design a necklace for her mother and decides it should resemble an eye. She makes a drawing as shown in Figure. 2.12. Identify the systems of quadratic inequalities that helped her draw the necklace.x

–3

–3

–3 –2.5

–2.5

–2.5 –2

–2

–2 –1.5

–1.5

–1.5 –1

–1

–1 –0.5

–0.5

–0.5 0.5 0.5 0.5 1 1 1 1.5 1.5 1.5 2 2 2 2.5 2.5 2.5 3 3 3 3.5 3.5 3.5 y

–4

–4

–4

–2

–2

–2 2 2 2 4 4 4 6 6 6 0 0 0

Figure 2.12: Necklace design

Solution

The blue quadratic inequality curve was drawn with y ≤ − 0.5x²+ 3 The brown quadratic inequality curve was drawn with y ≥ − 0.5x²+ 1.5 The magenta quadratic inequality curve was drawn with y ≤ 0.5x²− 1.5 The orange quadratic inequality curve was drawn with y ≥ 0.5x²− 3

Activity 2.9 - Research a Real-Life Problem

1. Identify a real-world scenario where quadratic inequalities can be applied. Use any of the following themes to base your research on:

a. Agriculture: Maximising the area of a crop field, given constraints such as fencing materials.

b. Sports: Determining the best angle and force to kick a ball in football for a successful shot.

c. Architecture: Designing the maximum height of an arch or doorway while ensuring structural safety.

d. Business: Managing production costs to ensure profit margins while minimising loss, such as determining the optimal number of products to produce.

e. Physics: Understanding projectile motion when an object is thrown into the air.

2. Research how quadratic functions and inequalities play a role in this real-world context. Use reliable online sources, textbooks, or interview experts if possible.

3. Once you’ve chosen your problem, model it using a quadratic inequality.

Identify the variables in your scenario (e.g., height, distance, time, cost) and set up the quadratic inequality that best represents the situation.

4. Solve the quadratic inequality you have written. Use algebraic methods or graphical tools to determine the solution set. You can use graphing software, calculators, or sketch graphs by hand to visualise the solution.

5. Discuss the real-life implications of your solution.

a. What do the solutions of the quadratic inequality mean in terms of the real-world problem?

b. What decisions can be made based on your results?

6. Create a visual presentation of your findings to share with your classmates or teacher. Your presentation should include:

a. The real-world problem you chose.

b. The quadratic inequality you modelled from the problem.

c. The steps you took to solve the inequality.

d. The real-world interpretation of the solution.

e. Any challenges or insights you encountered while working on this problem.

You can create a poster, a digital slide presentation or even a short video to present your work.

1. The sum of the 2nd and the 4th term of a geometric sequence is 10 and the sum of the 3nd and 5th term is 20. Find the first term and the common ratio.

2. a. Given that a₁ = 25 and aₙ = aₙ₋₁ + 4 for n > 1 Find:

(i) a₂ , a₃, and a₄

(ii) a formula for aₙ in terms of n

b. Use your formula for aₙ to deduce a formula for Sₙ, the sum of the first n terms of the sequence a₁, a₂, a₃, …aₙ and hence find S₁₀

3. The sum of the three consecutive term of an Arithmetic Progression is − 3 and the product is 24. Find the terms.

4. The nth term of an exponential sequence (G.P) is 2¹⁻ⁿ.

Calculate:

a. the sum Sₙ, of the first four terms of the sequence.

b. the sum of the sequence to infinity.

5. Calculate the difference between the sum to infinity of the series 1 + 1/3 + 1/9 + … , and the sum of the first 5 terms.

6. A sequence defined by U₁ = 2, Uₙ₊₁ = 2 Uₙ − 1, n = 2, 3, 4, …

a. Find the first five terms of the sequence

b. Show that Uₙ + 1 = 3 Uₙ − 2 Uₙ₋₁

7. The sum of the 5th and 9th terms of an A.P, is − 40 and and the 11th term is − 32 . Find the first term and the common difference.

8. A man’s salary is increased by GH¢24 000 each year. His total salary at the end of 14 years is GH¢65 184 000 .

Find:

a. his initial salary

b. his salary in the 14th year

9. The sum of the first n terms of a series is given by (n + 1)²− 1. Find the first three terms and an expression in terms of n for Uₙ, (n > 1).

10. A tennis ball is released from a height h cm, it rebounds one-third of the distance it has fallen after each fall. Find the height of;

a. the third bounce b. the nth bounce

11. Find the sum of the first n terms of a linear sequence whose common difference is 2 is 120 and the sum of the 2n term is 440. Calculate the first term.

12. Evaluate ∑ ᵣ₌₁ ⁶(r²− 2)

13. Renie’s outlet manufactures two types of clothing (shirts and trousers).

The profit is GH¢50.00 for each shirt and GH¢80.00 for each pair of trousers.

14. Due to limited resources, the company can produce a combined total of no more than 120 pieces. Additionally, to meet demand, they must produce at least 20 shirts and 10 pairs of trousers. How many of each of the clothing should be produced to make the most profit?

15. Find the solution(s) to:

a. 3x²+ 5x ≥ 2. b. x²− 2x − 3 ≥ 0.

16. A company determines that the cost in cedis C, of producing m bags of maize is C = 170m + 150.

17. The revenue R, in cedis, from selling all of the bags of maize is R = m²+ 530.

18. How many bags of maize should the company produce and sell if the company wishes to earn a profit of at least GH¢4 000.00?

19. Mr. Buabeng and Ms. Osei run business assembling Hyundai Toyota cars.

The cost of parts and labour needed for each type of car is shown in the

table below.

Car type Cost of Parts (GH¢ thousand) Labour (man-hours) Hyundai 18 15 Toyota 36 30 The business has GH¢150 (thousand) available to buy parts each week.

However, the total labour available each week is 85 man-hours. If they make h of Hyundai and t of Toyota each week:

a. Write down all the inequalities on h and t.

b. Determine the solution to the set of inequalities.

This lesson is continuation of what we learnt in year one functions. We will explore the properties, graphing techniques and basic theorems of polynomial functions. Polynomial functions are crucial in mathematics and have applications in science, engineering and economics. It covers factorising, finding zeros, graphing, Descartes’ Rule of Signs, Fundamental Theorem of Algebra and Linear and Quadratic Factor Theorems. By the end of the section, you will have a deeper understanding of polynomial functions and be prepared to solve a variety of mathematical problems.

KEY IDEAS

· Factorising polynomials involves linear and irreducible quadratic factors.

· Factors and zeros are variables that make a polynomial function equal to zero.

· Graphing polynomials with higher degrees reveal complex shapes, key features, and roots.

· The degree of a polynomial function determines the maximum number of solutions a function could have and the number of times a function will cross the x-axis when graphed.

· The degree of a polynomial function is the total number of factors.

Activity 3.1: Revision of finding quadratic factors In pairs or small groups or individually, perform these activities.

1. Factorise the following quadratic expressions:

a. y²+ 3y + 2

b. 3 y²+ 11y + 10

c. 6x²+ 5x + 1

d. x²− 9

2. Write down the factorising process clearly, showing each step taken to arrive at the factors.

3. Exchange your work with other groups or peers to compare answers.

4. Discuss any differences in methods used and validate the correctness of each group’s results.

An expression (x + y) is a factor if you can multiply it by another expression (e + f) to get the original expression (ex + fx + ey + fy). In simple terms, if you can find two or more expressions that multiply together to make a certain expression, then those are its factors.

For example, (x + 2)(x − 3) are factors of x²− x − 6 because expanding (x + 2)(x − 3) :

x(x − 3) + 2(x − 3) x²− 3x + 2x − 6 x²− x − 6 Therefore, we see that (x + 2)(x − 3) is indeed equal to x²− x − 6 confirming that they are factors of the expression. If you found activity 3.1 challenging, use the task below to help you.

To factorise quadratics of the form ax²+ bx + c :

Example

Task 2x²+ 7x + 6 Task 1 Find two numbers that multiply to give ac (the product of a and c) and add to give b.

a = 2, b = 7 and c = 6 ac = 2 × 6 = 12 4 + 3 = 7 and 4 × 3 = 12 Task 2 Rewrite the middle term using these two numbers as coefficients of x.

2x²+ 4x + 3x + 6 Task 3 If the polynomial has four terms, group the terms in pairs.

(2x²+ 4x) + (3x + 6) Task 4 Factorise out the common factor from each group. 2x(x + 2) + 3(x + 2) Task 5 Look for a common binomial factor and factorise it out and you have the factorised solution. (2x + 3)(x + 2) Factor Theorem and Remainder Theorem In addition to quadratic functions, there are several other types of polynomial functions that can be factorised using the Factor Theorem and the Remainder Theorem. Remember that if (x ± h) is a factor of a polynomial, then the remainder will be zero. Conversely, if the remainder is zero, then (x ± h) is a factor.

Here are some common types of polynomials:

Types of Polynomials

1. Cubic Polynomials: Polynomials of degree three, in the form:

f(x) = ax³+ bx²+ cx + d

2. Quartic Polynomials: Polynomials of degree four, in the form:

f(x) = ax⁴+ bx³+ cx²+ dx + e

3. Quintic Polynomials: Polynomials of degree five, in the form:

f(x) = ax⁵+ bx⁴+ cx³+ dx²+ ex + f

Activity 3.2: Revision on factorisation In groups or individually, use the idea of the factor and remainder theorems to find the factors or the remainders of the following polynomials.

1. Find the remainder when:

a. f(x) = 3x³− 4 x²+ 2x + 3 is divided by (x + 1)

b. f(x) = 8x³− 3 x²+ 6x + 4 is divided by (x − 2)

c. f(x) = 7x³+ 3 x²− x + 6 is divided by (x + 3)

d. f(x) = x³− x²+ x + 1 is divided by (x − 1) 2.

a. show that x + 2 is a factor of f(x) = x³− 2 x²− 5x + 6

b. show that x − 3 is a factor of f(x) = x³− 3 x²− 4x + 12

3. Compare your answers with other groups.

Zero-Product Property

This property of real numbers says that if you multiply two numbers together and the result is zero, then at least one of those numbers must be zero. In simpler terms, if you have two numbers, a and b, and when you multiply them (a×b), and you get zero, it means that either a is zero, or b is zero or both are zero. This property helps us to find the zeros or roots of functions of quadratic functions.

The zeros or roots of a function are the values of the input (usually represented as

x) that make the function equal to zero. In other words, if you have a function f(x), the zeros are the values of x for which f(x) = 0. For example, given the function f

(x) = 2x²− x − 6 , You:

1. equate f(x) = 0 2x²− x − 6 = 0

2. factorise the function 2x²− 4x + 3x − 6 = 0 2x(x − 2) + 3(x − 2) = 0 (2x + 3)(x − 2) = 0 Either (2x + 3) = 0 or (x − 2) = 0

3. find the solution for each factor (2x + 3) = 0 2x = − 3 x = −3_ 2 (x − 2) = 0 x = 0 − 2 x = 2 Therefore, the zeros, or roots, of the function f(x) = 2x²− x − 6 are x = −3/2 and x = 2 because both values make the function equal to zero.

We will apply the Zero-Product Property to solve polynomial equations in the next task.

Solving a Polynomial Equation

There are a number of ways that can be used to find the zeros, or the roots, of polynomial functions. We can use a calculator, plot a graph, use GeoGebra (software) and we can also calculate by using the zero property.

To find the zeros, or roots, by the calculation method:

Step 1: use the factor theorem approach to factorise the expression completely.

Step 2: equate each factor to 0 and solve.

In pairs or in groups, let’s work through the following examples using the steps above.

Example 3.1

Find the zeros of the cubic function f(x) = x³− 2 x²− 5x + 6

Solution

x³− 2 x²− 5x + 6 = 0 Possible factors of 6 = ± 1, ± 2, ± 3, ± 6 Testing with these factors, When x = 1, f(x) = (1)³− 2 (1)²− 5(1) + 6 f(x) = 1 − 2 − 5 + 6 f(x) = 0 ∴ (x − 1) is a factor. If x=1 was not a factor, try another option, eg x=-1.

Using the long division method:

x²− x − 6 x − 1 √_____________ x³− 2 x²− 5x + 6 −(x³− x²+ 0x + 0)__ −x²− 5x + 6 −(−x²+ x − 0)__ − 6x + 6 −(− 6x + 6)_ 0 Factorising, x²− x − 6 x²− 3x + 2x − 6 (x²− 3x) + (2x − 6) x(x − 3) + 2(x − 3) (x + 2)(x − 3) ∴ f(x) = (x − 1)(x + 2)(x − 3) f(x) = 0 (x − 1)(x + 2)(x − 3) = 0 For (x − 1) = 0 x = 1 For (x + 2) = 0 x = − 2 For (x − 3) = 0 x = 3 Hence the zeros are when x = − 2, 1, 3

Example 3.2

Find the roots, or zeros, of the cubic function f(x) = x³+ x²− 10x + 8

Solution

x³+ x²− 10x + 8 = 0 Possible factors of 8 = ± 1, ± 2, ± 4, ± 8 Testing with these factors, When x = 1, f(x) = (1)³+ (1)²− 10(1) + 8 f(x) = 1 + 1 − 10 + 8 f(x) = 0 ∴ (x − 1) is a factor.

Using the long division method:

x²+ 2x − 8 x − 1 √_____________ x³+ x²− 10x + 8 −(x³− x²+ 0x + 0)__ 2 x²− 10x + 8 −(2 x²− 2x − 0)__ − 8x + 8 −(− 8x + 8)_ 0 Factorising, x²+ 2x − 8 x²+ 4x − 2x − 8 (x²+ 4x) + (2x − 8) x(x + 4) − 2(x + 4) (x + 4)(x − 2) ∴ f(x) = (x − 1)(x − 2)(x + 4) f(x) = 0 (x − 1)(x − 2)(x + 4) = 0 For (x − 1) = 0 x = 1 For (x − 2) = 0 x = 2 For (x + 4) = 0 x = − 4 Hence the roots, or zeros, are when x = − 4, 2, 1

Example 3.3

Find the zeros of the cubic function. f(x) = x³− 3 x²− 4x + 12

Solution

x³− 3 x²− 4x + 12 = 0 Possible factors of 12 = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 Testing with these factors, When x = 1, f(x) = (1)³− 3 (1)²− 4(1) + 12 f(x) = 1 − 3 − 4 + 12 f(x) = 6 ∴ (x − 1) is not a factor and we need to try another potential factor.

When x = 2, f(x) = (2)³− 3 (2)²− 4(2) + 12 f(x) = 8 − 12 − 8 + 12 f(x) = 0 ∴ (x − 2) is a factor.

Using the long division method:

x²− x − 6 x − 2 √______________ x³− 3 x²− 4x + 12 −(x³− 2x²+ 0x + 0)__ −x²− 4x + 12 −(x²+ 2x − 0)__ − 6x + 12 −(6x + 12)_ 0 Factorising, x²− x − 6 x²− 3x + 2x − 6 (x²− 3x) + (2x − 6) x(x − 3) + 2(x − 3) (x + 2)(x − 3) ∴ f(x) = (x − 2)(x + 2)(x − 3) f(x) = 0 (x − 2)(x + 2)(x − 3) = 0 For (x − 2) = 0 x = 2 For (x + 2) = 0 x = − 2 For (x − 3) = 0 x = 3 Hence the zeros are when x = − 2, 2, 3 Following the remainder theorem is the Rational Zero Theorem which helps to identify exactly which of the factor will give us zero.

The Rational Zero Theorem

The Rational Zero Theorem helps us identify potential rational zeros of a polynomial by looking at the factors of two specific parts of the polynomial:

1. Factors of the Constant Term: This is the term without a variable (the last number in the polynomial when arranged in descending powers).

2. Factors of the Leading Coefficient: This is the coefficient (the number in front of the term) of the term with the highest power of x.

Steps to Find Possible Rational Zeros

1. Identify the Constant Term and Leading Coefficient.

For example, in the polynomial 4x³+ 7x²+ 2x + 5 :

The constant term is 5 .

The leading coefficient is 4 .

2. List the Factors.

Factors of the Constant Term (5): These are: ± 1 , ± 5 Factors of the Leading Coefficient (4): These are ± 1 , ± 2, ± 4 .

3. Create Possible Rational Zeros.

To find the possible rational zeros, take each factor of the constant term and divide it by each factor of the leading coefficient.

Possible rational zeros for our example would be:

From ±1: 1/1 , −1/1 , 1/2 , −1/2 , 1/4 , − 1/4 , = ± 1 , ±1/2 , ±1/4, From ±5: 5/1 − 5/1 , 5/2 , −5/2 , 5/4 , − 5/4 , = ± 5 , ±5/2 , ±5/4

4. List of Possible Rational Zeros:

Combining all these, the possible rational zeros for our polynomial are:

± 1 , ± 5 , ±1_ 2 , ±1_ 4 , , ±5_ 2 , ±5_ 4 Now that we have a list of possible rational zeros, we can use the Factor Theorem to test each one.

a. For each rational number c from the list, substitute it into the polynomial f(x)

b. If f(c) = 0, then c is a zero of the polynomial.

Example 3.4

In groups or individually, use the rational zeros theorem to list all the possible rational zeros.

a. 6x³+ 15x²− 8x + 2

b. x³+ x²− 2x − 5

c. 3x³+ x²+ x − 5

d. 6x³− 4x²+ 17

Solution

a. 6x³+ 15x²− 8x + 2 The constant term is 2 .

The leading coefficient is 6 .

Factors of the Constant Term (2): These are.± 1 , ± 2 Factors of the Leading Coefficient (6): These are ± 1 , ± 2, ± 3 , ± 6 .

Possible Rational Zeros:

From ±1: 1/1 , −1/1 , 1/2 , −1/2 , 1/3 , − 1/3 , 1/6, −1/6 = ± 1 , ±1/2 , ±1/3, ±1/6, From ±2: 2/2 , −2/2 , 2/3 , − 2/3 , 2/6, −2/6 = ± 1 , ±2/3 , ±1/6,

b. x³+ x²− 2x − 8 The constant term is − 8 .

The leading coefficient is 1 .

Factors of the Constant Term (-8): These are.± 1 , ± 2 , ± 4, ± 8 Factors of the Leading Coefficient (1): These are ± 1 .

Possible Rational Zeros:

From ±1: 1/1 , −1/1 = ± 1 From ±2: , 2/1 , −2/1 = ± 2 From ±4: 4/1 , −4/1 , = ± 4 From ±8: , 8/1 , −8/1 = ± 8, , Possible Rational Zeros (combined): ± 1, ± 2, ± 4, ± 8

c. 9x³+ x²+ x − 10 The constant term is − 10 .

The leading coefficient is 9 .

Factors of the Constant Term (-10): These are.± 1 , ± 2 , ± 5, ± 10 Factors of the Leading Coefficient (9): These are ± 1, ± 3, ± 9 .

Possible Rational Zeros:

From ±1: 1/1 , −1/1, 1/3, − 1/3, 1/9, −1/9 = ± 1, ±1/3 , ±1/9 From ±2: 2/1 , −2/1, 2/3, − 2/3, 2/9, −2/9 = ± 2, ±2/3 , ±2/9 From ±4: 5/1 , −5/1, 5/3, − 5/3, 5/9, −5/9 = ± 5, ±5/3 , ±5/9 From ±10: , 10/1 , −10/1 , 10/3 , − 10/3 , 10/9 , −10/9 = ± 10, ±10/3 , ±10/9 Possible Rational Zeros (combined): ± 1, ± 2, ± 5, ± 10 ±1/3 , ±1/9 , ±2/3 , ±2/9, ±5/3 , ±5/9, ±10/3 , ±10/9

d. 6x³− 4x²+ 17 The constant term is 17 .

The leading coefficient is 6 .

Factors of the Constant Term (17): These are.± 1 , ± 17 Factors of the Leading Coefficient (6): These are ± 1, ± 2, ± 3, ± 6 .

Possible Rational Zeros:

From ±1: 1/1 , −1/1, 1/2, −1/2, 1/3 , −1/3, 1/6, −1/6 = ± 1, ± 1/2, ±1/3 , ±1/6 From ±17: 17/1 , −17/1 , 17/2 , −17/2 , 17/3 , −17/3 , 17/6 , −17/6 = = ± 17, ± 17/2 , ±17/3 , ±17/6 Possible Rational Zeros (combined): ± 1, ± 17, ± 1/2, ±1/3 , ±1/6, ± 17/2 , ±17/3 , ±17/6 Using the rational zero theorem to find the zeros of a polynomial function Steps to Follow:

1. Identify Possible Rational Zeros

Use the Rational Zero theorem to list all possible rational zeros. These are of the form p and q, where p is a factor of the constant term (the last term of the polynomial) and q is a factor of the leading coefficient (the coefficient of the term with the highest degree).

2. Substitute and Evaluate

a. Substitute each possible rational zero into the polynomial function.

b. Calculate the value of the polynomial for each substitution.

3. Determine Actual Zeros

a. Identify which substitutions result in a value of zero.

b. The values for which the polynomial equals zero are the rational zeros of the function.

Example 3.5

Given the polynomial 3x³+ 8x²+ 7x + 2

1. List the factors of the constant term and leading coefficient.

2. Determine possible rational zeros

3. Substitute each value into the polynomial to find which ones yield zero.

Solution

The constant term is 2 .

The leading coefficient is 3 .

Factors of the Constant Term (2): These are.± 1 , ± 2 Factors of the Leading Coefficient (3): These are ± 1, ± 3 .

Possible Rational Zeros:

From ±1: 1/1 , −1/1, 1/3, −1/3 = ± 1, ± 1/3 From ±2: 2/1, −2/1 , 2/3, −2/3, = = ± 2, ± 2/3 Possible Rational Zeros (combined): ± 1, ± 2, ± 1/3, ±2/3 Substitute and evaluate:

3x³+ 8x²+ 7x + 2 For x = ± 1 :

x = 1 3(1)³+ 8(1)²+ 7(1) + 2 3 + 8 + 7 + 2 = 20 x = − 1 3( − 1)³+ 8( − 1)²+ 7( − 1) + 2 − 3 + 8 − 7 + 2 = 0 For x = ± 2 x = 2 3(2)³+ 8(2)²+ 7(2) + 2 24 + 32 + 14 + 2 = 71 x = − 2 3( − 2)³+ 8( − 2)²+ 7( − 2) + 2 − 24 + 32 − 14 + 2 = − 4 For ± 1/3 x = 1_ 3 3(1_ 3) 3 + 8 (1_ 3) 2 + 7(1_

3) + 2 1_ 9 + 8_ 9 + 7_ 3 + 2 = 16_ 3 x = −1_ 3 3(−1_ 3) 3 + 8 (−1_ 3) 2 + 7(−1_

3) + 2 −1_ 9 + 8_ 9 − 7_ 3 + 2 = 4_ 9 For ± 2/3 x = 2_ 3 3 (2_ 3) 3 + 8 (2_ 3) 2 + 7(2_

3) + 2 8_ 9 + 32_ 9 + 14_ 3 + 2 = 100_ 9 x = −2_ 3 3 (−2_ 3) 3 + 8 (−2_ 3) 2 + 7(−2_

3) + 2 −8_ 9 + 32_ 9 − 14_ 3 + 2 = 0 The values 1 , 2, − 2, 1/3, −1/3, 2/3 did not yield zero and are not zeros of the function.

However, − 1 and −2/3 , yielded zero and therefore represent the zeros of the function

1. In mathematics, the terms “sketch” and “draw” have distinct meanings.

2. Sketch is a rough representation of a graph or function without precise measurements, used to illustrate its behaviour. It often lacks scales and coordinates.

3. Draw, on the other hand, creates a more accurate representation with specific values, scales, and tools, providing a detailed visualization of a mathematical function or data set.

We will be sketching polynomial functions with a degree higher than 2.

These will be cubics, quartics etc.

To sketch a polynomial of degree greater than 2, for example, ax³+ bx²+ cx + d , we:

Step 1: Determine the sign ( + or -) of a; if a > 0 or a < 0 If a < 0, the curve will have the shape:

Figure 3.1: Curve of a polynomial If a < 0, the curve will have the shape:

Figure 3.2: Curve of a polynomial

Step 2: Use the factor theorem or any known method to find the zeros of the polynomial

Step 3: Find the y- intercept of the function by substituting x = 0 into the given polynomial function. (Note that this equates to the constant.)

Step 4: Find the intervals of the curve based on the zeros obtained

Step 5: Sketch the curve.

Individually, or in groups, let’s use the steps for sketching polynomials to solve the examples below.

Example 3.6

Sketch the curve 2x³+ 3x²− 5x − 6

Solution

Step 1: a is 2, which is greater than 0. Therefore, we know the shape that the curve will take. The extremes will go from bottom left to top right.

Step 2: Using the factor theorem, the factors of the constant, - 6 are ± 1, ± 2, ± 3, ± 6 Testing when x = - 1, we have:

2( − 1)³+ 3(− 1)²− 5(− 1) − 6 = − 2 + 3 + 5 − 6 = 0 Which means (x + 1) is a factor.

We can use long division to find the other factors.

2 x²+ x − 6 x + 1 √______________ 2x³+ 3x²− 5x − 6 −(2 x³+ 2x²+ 0 + 0)__ x²− 5x − 6 −(x²+ x + 0)__ (− 6x − 6) − 6x − 6_ Factorising:

2 x²+ x − 6 2x²+ 4x − 3x − 6 2x(x + 2) − 3(x + 2) (2x − 3)(x + 2) Factors are:

(x + 1)(2x − 3)(x + 2) The zeros:

(x + 1)(2x − 3)(x + 2) = 0 x = − 1, x = − 2, x = 3_ 2

Step 3: y intercept When x = 0 2(0)³+ 3(0)²− 5(0) − 6 = − 6

Step 4: finding the intervals using the zeros The intervals are:

1. − 2 < x

2. − 2 < x < − 1

3. − 1 < x < 1.5

4. x > 1.5 Next is to determine the signs or test for the signs in each of the intervals obtained Interval − 2 > x − 2 < x < − 1 − 1 < x < 1.5 x > 1.5 Chosen values (you can choose any number within the interval)

-3 -1.5 1 2 y value (substitute your chosen number into the given function) 2( − 3)³+ 3( − 3)²− 5 (− 3) − 6 = -18 2( − 1.5)³+ 3( − 1.5)²− 5(− 1.5) − 6 = 1.5 2(1)³+ 3(1)²− 5(1) − 6 = -6 2(2)³+ 3(2)²− 5(2) − 6 = 12 Sign Negative Positive Negative Positive Behaviour of curve Below x - axis Above x - axis Below x - axis Above x - axis

Step 5: sketch the curve

Figure 3.3: Curve of 2x³+ 3x²− 5x − 6

Example 3.7

Sketch the curve x³− 6x²+ 5x + 12

Solution

Step 1: a is 1 which is greater than 0. This means we know the shape the curve will take.

Step 2: Using the factor theorem, the factors of the constant, 12 are ± 1, ± 2, ± 3, ± 4 ± 6, ± 12 Testing when x = - 1, we have:

( − 1)³− 6 (− 1)²+ 5(− 1) + 12 − 1 − 6 − 5 + 12 = 0 Which means (x + 1) is a factor We can use long division to find the other factors.

x²− 7x + 12 x + 1 √______________ x³− 6x²+ 5x + 12 −(x³+ x²+ 0 + 0)__ − 7 x²+ 5x + 12 (− 7x²− 7x + 12) 12x + 12 −(12x + 12)__ 0 Factorising:

x²− 7x + 12 x²− 4x − 3x + 12 x(x − 4) − 3(x − 4) (x − 3)(x − 4) Factors are (x + 1)(x − 3)(x − 4) The zeros (x + 1)(x − 3)(x − 4) = 0 x = − 1, x = 3, x = 4

Step 3: y intercept At x = 0 (0)³− 6(0)²+ 5(0) + 12 = 12

Step 4: finding the intervals using the zeros The interval are

1. − 1 < x

2. − 1 < x < 3

3. 3 < x < 4

4. x > 4 Next is to determine the signs or test for the signs in each of the intervals obtained

Table 3.1: Test for signs in each of the intervals obtained Interval − 2 > x − 2 < x < − 1 − 1 < x < 1.5 x > 1.5 Chosen values (you can choose any number within the interval)

-2 1 3.5 5 y value (substitute your chosen number into the given function) ( − 2)³− 6 (− 2)²+ 5 (− 2) + 12 = -30 (1)³− 6 (1)²+ 5(1) + 12 = 12 (3.5)³− 6 (3.5)²+ 5 (3.5) + 12 = -1.125 2(5)³+ 3(5)²− 5(5) − 6 = 294 Sign Negative Positive Negative Positive Behaviour of curve Below x - axis Above x - axis Below x - axis Above x - axis

Figure 3.4: Curve of x³− 6x²+ 5x + 12

Example 3.8

Sketch the curve −x³+ 4x²− x − 6

Solution

Step 1: a is -1 which is less than 0. This means that the extremes of the graph go from top left to bottom right.

Step 2: Using the factor theorem, the factors of the constant, -6 are ± 1, ± 2, ± 3, ± 6 Testing when x = 2, we have:

−(2)³+ 4 (2)²− 2 − 6 = − 8 + 16 − 2 − 6 = 0 Which means (x − 2) is a factor.

We can use long division to find the other factors.

− x²+ 2x + 3 x − 2 √_____________ − x³+ 4x²− x − 6 −(−x³+ 2 x²+ 0 + 0)__ 2 x²− x − 6 −(2x²− 4x − 6) 3x − 6 −(3x − 6)__ 0 Factorising:

−x²+ 2x + 3 − x²+ 3x − x + 3 − x(x − 3) − 1(x − 3) ( − x − 1)(x − 3) Factors are (x − 2)(x − 3)( − x − 1) The zeros (x − 2)(x − 3)(− x − 1) = 0 x = − 1, x = 3, x = 2

Step 3: y intercept:

When x = 0 −(0)³+ 4 (0)²− (0) − 6 = − 6

Step 4: finding the intervals using the zeros The intervals are

1. − 1 < x

2. − 1 < x < 2

3. 2 < x < 3

4. x > 3 Next is to determine the signs or test for the signs in each of the intervals obtained

Table 3.2: Test for signs in each of the intervals obtained Interval − 1 > x − 1 < x < 2 2 < x < 3 x > 3 Chosen values (you can choose any number within the interval)

-3 1 2.5 4 y value (substitute your chosen number into the given function) −( − 3)³+ 4 (− 3)²− (− 3) − 6 = 60 −(1)³+ 4 (1)²− (1) − 6 = -4 −(2.5)³+ 4 (2.5)²− (2.5) − 6 = 0.875 −(4)³+ 4 (4)²− (4) − 6 = -10 Sign Positive Negative Positive Negative Behaviour of curve Above x - axis Below x - axis Above x - axis Below x - axis

Figure 3.5: Curve of −x³+ 4x²− x − 6 What polynomial function is represented in the graph below?

a. How many factors does it have?

b. Share your answers with the class.

Figure 3.6: Graph of a polynomial function Remember that it is an important skill to be able to sketch graphs by hand.

However, when appropriate technology is there to assist. For example, you can use apps or web-based programs such as GeoGebra, Demos, PhET Simulations and Geometer’s Sketch Pad.

There is another way that can be used to determine how many positive and negative real zeros a polynomial function might have. To do this, first write the polynomial in descending order (from the highest degree to the lowest) if it is not already. We use Descartes’ Rule of Signs. This rule helps you see how the number of times the signs change in the polynomial is related to the number of positive real zeros.

For example, if you look at the polynomial function below, you will notice it has one sign change, which gives you information about its positive zeros.

Let us go through an example using Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros.

Example 3.9

Consider the polynomial:

f(x) = 3x³− 2 x²+ 4x − 5

Solution

Step 1: Identify Sign Changes for Positive Real Zeros

1. Write the polynomial in descending order:

2. f(x) = 3x³− 2 x²+ 4x + 5 Look at the coefficients:

3 (positive)

–2 (negative) 4 (positive) −5 (negative)

3. Determine the sign changes:

From 3 to -2 (1 change) From −2 to 4 (2 changes) From 4 to −5 (3 changes) Total sign changes for positive zeros = 3. This means there could be 3, (3

– 2 = 1) positive real zeros. Note, the number of possible roots is found by subtracting 2 each time until we would get a negative number.

Step 2: Identify Sign Changes for Negative Real Zeros Now, let’s find the negative real zeros by evaluating f(−x):

f(− x) = 3( − x)³− 2 (− x)²+ 4( − x) − 5 f(− x) = − 3 x³− 2 x²− 4x − 5

Step 3: Look at the Signs of f(−x) The coefficients are:

− 3 (negative) − 2 (negative) − 4 (negative) − 5 (negative)

Step 4: Determine Sign Changes for Negative Real Zeros There are no sign changes in f(−x) Total sign changes for negative zeros = 0. This means there are no negative real roots or zeros.

This example illustrates how to use Descartes’ Rule of Signs to analyse a polynomial and determine the possible numbers of positive and negative real zeros.

Let us solve more examples.

Example 3.10

Use Descartes’ Rule of Signs to determine how many possible positive and negative real zeros of the following:

a. g(x) = 2x⁴− 11 x³+ 4x²− 8x − 40

b. g(x) = 9 x³− 10 x²+ 5x + 3

c. f(x) = −x⁵+ 3 x⁴− 5x³− 12 x²+ 6x + 4

Solution

a. g(x) = 2x⁴− 11 x³+ 4x²− 8x − 40

Step 1: Identify Sign Changes for Positive Real Zeros Write the polynomial in descending order:

g(x) = 2x⁴− 11 x³+ 4x²− 8x − 40 Look at the coefficients:

2 (positive)

–11 (negative) 4 (positive) −8 (negative) −40 (negative) Look at the coefficients:

Determine the sign changes:

From 2 to -11 (1 change) From −11 to 4 (2 changes) From 4 to −8 (3 changes) From -8 to −40 (no change) Total sign changes for positive zeros = 3. This means there could be 3 or 1 positive real zeros.

Step 2: Identify Sign Changes for Negative Real Zeros Now, let’s find the negative real zeros by evaluating f(−x):

g(x) = 2( − x)⁴− 11( − x)³+ 4(− x)²− 8( − x) − 40 g(x) = 2x⁴+ 11 x³+ 4x²+ 8x − 40

Step 3: Look at the Signs of f(−x) The coefficients are:

2 (positive) 11 (positive) 4 (positive) 8 (positive) − 40 (negative)

Step 4: Determine Sign Changes for Negative Real Zeros Determine the sign changes:

From 2 to 11 (no change) From 11 to 4 (no changes) From 4 to 8 (no changes) From 8 to −40 (1 change) Total sign changes for negative zeros = 1. This means there is 1 negative real zero.

b. g(x) = 9 x³− 10 x²+ 5x + 3

Step 1: Identify Sign Changes for Positive Real Zeros Write the polynomial in descending order: g(x) = 9 x³− 10 x²+ 5x + 3 .

Look at the coefficients:

9 (positive) − 10 (negative) 5 (positive) 3 (negative) Determine the sign changes:

From 9 to -10 (1 change) From −10 to 5 (2 changes) From 5 to 3 (no change) Total sign changes for positive zeros = 2. This means there could be 2 or 0.

Step 2: Identify Sign Changes for Negative Real Zeros Now, let’s find the negative real zeros by evaluating g(−x):

g(x) = 9( − x)³− 10( − x)²+ 5( − x) + 3 g(x) = − 9 x³− 10 x²− 5x + 3

Step 3: Look at the Signs of f(−x) The coefficients are:

− 9 (negative) − 10 (negative) − 5 (negative) 3 (positive)

Step 4: Determine Sign Changes for Negative Real Zeros Determine the sign changes:

From -9 to -10 (no change) From -10 to -5 (no changes) From -5 to 3 (1 change) Total sign changes for negative zeros = 1. This means there is 1 negative real zero.

c. f(x) = −x⁵+ 3 x⁴− 5x³− 12 x²+ 6x + 4

Step 1: Identify Sign Changes for Positive Real Zeros Write the polynomial in descending order:

f(x) = −x⁵+ 3 x⁴− 5x³− 12 x²+ 6x + 4 Look at the coefficients:

−1 (negative) 3 (positive) −5 (negative) −12 (negative) 6 (positive) 4 (positive) Determine the sign changes:

From −1 to 3 (1 change) From 3 to −5 (2 changes) From −5 to −12 (no changes) From −12 to 6 (3 changes) From 6 to 4 (no change) Total sign changes for positive zeros = 3. This means there could be 3 or 1 positive real zeros.

Step 2: Identify Sign Changes for Negative Real Zeros Now, let’s find the negative real zeros by evaluating f(−x):

f(x) = −( − x)⁵+ 3( − x)⁴− 5(− x)³− 12( − x)²+ 6( − x) + 4 f(x) = x⁵+ 3 x⁴+ 5x³− 12 x²− 6x + 4

Step 3: Look at the Signs of f(−x) The coefficients are:

1 (positive) 3 (positive) 5 (positive)

–12 (negative)

–6 (negative) 4 (positive)

Step 4: Determine Sign Changes for Negative Real Zeros Determine the sign changes:

From 1 to 3 (no change) From 3 to 5 (no changes) From 5 to −12 (2 changes) From −2 to −6 no changes) From −6 to 4 (3 change) Total sign changes for negative zeros = 2. This means there is 2 or 0 negative real zeros.

The Fundamental Theorem of Algebra tells us something very important about polynomial functions. It says: ‘Every polynomial function of degree n (where n >

0) has at least one complex zero.’

This might sound complicated at first, but here is what it means:

1. The degree of a polynomial is the highest power of x in the expression.

2. For example, x³− 2x²+ 4x − 8 has a degree of 3.

3. Complex numbers include real numbers and imaginary numbers (numbers with i, where i = √− 1)

4. In factorising the polynomial, each root cᵢcorresponds to a linear factor of the form (x − cᵢ). This means we can write the polynomial as f(x) = a (x − c₁) (x − c₂) ⋯ (x − cₙ).

The, a is a non-zero number that affects how the polynomial behaves, and c₁, c₂, …, cₙ, are the roots, or zeros.

Let us go through these examples to show how it works.

Example 3.11

Factorise the polynomial f(x ) = 3 x⁵− 48x completely and find all its zeros.

State the multiplicity of each zero.

Solution

f(x ) = 3 x⁵− 48x Factorising:

3x( x⁴− 16) 3x[(x²)²− 4²] 3x[(x²+ 4)(x²− 4)] 3x[x²− ( − 4)( x²− 4)] 3x[(x + 2i)(x − 2i)(x + 2)(x − 2)] To find the zeros, equate the factors to zero:

3x[(x + 2i)(x − 2i)(x + 2)(x − 2)] = 0 3x = 0 , x = 0 x + 2i = 0 , x = − 2i x − 2i = 0 , x = 2i x + 2 = 0 , x = − 2 x − 2 = 0 , x = 2 Therefore, the zeros of f(x) are 0, 2, – 2, 2i and – 2i . Since each factor occurs only once, all the zeros are of multiplicity 1 and the total number of zeros is five.

Example 3.12

Factorise the polynomial f(x ) = x⁴− 625 completely and find all its zeros.

State the multiplicity of each zero.

Solution

f(x ) = x⁴− 625 Factorising:

( x⁴− 625) [(x²)²− 25²] [(x²+ 25)(x²− 25)] 3x[x²− ( − 25)( x²− 25)] [(x + 5i)(x − 5i)(x + 5)(x − 5)] To find the zeros, equate the factors to zero:

[(x + 5i)(x − 5i)(x + 5)(x − 5)] = 0 x + 5i = 0 , x = − 5i x − 5i = 0 , x = 5i x + 5 = 0 , x = − 5 x − 5 = 0 , x = 5 Therefore, the zeros of f(x) are 5, – 5, 5i and – 5i . Since each factor occurs only once, all the zeros are of multiplicity 1 and the total number of zeros is four.

Example 3.13

Factorise the polynomial f(x ) = 15 x⁴+ 8x²+ 1 completely and find all its zeros.

State the multiplicity of each zero.

Solution

f(x ) = 15 x⁴+ 8x²+ 1 in this scenario, we can represent x²by any other letter, for example, u f(x) = 15 (x²)²+ 8x²+ 1 f(x) = 15 (u)²+ 8u + 1 15u²+ 8u + 1 Factorising:

15u²+ 5u + 3u + 1 5u(3u + 1) + (3u + 1) (5u + 1)(3u + 1) Substituting x²back, ( 5x²+ 1)( 3x²+ 1) To find the zeros, equate the factors to zero:

( 5x²+ 1)( 3x²+ 1) = 0 5x²+ 1 = 0 or 3x²+ 1 = 0 For 5x²+ 1 = 0 5x²= − 1 x²= −1_ 5 x = ±√−1_ 5 x = −√__ −1/5 or x = √__ −1/5 For x = −√__ −1/5 x = −√1_ 5 × √− 1 x = − i √1_ 5 x = −√5_ 5 i For x = √__ −1/5 x = √1_ 5 × √− 1 x = i √1_ 5 x = √5_ 5 i For 3x²+ 1 = 0 3x²= − 1 x²= −1_ 3 x = ±√−1_ 3 x = −√__ −1/3 or x = √__ −1/3 For x = −√__ −1/3 x = −√1_ 3 × √− 1 x = − i √1_ 3 x = −√3_ 3 i For x = √__ −1/3 x = √1_ 3 × √− 1 x = i √1_ 3 x = √3_ 3 i Therefore, the zeros of f(x) √3/3 i , −√3/3 i , −√5/5 i , √5/5 i . Since each factor occurs only once, all the zeros are of multiplicity 1 and the total number of zeros is four.

Example 3.14

Factorise the polynomial f(x ) = x⁴− 4 x²− 21 completely and find all its zeros.

State the multiplicity of each zero.

Solution

f(x ) = x⁴− 4 x²− 21 in this scenario, we can represent x²by u f(x) = (x²)²− 4 x²− 21 f(x) = (u)²− 4u − 21 u²− 4u − 21 Factorising:

u²− 7u + 3u − 21 u(u − 7) + 3(u − 7) (u + 3)(u − 7) Substituting x²back, ( x²+ 3)( x²− 7) To find the zeros, equate the factors to zero:

( x²+ 3)( x²− 7) = 0 x²+ 3 = 0 or x²− 7 = 0 For x²+ 3 = 0 x²= − 3 x²= − 3 x = ±√− 3 x = −√− 3 or x = √− 3 For x = −√− 3 √− 3 = −√3 × √− 1 x = −√3 × √− 1 x = − i √3 For x = √− 3 x = √3 × √− 1 x = i √3 For x²− 7 x²= 7 x²= 7 x = ±√7 x = −√7 or x = √7 For x = −√7 x = −√7 For x = √7 x = √7 x = √7 Therefore, the zeros of f(x) are x = i √3 , x = − i √3, x = −√7, x = √7. Since each factor occurs only once, all the zeros are of multiplicity 1 and the total number of zeros is four.

Note: Multiplicity refers to the number of times a particular zero (or root) appears in the factorisation of a polynomial. It indicates how many times a specific value x = r is a solution to the polynomial equation P(x) = 0.

Key Points about Multiplicity:

1. Simple Roots: If a root appears once, it has a multiplicity of 1. For example, in the polynomial (x − 2), the root x = 2 has multiplicity 1.

2. Repeated Roots: If a root appears more than once, it has a higher multiplicity.

For example, in the polynomial (x − 2)², the root x = 2 has a multiplicity of 2.

3. Multiplicity and Behaviour: This can be even or odd.

Odd Multiplicity: If a root has an odd multiplicity, the graph of the polynomial crosses the x - axis at that root.

Even Multiplicity: If a root has an even multiplicity, the graph of the polynomial touches the x-axis at that root but does not cross it.

For instance, for the polynomial P(x) = (x − 1) (x − 1) (x + 2):

The root x = 1 has a multiplicity of 2 (it appears twice), so the graph just touches the x-axis at x = 1 and does not cross it.

The root x = −2 has a multiplicity of 1 (it appears once), so the graph crosses the x-axis at x = -2.

Complex Numbers: A complex number is expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit defined as i²= −1, or i = √− 1 Roots of Polynomials: When dealing with polynomial equations, the Fundamental Theorem of Algebra states that every non-constant polynomial has at least one complex root. This means that if a polynomial has real coefficients, any non-real roots must occur in conjugate pairs.

If a + bi (where b ≠ 0) is a root of a polynomial, then its conjugate a − bi is also a root.

Example: For a polynomial like x²+ 4:

The roots can be found by rearranging to x²= −4.

Taking the square root gives x = ± 2i Here, 2i and −2i are complex roots that come in conjugate pairs.

Note that:

Complex roots always come in conjugate pairs.

If a + bi is a root of a polynomial, then a−bi is also a root.

This property helps us find all the roots of polynomials, ensuring we account for both real and complex solutions.

The Complex Conjugate Theorem states that if a polynomial has a complex root, its conjugate must also be a root. This is an important concept in algebra that helps us understand the nature of polynomial equations and their solutions.

Let us go through the following examples to explore how to use the Complex conjugate theorem.

For example, if we are given one complex factor of a polynomial, we know another factor must be the complex conjugate. This gives us two factors. We can then form a quadratic equation and use algebraic division to find another factor.

Or we may be given two real factors, for example, x = 2 and x = − 3 . This means that x − 2 = 0 or x + 3 = 0 ∴ (x − 2)(x + 3) = 0 x²+ 3x − 2x − 6 = 0 x²+ x − 6 = 0

Example 3.15

Find all the roots of f(x) = x³+ 8 x²+ 21x + 20 if one root is -2 − i .

Solution

f(x) = x³+ 8 x²+ 21x+ 20 Since the factor given ( − 2 − i) is complex in nature, we need the conjugate as well: − 2 + i [x − ( − 2 − i)[(x − ( − 2 + i)] [x + 2 + i][x + 2 − i] (x + 2)²− (i)²x²+ 4x + 4 − i²NB: −i²= 1 x²+ 4x + 4 − ( − 1) x²+ 4x + 5 We can use long division to find the other factors:

x + 4 x²+ 4x + 5 √_______________ x³+ 8 x²+ 21x + 20 −(x³+ 4x²+ 5x + 20)__ 4 x²+ 16x + 20 −(4x²+ 16x + 20)__ 0 x + 4 = 0 x = − 4 Therefore, the roots of (x ) are −4, − 2 + i and −2 – i

Example 3.16

Find all the roots of f(x) = 2x³− 13x²+ 26x − 10 if one root is 3 + i .

Solution

f(x) = 2x³− 13x²+ 26x − 10 Since the factor given 3 + i . is complex in nature, we need the conjugate as well:

3 − i [x − (3 + i)[(x − (3 − i)] [x − 3 − i][x − 3 + i] (x − 3)²− (i)²x²− 6x + 9 − i²NB: −i²= − 1 x²− 6x + 9 − ( − 1) x²− 6x + 10 We can use long division to find the other factor.

2x − 1 x²− 6x + 10 √__________________ 2 x³− 13 x²+ 26x − 10 −(2x³− 1 2x²+ 20x + 0)___ −x²+ 6x − 10 −(− x²+ 6x − 10)__ 0 2x − 1 = 0 x = 1_ 2 Therefore, the roots of (x ) are = 1/2 , 3 + i and 3 – i

The linear quadratic factor theorem is another method which can be used to solve polynomial functions. This is when we break polynomial functions into linear and quadratic factors to make solving it simpler.

For example, to use the linear and quadratic factor approach to find the factors of x⁴− 9 x²+ 14 , we:

Step 1: Let x²= u

Step 2: Using this substitution it means x⁴− 9 x²+ 14 = (u)²− 9(u) + 14

Step 3: Factorise u²− 9u + 14 :

u²− 7u − 2u + 14 u(u − 7) − 2(u − 7) (u − 2)(u − 7)

Step 4: Substitute x²back in for u :

( x²− 2)( x²− 7) x²− 2 = 0 and x²− 7 = 0 For x²− 2 = 0 x²= 2 x = ±√2 x = −√2 or x = √2 x + √2 and x − √2 are factors For x²− 7 = 0 x²= 7 x = ±√7 x = −√7 or x = √7 x + √7 and x − √7 are factors Hence the factors are +√2, x − √2, x + √7 and x − √7

Example 3.17

Given f(x) = x⁴− 1 , find all the factors using the linear quadratic factor approach.

Solution

Let u = x²f(x) = (x²)²− 1 (u)²− 1 u²− 1 Factorising:

(u − 1)(u + 1) Substituting x²back in:

( x²− 1)( x²+ 1) ( x²− 1)( x²+ 1) = 0 x²− 1 = 0 or x²+ 1 = 0 For x²− 1 = 0 x²= ± 1 x = 1 or x = − 1 For ( x²+ 1) x²+ 1 = 0 x²= − 1 x = ±√− 1 NB: √− 1 = i x = ± i x = − i or x = i Hence the factors are (x + 1), (x − 1), (x − i)(x + i)

Example 3.18

Given f(x) = 2x⁴+ x²− 10 , find all the factors using the linear quadratic factor approach

Solution

Let u = x²f(x) = 2(x²)²+ u − 10 2(u)²+ u − 10 2 u²+ u − 10 Factorising:

2 u²− 4u + 5u − 10 2u(u − 2) + 5(u − 2) (2u + 5)(u − 2) Substituting:

u = x²(2 x²+ 5)( x²− 2) (2 x²+ 5)( x²− 2) = 0 2x²+ 5 = 0 or x²− 2 = 0 For 2 x²+ 5 = 0 x²= −5_ 2 x = ±√−5_ 2 x = 5/2 i or x = −5/2 i For ( x²− 2) x²− 2 = 0 x²= 2 x = ±√2 x = −√2 or x = √2 Hence the factors are (x + 5/2 i) (x − 5/2 i) (x − √2) (x + √2) We can use what we have learned to help in determining the maximum or minimum profit or productivity in the real world.

In small groups, or individually, discuss how the following problem can be solved.

Example 3.19

A water production company made sales of x bags of water and the profit was given by 2x²+ 7x − 15 where 0 ≤ x ≤ 10 .

Help the company determine their maximum profit and loss.

Solution

Since the range was given (0 ≤ x ≤ 10 ), we do our calculation within this range.

Evaluating the function in the given interval, we obtain:

x 0 1 2 3 4 5 6 7 8 9 10 2x²+ 7x − 15 -15 -6 7 24 45 70 99 132 169 210 255 The maximum profit occurs when x=10 with a profit of 255 while the loss will occur when there is no production, i.e., x = 0 at a loss of 15 ( -15).

1. Use the Rational Zero Theorem to find the rational zeros of:

a. 6x³− 13x²− 14x − 3

b. 5x³+ 2 x²− 5x − 2

2. Carefully, study the graph above which represents a polynomial function f(x).

a. State the zeros

b. Find the factors

c. State the interval on the curve.

d. Find the polynomial function f(x)

3. Sketch the following curves.

a. f(x) = x³+ 2x²− 9x − 18

b. f(x) = x³− 2x²− x + 2

c. f(x) = x³+ 3x²− 4x − 12

d. f(x) = x³− 5x²− x + 5

e. f(x) = 1/5(x − 2)(x − 4)(x + 5)

4. Factorise:

a. x³+ 5x²+ 2x − 8

b. 2x³+ 3x²− 17x − 30

c. 4x³− 4x²− 11x + 6

d. 6x³+ 13x²− 4

e. 3x³+ 16x²− 13x − 6

5. Find the zeros of the following:

a. x³+ 4x²+ x − 6

b. 2x³+ 7x²− 14x + 5

c. 2x³+ x²− 5x + 2

d. 4x³− 8x²− 9x + 18

e. 4x³− 4 x²− x + 1

f. 4x³− 12x²+ 11x − 3

6. Find the roots of:

a. x³+ 6x

b. 2x⁴+ 17 x²+ 35

c. x⁴− 14 x²+ 36

d. x⁴− 1

e. x³+ 27x

7. Find the polynomial function f(x) as represented by the curve above.

8. Q(x) = 10x³+ ax²− 10x + b , where a and b are integers, is divisible by (2x + 1). When Q(x) is divided by x + 1 , the remainder is -2.

a. Find the values of a and of b

b. find the expression for Q(x) as a product of the three linear factors

9. Given that x₁ = 1 + √2 i and x₂ = 2 + 3i are roots of a polynomial function P(x), determine the P(x) = 0

10. Applying Descartes’ Rule of Signs, identify the potential number of positive and negative roots.

a. f(x) = 10x⁶− 5x⁵+ 3 x⁴− 6x³+ 24x²− 38x

b. f(x) = 19x⁵+ 6 x⁴+ 17 x³+ 18x²− 38x − 70

c. f(x) = x⁶− 729

d. f(x) = 12x⁶+ 13 x³+ 1

e. f(x) = 8x⁸− 13 x⁴+ 100

11. Write down the equation having only the following roots.

a. -5 , -1, 3

b. 3, −1/5, −1/3

c. 2, -2, 2 + √3, 2 − √3

d. 0, 1 + 4i, 1- 4i

12. Find the four roots of y⁴+ 3x²+ 2

13. Study the graph below and use it to answer the following questions:

a. state the factors

b. find the polynomial function

c. how many turning points

d. what is the relationship between the degree of the function and the number of turning points

14. A new bakery in Accra specialises in creating various types of bread. The bakery wants the volume of a small rectangular loaf of bread to be 4 800 cm³. The loaf is to be shaped like a rectangular solid. The length of the loaf is to be 4cm longer than the width. The height of the loaf is one-half of the width. Determine the dimensions of the loaf.

Ever wondered how your favourite animation movie characters get to move from one place to another in the same scene? Well transformation of matrices helps the computer graphic designers with that. The matrix concept is important to study because it is applicable in fields such as genetics and machine learning, organising data for pattern recognition and supporting advancements in medicine. In year one we discussed the concept of matrices, especially 2 × 2 matrices, and the varied arithmetic operations that can be performed on them. This section is going to expose you to the determinant of 3 × 3 matrices, inverse of 2 × 2 matrices, solving of systems of linear equations using matrices and how to solve and model real-life problems using matrices.

Next time you watch a cartoon animation, think about matrices!

KEY IDEAS

• A cofactor is obtained by multiplying the element’s place sign and minor

• A matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions arranged in rows and columns.

• In a 3 × 3 matrix, a minor of an element is the determinant of the 2 × 2 submatrix that remains after removing the row and column containing that element.

• Transposing a matrix is obtained by flipping the matrix over its diagonal, effectively switching its rows with its columns.

A matrix (plural matrices) is a rectangular array of numbers, symbols, or expressions, organised in rows and columns. Each entry in a matrix is called elements or entries and it is identified by its row and column indices.

For example, in the 3 x 3 matrix below you can see the labelling of each element according to its row and column:

A = [ a₁₁ a₁₂ a₁₃ a₂₁ a₂₂ a₂₃ a₃₁ a₃₂ a₃₃] Everyday situations that exemplify the concept of matrices include:

a. Image compression (matrix transformation)

b. Electrical circuit analysis (matrix equation)

c. 3D transformations (matrix multiplication) Let us look at some types of matrices.

1. A square matrix is matrix that has the same number of rows and the same number of columns. For instance, B = [2 7 5 3] and E = [ 4 2 1 3 5 4 2 6 3] are examples of square matrices. B is a 2 × 2 matrix and E is a 3 x 3 matrix.

2. A rectangular matrix is a matrix in which the number of rows is not equal to the number of columns. For example, B = [2 3 5 1 4 6] and D = ( 4 2 3 5 2 6) are rectangular matrices. B is a 2 × 3 matrix and D is a 3 × 2 matrix. The number of rows are put first, followed by the number of columns.

3. A zero matrix is a matrix whose entries are all zeros or are equivalent to zero. Examples are, A = [0 0 0 0], B = [0 0 0 0 0 0], (0 0), ( 0 0

0) and (0 − i + i 0 0 0 0 0 0 x − x).

4. A diagonal matrix is a square matrix in which all elements outside the diagonal (top left corner to bottom right corner) are zero. It has non-zero elements on its main diagonal. For example, D = [ 4 0 0 0 5 0 0 0 3].

5. An identity/unit matrix is a unique square matrix with the element one (1) as entries on the main diagonal and zero (0) everywhere else.

6. For a 2 × 2 matrix, the identity matrix is [1 0 0 1]. For a 3 × 3 matrix, the identity matrix is [ 1 0 0 0 1 0 0 0 1]. It is denoted by I.

7. A triangular matrix has the entries above and/or below the diagonal as zeros.

When only the entries above the diagonal are zero, we refer to the matrix as a lower triangular matrix. On the other hand, if the entries below the diagonal are zero, we refer to them as upper triangular matrices. For

example, [ 4 3 6 0 5 1 0 0 3] is an upper triangular matrix while [ 1 0 0 7 3 0 3 4 3] is a lower triangular matrix.

Now, work through these examples to consolidate your knowledge. You can work individually, or in pairs.

Example 6.1

Which of the following matrices are zero matrices?

a. A = (x − x 0 0 0 − b + b 0)

b. B = ( 1 − 1 0 0 0 0 )

c. C = (m − m 0 0 0)

Solution

a. Simplifying the entries in matrix A gives (0 0 0 0 0 0), thus A is a zero matrix

b. Matrix B has two (2) non-zero entries, thus B is not a zero matrix

c. Matrix C can be simplified as (0 0 0 0) making it a zero matrix Hence options A and C are zero matrices.

Example 6.2

If P = ( 3/2 x y − x − 3) and Q = ( x 1.5 3.5 − 3) are two equal matrices, find the values of x and y .

Solution

Since P = Q , the entries for each column and row need to be the same.

3/2 = x and y − x = 3.5 y − 3/2 = 3.5 y = 5

Example 6.3

Find the product of the two matrices, A = (2 3 4 5) and I = (1 0 0 1).

Solution

AI = | 2 × 1 + 3 × 0 2 × 0 + 3 × 1 4 × 1 + 5 × 0 4 × 0 + 5 × 1 |= AI = |²³4 5| = A

Example 6.4

If A = ( − 5 3 2 − 1), B = ( 4 − 3 7 − 5) and C = (− 1 3 5 4), find;

a. A + B

b. B + A

c. A + (B + C)

d. (A + B) + C

Solution

a. A + B = (− 5 3 2 − 1) + (4 − 3 7 − 5) = (− 5 + 4 3 − 3 2 + 7 − 1 − 5) = (− 1 0 9 − 6)

b. B + A = (4 − 3 7 − 5) + (− 5 3 2 − 1) = (4 − 5 − 3 + 3 7 + 2 − 5 − 1) = (− 1 0 9 − 6)

c. B + C = (4 − 3 7 − 5) + (− 1 3 5 4) = (4 − 1 − 3 + 3 7 + 5 − 5 + 4) = ( 3 0 12 − 1) A + (B + C) = (− 5 3 2 − 1) + ( 3 0 12 − 1) = ( − 5 + 3 3 + 0 2 + 12 − 1 − 1) = (− 2 3 14 − 2 )

d. (A + B) + C = (− 1 0 9 − 6) + (− 1 3 5 4) = (− 1 − 1 0 + 3 9 + 5 − 6 + 4) = (− 2 3 14 − 2)

Example 6.5

If A = (− 5 3 2 − 1), B = (4 − 3 7 − 5) and C = (− 1 3 5 4), evaluate the following;

a. A − B

b. B − A

c. A − (B − C)

d. (A − B) − C

Solution

a. A − B = (− 5 3 2 − 1) − (4 − 3 7 − 5) = ( − 5 − 4 3 − ( − 3) 2 − 7 − 1 − ( − 5)) = (− 9 6 − 5 4)

b. B − A = (4 − 3 7 − 5) − (− 5 3 2 − 1) = ( 4 − ( − 5) − 3 − 3 7 − 2 − 5 − ( − 1)) = (9 − 6 5 − 4)

c. B − C = (4 − 3 7 − 5) − (− 1 3 5 4) = (4 − ( − 1) − 3 − 3 7 − 5 − 5 − 4) = (5 − 6 2 − 9) A − (B − C) = (− 5 3 2 − 1) − (5 − 6 2 − 9) = ( − 5 − 5 3 − ( − 6) 2 − 2 − 1 − ( − 9)) = (− 10 9 0 8)

d. (A − B) − C = (− 9 6 − 5 4) − (− 1 3 5 4) = (− 9 − ( − 1) 6 − 3 − 5 − 5 4 − 4) = (− 8 3 0 0)

Example 6.6

Given that A = ( 5 3

2) and the matrix P = − 2A , write out the matrix P.

Solution

P = − 2A = − 2( 5 3

2) = ( − 10 − 6 − 4 )

Example 6.7

If A = (− 3 0 7 − 4), B = ( 2 − 1 − 7 4 ) and C = ( 1 0 − 2 − 4), find 2A – 3B + 4C.

Solution

2A = 2(− 3 0 7 − 4) = (− 6 0 14 − 8), 3B = 3( 2 − 1 − 7 4 ) = ( 6 − 3 − 21 12 ) 4C = 4( 1 0 − 2 − 4) = ( 4 0 − 8 − 16) ∴ 2A – 3B + 4C = (− 6 0 14 − 8) − ( 6 − 3 − 21 12 ) + ( 4 0 − 8 − 16) = (− 8 3 27 − 36)

Example 6.8

Given that A = (4 − 2 1 3 ) and B = (− 2 3 − 2 − 7) Evaluate the following:

a. AB

b. BA

Solution

a. AB = (4 − 2 1 3 )(− 2 3 − 2 − 7) = (4(− 2) + (− 2)(− 2) 4(3) + (− 2)(− 7) 1(− 2) + 3(− 2) 1(3) + 3(− 7) ) = (− 4 26 − 8 − 18)

b. BA = (− 2 3 − 2 − 7)(4 − 2 1 3 ) = ( ( − 2)(4) + (3)(1) (− 2)(− 2) + (3)(3) (− 2)(4) + (− 7)(1) (− 2)(− 2) + (− 7)(3)) = ( − 5 13 − 15 − 17)

Example 6.9

If A = (1 2 3 4) and B = (− 1 2 − 3 1), find p and q if AB = ( p − 2 − 6 4 ) + 3( 4 2 − 3 q)

Solution

AB = (1 2 3 4)(− 1 2 − 3 1) = ( − 7 4 − 15 10) And AB = ( p − 2 − 6 4 ) + 3( 4 2 − 3 q) = ( p − 2 − 6 4 ) + ( 12 6 − 9 3q) = ( p + 12 4 − 15 4 + 3q) Equating corresponding entries, it follows that:

( − 7 4 − 15 10) = ( p + 12 4 − 15 4 + 3q) − 7 = p + 12 ⟹ p = − 19 10 = 4 + 3q ⟹ 3q = 6 ∴ q = 2

Example 6.10

If A = (2 3 1 4), find the determinant.

Solution

Using the formula, A = (a b c d), determinant of A, det(A) = ad − bc ∴ det(A) = (2 × 4)− (3 × 1) = 8 − 3 = 5

Example 6.11

Evaluate the determinants of the following matrices:

a. |²¹⁴17 9| _(b). | ²− 8 − 3 6 |

c. |ᵃ+ 3 7 − a a 7 | _(Solution)

a. det|²¹⁴17 9| = (21 × 9) − (4 × 17) = 121

b. det| ²− 8 − 3 6 | = ((2 × 6) − (− 8 × − 3)) = − 12

c. det|ᵃ+ 3 7 − a a 7 | = (7(a + 3) − a(7 − a)) = 7a + 21 − 7a + a²= a²+ 21 DETERMINANT OF 3 × 3 MATRICES In year one we learnt and practised finding the determinant of a 2 × 2 matrix:

Determinant = ad − bc given the matrix (a b c d). Now, we will apply that concept to help find the determinant of a 3× 3 matrix. In order to work out the determinant of a 3 × 3 matrix we need to define minors and cofactors of a square matrix.

Minor of a Square Matrix

In a 3 × 3 matrix, a minor of an element is the determinant of the 2 × 2 submatrix that remains after removing the row and column containing that element. If we let B = (bᵢⱼ) be a matrix of order n × n. Then, the minor of element (bᵢⱼ) (denoted by (Bᵢⱼ) is the determinant of the (n − 1) × (n − 1) matrix obtained after removing row i and column j from B.

Taking a 3 × 3 square matrix, B = ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ , the minor of B can be found by crossing out the row and column of b₁₁, b₁₂, b₁₃ ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ Illustration 1 Illustration 2 Illustration 3 From the illustrations, the minor of b₁₁ is ( b₂₂ b₂₃ b₃₂ b₃₃), b₁₂ is ( b₂₁ b₂₃ b₃₁ b₃₃) and b₁₃ is ( b₂₁ b₂₂ b₃₁ b₃₂).

Note that each element in the square matrix has its own minor.

Example 6.12

Given M = ( 4 8 − 3 10 5 11 − 6 7 4 ), determine the minor of M₁₂.

Solution

The minor of M₁₂ is the determinant of the matrix formed by crossing out row one and column two from M.

The minor of M₁₂ = ( 10 11 − 6 4 ) = (10 × 4) − (11 × − 6) = 106 .

Cofactors of a square matrix Every entry in a square matrix has its own cofactor. The cofactor is obtained by multiplying the element’s place sign and minor. The place sign of Nᵢⱼ is given by ( − 1)ⁱ+j. Thus, the cofactor of an element, Nᵢⱼ in a square matrix is given ( − 1)ⁱ+j .

Nᵢⱼ. Each entry in the square matrix has its own cofactor.

Let us go through these examples to practise how to determine a cofactor. You can work individually or in pairs.

Example 6.13

Given B = ( 4 8 − 3 10 5 11 − 6 7 4 ), determine the cofactor of 7 ( B₃₂) and 10 (B₂₁).

Solution

The minor of 7 = ( 4 − 3 10 11 ) = (11 × 4) − (10 × − 3) = 74 The place sign of 7 is ( − 1)³⁺²= − 1 ∴ the cofactor of 7 is (− 1)(74) = − 74 The cofactor of 10 is ( − 1)²⁺¹× detof(8 − 3 7 4 ) (− 1)[(8 × 4) − (− 3 × 7)] = − 53 ∴ the cofactor of 10 = − 53 Having defined the minor and cofactor of square matrices, let’s talk about using the theorem of expanding by cofactors to find the determinant of a matrix.

The theorem states that the value of a determinant can be found by expanding by cofactors of any row or column.

Note: No matter which row or column is chosen, the value for the determinant will remain the same. It is advised to use the row or column which has the most zeros simply because it makes the calculations easier.

Generally, to evaluate the determinant of a matrix B = ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ , by expanding the cofactors.

Choose, for example, the second row then |B| = b₂₁ (− 1)²⁺¹(M₂₁) + b₂₂ (− 1)²⁺²(M₂₂) + b₂₃ (− 1)²⁺³(M₂₃), where M represents the minor.

Example 6.14

Evaluate the determinant of C = ( 2 5 1 3 8 3 0 − 1 − 3) by expanding the cofactors.

Solution

Finding the determinant of C from row one, C = ( 2 5 1 3 8 3 0 − 1 − 3) |^(C)| = 2 (− 1)¹⁺¹(C₁₁) + 5 (− 1)¹⁺²(C₁₂) + 1 (− 1)¹⁺³(C₁₃) |C| = 2| ⁸³− 1 − 3| − 5| ³³− 0 − 3| + 1|³⁸0 − 1| |C| = 2[(8 × − 3) − (− 1 × 3)] − 5[(3 × − 3) − (0 × 3)] + 1[(3 × − 1) − (0 × 8)] |C| = 2 × − 21 − 5 × − 9 − 3 |C| = 0

Activity 6.1 – Verifying the determinant using different rows/columns

1. Working in small groups, find the determinant of C by choosing different rows/columns.

2. Did you get the same determinant or not? Check each other’s workings as each route to finding the determinant should give the same answer.

3. Discuss the results with your group.

Using the Sarrus rule to find the determinant Another way to find the determinant of matrices is to use the Sarrus rule.

To find the determinant of B = ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ , extend the matrix by writing the first and second columns to the right-hand side of B .

i.e., B = ⎛ ⎜ ⎝ b₁₁ b₁₂ b₁₃ b₂₁ b₂₂ b₂₃ b₃₁ b₃₂ b₃₃ ⎞ ⎟ ⎠ , b₁₁ b₁₂ b₂₁ b₂₂ b₃₁ b₂₂ Add the product of the diagonals b₁₁ b₂₂ b₃₃ + b₁₂ b₂₃ b₃₁ + b₁₃ b₂₁ b₃₂ Subtract the product of the opposite diagonals −b₁₂ b₂₁ b₃₃ − b₁₁ b₂₃ b₃₂ − b₁₃ b₂₂ b₃₁ Put the two formulae together to find the determinant.

det(B) = b₁₁ b₂₂ b₃₃ + b₁₂ b₂₃ b₃₁ + b₁₃ b₂₁ b₃₂ − b₁₂ b₂₁ b₃₃ − b₁₁ b₂₃ b₃₂ − b₁₃ b₂₂ b₃₁

Example 6.15

Evaluate the determinant of C = ( 2 5 1 3 8 3 0 − 1 − 3).

Solution

Using the Sarrus rule:

( 2 5 1 3 8 3 0 − 1 − 3) 2 5 3 8 0 − 1 det(A) = (2)(8)(−3) + (5)(3)(0) + (1)(3)(−1) − (5)(3)(−3) − (2)(3)(−1) − (1)(8)(0) = − 48 + 0 − 3 − (− 45) − (− 6) − 0 = 0

Example 6.16

Evaluate the determinant of the matrix D = ( 2 4 3 6 2 3 2 5 4)

Solution

D = ( 2 4 3 6 2 3 2 5 4) = 2|²³5 4| − 4|⁶³2 4| + 3|⁶²2 5| = 2(8 − 15) − 4(24 − 6) + 3(30 − 4) = − 14 − 72 + 78 = − 8 Singular and Non-Singular Matrices A square matrix is classified as a singular matrix if it has a determinant of zero and it has no inverse. On the other hand, a non-singular matrix is a square matrix with a non-zero determinant and it has an inverse.

Tabular representation on the differences between singular and non-singular matrix Non-singular Singular B is invertible Not invertible Rows independent dependent Columns independent dependent det(B) ≠ 0 = 0 Bx = 0 one solution x = 0 infinitely, many solutions Bx = c one solution No solution To find out if a square matrix is singular or non-singular find the determinant of the given matrix. If the determinant is zero, it is a singular matrix and if the determinant is not zero, the matrix is non-singular.

Let us work through some examples.

Example 6.17

Find the value of x, if A = (4 12 x 6 ) is a singular matrix.

Solution

If A is a singular matrix then ad − bc = 0 (4)(6) − (x)(12) = 0 24 − 12x = 0 24 = 12x x = 2

Example 6.18

Find the value of x, if B = |⁵ˣ+ 2 6x − 3 4 3 | is a singular matrix.

Solution

B = |⁵ˣ+ 2 6x − 3 4 3 | ₍₅ₓ + 2)(3) − (6x − 3)(4) = 0 3(5x + 2) − 4(6x − 3) = 0 15x + 6 − 24x + 12 = 0 18 − 9x = 0 x = 2 Transpose of a matrix In section 5 we looked at transposing vectors. We can also transponse matrices in a similar manner. The transpose of a matrix is determined by interchanging its rows into columns or columns into rows. For example, given G = (3 2 4 5 1 2), then Gᵀ= ( 3 5 2 1 4 2).

Let us work through these examples.

Example 6.19

Given that P = ( 1 − 1 2 2 1 1 3 2 − 1), find Pᵀ.

Solution

Pᵀ= ( 1 2 3 − 1 1 2 2 1 − 1)

Example 6.20

What is the transpose of a 4 × 3 matrix?

Solution

The transpose of a 4 × 3 is a 3 × 4 matrix.

Activity 6.2

1. Given that Q = (− 2 − 1 4 4 3 − 9), find Qᵀ.

2. Now find the transpose of Qᵀ.

3. What is your observation?

4. Discuss with your peers.

Hopefully you found that (Qᵀ)ᵀ= Q Addition Property of Transpose If R = ( 3 2 − 5 8) and S = ( 7 1 − 6 3), find (R + S)ᵀand Rᵀ+ Sᵀ.

What is the relationship between (R + S)ᵀand Rᵀ+ Sᵀ?

Solution

(R + S) = ( 3 2 − 5 8) + ( 7 1 − 6 3) = ( 10 3 − 11 11) (R + S)ᵀ= (10 − 11 3 11 ) Rᵀ= (3 − 5 2 8 ) an d Sᵀ= (7 − 6 1 3 ) ∴ Rᵀ+ Sᵀ= (3 − 5 2 8 ) + (7 − 6 1 3 ) = (10 − 11 3 11 ) The sum of (R + S)ᵀand Rᵀ+ Sᵀare the same.

This communicates that adding the transpose of two individual matrices say M and N i.e., Mᵀ+ Nᵀis equal to the transpose of the sum of the two matrices (M + N)ᵀ.

INVERSES OF 2 × 2 MATRICES The real numbers 6 and ¹_ ₆ are referred to as multiplicative inverses because their product is a multiplicative identity 1 (i.e., 6 × 1/6 = 1 ). If we have the matrix (1 2 3 4 ) multiplying by ( 4 − 2 − 3 1 ), results in (1 0 0 1) which is the 2 x 2 identity matrix (I).

Hence, we call (1 2 3 4) and ( 4 − 2 − 3 1 ) multiplicative inverses of each other.

Let us go through this activity to know how to find the inverse of a 2 × 2 matrix.

Activity 6.3 – Proof of inverse of a matrix formula Work through these steps in small groups where possible.

1. Choose a 2 × 2 matrix such as M = (a b c d).

2. Assume the inverse of M is M⁻¹= ( m p n v) .

3. Multiply M by its inverse

4. Equate the result to a 2 × 2 Identity matrix.

5. Using your idea in equality of matrices, equate each entry in the matrix on the LHS to the corresponding entry on the RHS.

6. Solve the equations containing m and n simultaneously.

7. Now, solve the equations containing p and v simultaneously.

8. Create a new matrix using the results for m, n, p and v.

9. Factorise ( 1______ ad − bc) from the new matrix.

10. Hopefully you have 1/d − bc( d − b − c a )?

11. Congratulations, you have derived the formula for finding the inverse of a 2 x 2 matrix.

Note: ( d − b − c a ) is what is referred to as the adjoint of a matrix while ad − bc is the determinant for the matrix M.

Let us work through these examples to practise finding adjoints and inverses of a matrix.

Example 6.21

Find the inverse of matrix A = (2 3 3 1)

Solution

A⁻¹= 1___ etA ( d − b − c a ) A⁻¹= 1___________ (1)(2) − (3)(3) ( 1 − 3 − 3 2 ) − 1/7 ( 1 − 3 − 3 2 ) = ( −1/7 3/7 3/7 −2/7)

Example 6.22

Given the matrix B = (1 2 3 4), find the inverse matrix B⁻¹.

Solution

B = (1 2 3 4) B⁻¹= −1/2( 4 − 2 − 3 1 ) = ( − 2 1 3/2 −1/2)

Example 6.23

Find the inverse of the matrices of the following:

C = (2 3 3 1), D = (1 4 0 2) Hence, find:

a. C C⁻¹b. D⁻¹D

Solution

detC = 2 − 9 = − 7 ∴ C⁻¹= −1/7( 1 − 3 − 3 2 ) = ( −1/7 3/7 3/7 −2/7) Hence:

a. C C⁻¹= (2 3 3 1) × −1/7( 1 − 3 − 3 2 ) = −1/7(2 3 3 1)( 1 − 3 − 3 2 ) = −1/7(2 − 9 − 6 + 6 3 − 3 − 9 + 2) = −1/7(− 7 0 0 − 7) = ( − 7___ − 7 0/7 0___ − 7 − 7___ − 7) = (1 0 0 1) ∴ C C⁻¹= I

b. D⁻¹D D⁻¹= 1/2(2 − 4 0 1 )) = ( 1 − 2 0 1_ 2 ) D⁻¹D = ( 1 − 2 0 1/2 )(1 4 0 2) = (1 4 − 4 0 1 ) = (1 0 0 1) ∴ D⁻¹D = I .

In matrices, a system of linear equations can be written in the form MX = C where M is the matrix of coefficients, X is the unknown column matrix and C is a column matrix of constants. The matrix method for solving systems of linear equations is more helpful when the number of equations is many. We will explore how the matrix method and Cramer’s rule can be used to solve systems of linear equations.

Let us start with examples on equations involving 2 × 2 matrices.

Example 6.24

Find the value of x and y in (2x + 2 3 4 6x − 3) = (10 3 4 9).

Solution

(2x + 2 3 4 6y − 3) = ( 10 3 − 3 9) 2x + 2 = 10 x = 4 Also, 6y − 3 = 9 6y = 12 y = 2

Example 6.27

(3x − 2 2 1 2y + 1) = (− 7 2 1 − 6) 3x − 2 = − 7 x = − 5 __ Also, 2y + 1 = − 6 y = −7/2

Note: In Examples 6.26 and 6.27, the entries of the matrices on the left hand and the right hand were compared to each other. Since the matrices are equal, the entries with the variables were equated and a solution was found.

Example 6.28

Given that (4 − 1 3 x )( 4 − 3) = (19 8 ), find x.

Solution

( (4 × 4)+ ( − 1 × − 3) (3 × 4)+ ( − 3 × x) ) = (19 8 ) ( 19 12 − 3x) = (19 8 ) 12 − 3x = 8 3x = 8 + 12 3x = 20 x = 20/3

Example 6.29

Given that (− 3 4 y 2)(2

3) = ( 6 12), find y

Solution

(− 3 4 y 2)(2

3) = (− 5 12 ) ( 6 2y + 6) = ( 6 12) 2y + 6 = 12 2y = 12 − 6 2y = 6 y = 3

Note: In Example 6.28 and 6.29, the concept of multiplication of matrices were applied before equating the entry with the variable to its corresponding entry.

Using Cramer’s rule for involving 2 × 2 matrix In using the Cramer’s rule, you need to know the determinant of a matrix to find the solution of the system MX = C.

Given two linear equations a₁ x + b₁ y = c₁ and a₂ x + b₂ y = c₂ , rewrite the equations in the form MX = C.

( a₁ b₁ a₂ b₂)( x y) = ( c₁ c₂) Keeping in mind that M is the coefficient matrix, X is the variable and matrix C is the constant matrix, you have to find the determinants so that det(M) = a₁ b₂ − b₁ a₂.

Thus, det(Mₓ) = ( c₁ b₁ c₂ b₂) = c₁ b₂ − b₁ c₂ and det(M_(y)) = ( a₁ c₁ a₂ c₂) = a₁ c₂ − c₁ a₂ Therefore x = det(Mₓ)______ det(M) , and y = det(M_(y))______ det(M) .

Example 6.30

Solve for x and y in the equations

a. 3x − 4y = 2 and x + 2y = 4

b. 3x − y = 7 and 5x − 4y = 0

Solution

a. (3 − 4 1 2 )( x y) = (2 4) Find the determinant det(M) = 6 + 4 = 10 To find the determinant of Mₓ , use the constant (2

4) and the last column in the matrix (− 4 2 ) Mₓ = (2 − 4 4 2 ) = 4 + 16 = 20 Repeat the same process to find M_(y) To find M_(y) = (3 2 1 4) = 12 − 2 = 10 x = det (Mₓ)_______ det(M) = 20/10 = 2 y = det (M_(y))_______ det(M) = 10/10 = 1 ∴ The solution to the simultaneous equation is x = 2 and y = 1.

b. (3 − 1 5 − 4)( x y) = (7 0) Find the determinant det(M) = − 12 + 5 = − 7 To find the determinant of M_(y) use the constant (7

0) and the last column in the matrix (− 1 − 4) Mₓ = (7 − 1 0 − 4) = − 28 − 0 = − 28 M_(y) = (3 7 5 0) = 0 − 35 = − 35 Now, x = ᵈᵉᵗ(Mₓ)_ _(det)(M) = − 28____ − 7 = 4 y = det (M_(y))_______ det(M) = − 35____ − 7 = 5 ∴ The solution to the simultaneous equation is x = 4and y = 5.

Now, let us talk about solving equations involving 3 × 3 matrices.

Using Cramer’s rule for equations involving a 3 × 3 matrix In using the Cramer’s rule, you need to know the determinant of a matrix to find the solution of the system MX = C.

Given three linear equations, say a₁ x + b₁ y + c₁ z = e₁ , a₂ x + b₂ y + c₂ z = e₂, and a₃ x + b₃ y + c₃ z = e₃, rewrite the equations in the form MX = C.

Where, M = ⎛ ⎜ ⎝ a₁ b₁ c₁ a₂ b₂ c₂ a₃ b₃ c₃ ⎞ ⎟ ⎠ , X = ( x y z) and C = ( e₁ e₂ e₃).

Now you have to find det(M).

Using ⎛ ⎜ ⎝ e₁ b₁ c₁ e₂ b₂ c₂ e₃ b₃ c₃ ⎞ ⎟ ⎠ find det( Mₓ), ( a₁ e₁ c₁ a₂ e₂ c₂ a₃ e₃ c₃) to find det( M_(y)) and ⎛ ⎜ ⎝ a₁ b₁ e₁ a₂ b₂ e₂ a₃ b₃ e₃ ⎞ ⎟ ⎠ to find det( M_(z)). You will notice that you are swapping the constant column with the corresponding, x, y or z column for Mₓ, M_(y), M_(z).

From Cramer’s rule, x = det( Mₓ)_______ det(M) , y = det( M_(y))_______ det(M) and z = det( M_(z))_______ det(M) .

Let us go through these examples to practise solving systems of linear equations.

Example 6.31

Solve for x, y and z in the equations:

a.

5x + 3y + z = 2 4x + 3y + 2z = 1 6x + 4y + z = 1 b.

2x+ y − 4z = 4 x+ y+ z = 6 x − 3y+ 2z = 8

Solution

a. ( 5 3 1 4 3 2 6 4 1)( x y z) = ( 2 1 1) Let A = ( 5 3 1 4 3 2 6 4 1), and find its determinant.

det(A) = ( 5 3 1 4 3 2 6 4 1) = 5(3 − 8) − 3(4 − 12) + 1(16 − 18) = − 25 + 24 − 2 = − 3 To find the det(Aₓ) replace the first column matrix ( 5 4

6) by the constant ( 2 1 1 ) det(Aₓ) = ( 2 3 1 1 3 2 1 4 1) = 2(3 − 8) − 3(1 − 2) + 1(4 − 3) = − 10 + 3 + 1 = − 6 To find the det(A_(y)) replace the second column matrix ( 3 3

4) by the constant ( 2 1 1) det(A_(y)) = ( 5 2 1 4 1 2 6 1 1) = 5(1 − 2) − 2(4 − 12) + 1(4 − 6) = − 5 + 16 − 2 = 9 To find the det(A_(z)) replace the third column matrix ( 1 2

1) by the constant ( 2 1 1) det(A_(z)) = ( 5 3 2 4 3 1 6 4 1) = 5(3 − 4) − 3(4 − 6) + 2(16 − 18) = − 5 + 6 − 4 = − 3 Hence; x = det(Aₓ)______ det(A) = − 6___ − 3 = 2, y = det(A_(y))______ det(A) = 9__

–3 = − 3 , z = det(A_(z))______ det(A) = − 3___ − 3 = 1 ∴ The solution to the simultaneous equation is x = 2, y = − 3 and z = 1.

b. ( 2 1 − 4 1 1 1 4 − 3 2 )( x y z) = ( 4 6 8) Let A= ( 2 1 − 4 1 1 1 4 − 3 2 ) det A = 2(2 + 3) − 1(2 − 4) − 4(− 3 − 4) = 10 + 2 + 28 = 40 det(Aₓ) = ( 4 1 − 4 6 1 1 8 − 3 2 ) = 4(2 + 3) − 1(12 − 8) − 4(− 18 − 8) = 20 − 4 + 104 = 120 det(A_(y)) = ( 2 4 − 4 1 6 1 4 8 2 ) = 2(12 − 8) − 4(2 − 4) − 4(8 − 24) = 8 + 8 + 64 = 80 det(A_(z)) = ( 2 1 4 1 1 6 4 − 3 8) = 2(8 + 18) − 1(8 − 24) + 4(− 3 − 4) = 52 + 16 − 28 = 40 Hence, x = det(Aₓ)______ det(A) = 120/40 = 3, y = det(A_(y))______ det(A) = 80/40 = 2, z = det(A_(z))______ det(A) = 40/40 = 1 ∴ The solution to the simultaneous equation is x = 3, y = 2 and z = 1.

Just as we learnt in the section on vectors, knowing how the mathematical concepts we learn are applicable in solving real life problems is very important. Matrices are useful in industries that deal with input-output analysis. Considering this, let us go through examples showing how matrices are applied in real life.

Activity 6.4 – Group Investigation

1. In small groups, discuss and come up with circumstances in real life where matrices can be used in solving problems.

2. Prepare a presentation (use PowerPoint if accessible) and share with your classmates and your teacher.

Example 6.32

A dietitian wants to create a daily plan for a patient using three food supplements, x, y and z. The patient requires exactly 22 units of calcium, 18 units of iron and 40 units of vitamin A.

The nutrient content per ounce of supplements is represented on the table below.

X Y Z Calcium (units) 2 3 1 Iron (units) 1 2 3 Vitamin A (units) 4 5 2 How much of each supplement (x, y, z) should the patient consume daily to meet the nutrient requirements exactly.

Solution

Let x, y and z represent the amount of supplement to be taken daily. The following can be derived from the table.

( 2x+ 3y+ z = 22 x+ 2y+ 3z = 18 4x+ 5y+ 2z = 40) = ( 2 3 1 1 2 3 4 5 2)( x y z) = ( 22 18 40) By finding the determinant Let A = ( 2 3 1 1 2 3 4 5 2) det(A) = 2(4 − 15) − 3(2 − 12) + 1(5 − 8) = − 22 + 30 − 3 = 5 Replace x-column with constants:

Let Aₓ = ( 22 3 1 18 2 3 40 5 2) det(Aₓ) = 22(4 − 15) − 3(36 − 120) + 1(90 − 80) = − 242 + 252 + 10 = 20 Replace y-column with constants:

Let A_(y) = ( 2 22 1 1 18 3 4 40 2) det(A_(y)) = 2(36 − 120) − 22(2 − 12) + 1(40 − 72) = − 168 + 220 − 32 = 20 Replace z-column with constants:

Let A_(z) = ( 2 3 22 1 2 18 4 5 40) det(A_(z)) = 2(80 − 90) − 3(40 − 72) + 22(5 − 8) = − 20 + 96 − 66 = 10 det (Aₓ)______ det(A) = 20/5 = 4 , det (A_(y))_______ det(A) = 20/5 = 4, det (A_(z))______ det(A) = 10/5 = 2 ∴ The dietitian’s solution is to have 4 lots of x, 4 lots of y and 2 lots of z.

Example 6.33

A Company produces two products, A and B. The ideal combination is when:

2A + 3B = 85 and A + 2B = 50 .

Find the values of A and B which makes both of these true.

Solution

Coefficient matrix (2 3 1 2)(A B) = (85 50) Let M= (2 3 1 2) Determinant M = (2)(2) − (3)(1) = 1 Let M_(A) = (85 3 50 2) det(M_(A)) = (85)(2) − (3)(50) = 170 − 150 = 20 Let M_(B) = (2 85 1 50) det( M_(B)) = (2)(50) − (85)(1) = 100 − 85 = 15 Hence, det( M_(A))_______ det(M) = 20/1 = 20 and det( M_(B))_______ det(M) = 15/1 = 15 ∴ A should be 20 and B should be 15.

Example 6.34

Forces F₁ and F₂ act on an object with the following force equations:

F₁ + F₂ = 8 and 2 F₁ − F₂ = 7.

Find the values of F₁ and F₂.

Solution

Coefficient matrix (1 1 2 − 1)( F₁ F₂) = (8 7) Let A = (1 1 2 − 1) Determinant A = (1)(− 1) − (1)(2) = − 3 Determinant (Aₓ) = (8 1 7 − 1) = − 8 − 7 = − 15 Determinant of (Aₓ) = (1 8 2 7) = 7 − 16 = − 9 Hence,F₁ det(Aₓ)_____ det(A) = − 15____ − 3 = 5 F₂ det(A_(y))______ det(A) = − 9___ − 3 = 3 ∴ F₁ = 5N and F₂ = 3N

1. If T = ( 4 − 2 3x 5 ), find the value of x , if the determinant of T = 32.

2. If A = (1 − 1 2 4 ), find A⁻¹.

3. Evaluate the determinant of P = ( − 2 1 3 1 − 1 0 − 3 1 1)

4. Determine whether Q = ( 2 − 2 − 3 1 4 3 3 2 − 1), is singular or not.

5. B = ( 2 x + 5 2x − 2 − 9 ) Given that B is a singular matrix, find the value of x .

6. Find the determinant of the matrix ( 2 5 1 3 8 3 0 − 1 − 3)

7. a. If A = (3 1 x 2) and B = (6 2 4 y), find the values of x and y given that AB = BA

b. Show that the inverse of the matrix AB is B⁻¹A⁻¹.

8. If A = ( 5 1 − 2 2), find A⁻¹.

9. Hence find the matrix B such that BA = C, where C = (12 0 1 5).

9. Use Cramer’s rule to find the value x and y in the equations:

3x + 2y = 10 and 7x + 5y = 23.

10. Use Cramer’s rule to find the value x, y and z in the equations:

2x − 3y + 4z = 10, x + y − z = 1 and x − 6y + 3z = − 1 .

11. Use Cramer’s rule to find the value x, y and z in the equations:

2x + 3z = − 1, y + 2z = 5 and x + y = 1.

12. A financial advisor wants to invest in three stocks A, B and C. The advisor requires a portfolio with a total value of exactly GH¢ 1920, annual returns of exactly GH¢ 66 and Beta value (market risk) of exactly 4.2. The characteristics of each stock value are:

A B C Stock value 100 60 80 Annual returns 2 3 4 Beta value 0.1 0.3 0.2 How many shares of each stock A, B and C should the advisor buy to meet the portfolio requirement exactly.

Some phenomena in everyday life, like population growth and decay, asset depreciation and the loudness of sound, can be modelled with exponential or logarithmic functions. In year one, we covered logarithms and their inverse functions (exponential functions). Now we will reinforce your understanding and introduce you to how they are applied in real life. Before we look at the applications of Indices and logarithms, let us first review the laws associated with these concepts.

Laws of indices They provide rules for simplifying calculations or expressions involving powers of the same base.

1. Multiplying indices: aᵐ× aⁿ= aᵐ+n

2. Dividing indices: aᵐ÷ aⁿ= aᵐ−n

3. Multiplying exponents: (aᵐ)ⁿ= aᵐ×n Laws of logarithms They provide rules for combining, splitting and evaluating logarithms.

1. Product Law: loga + logb = logab

2. Quotient Law: loga − logb = logᵃ_ _(b)

3. Power Law: log aⁿ= nloga

KEY IDEAS

• Exponential Decay can be modelled by the equation A = P (1 − r)ᵗ.

• Natural Phenomena that follow this relation include depreciation, population decay and radioactive decay.

• Exponential Growth can be modelled by the equation A = P (1 + r)ᵗ.

• Natural phenomena that follow this relationship include inflation, compound interest and population growth.

• A in the equation represents future value,

• r represents the rate of growth,

• P is the starting value and

• t is the time duration over which the natural phenomenon is being considered.

• To draw a linear graph for Y = a b^(X), plot LogY against X

• To reduce an exponential function of the form Y = a Xⁿto linear, plot LogY against LogX. The gradient of the resulting line = n .

Solving Equations Involving Logarithms

Express terms on both sides of the equation as a single logarithm to the same base. Equate the numbers and solve for the unknown variable or variables.

In general, if log_(b)x = log_(b)y ⟹ x = y Also, if logₓa = log_(y)a ⟹ x = y Remember the laws of logarithms:

log_(b)x + log_(b)y = log_(b)xy log_(b)x − log_(b)y = log_(b) x_ y log_(b)xⁿ= n log_(b)x log_(b)a = ^(logxa)_ _(log)ₓ_(b)

Example 8.1

Solve log₁₀(2x + 1) − log₁₀(1 − x) = log₁₀2

Solution

log₁₀(2x + 1) − log₁₀(1 − x) = log₁₀2 log₁₀(2x + 1_ 1 − x ) = log₁₀2 As we now have single logarithmic terms on both side to the same base, we can equate the terms.

2x + 1_ 1 − x = 2 2x + 1 = 2(1 − x) 2x + 1 = 2 − 2x 2x + 2x = 2 − 1 4x = 1 x = 1/4

Example 8.2

Solve simultaneously log₂y = 4 + log₂x and y log₄8 = 2x + 3

Solution

Simplify equation 1:

log₂y = 4 + log₂x log₂y = 4 log₂2 + log₂x log₂y = log₂2⁴+ log₂x log₂y = log₂16x y = 16x …………………..equation (1) Simplify equation 2:

y log₄8 = 2x + 3 y log₂8_ log₂4 = 2x + 3 3/2 y = 2x + 3 …………….equation (2) Substitute (1) into (2) to solve the simultaneous equations:

3_ 2 × 16x = 2x + 3 24x = 2x + 3 22x = 3 x = 3/22 Substitute x = 3/22 into (1) y = 16 × 3/22 y = 24/11 Confirm you have done this correctly by substituting both values into (2).

Example 8.3

Find the truth set of the equation log₁__ 2 x + log₂(¹_ₓ) + 6 = 0

Solution

log₁__ 2 x + log₂(¹_

x) + 6 = 0 The base of the logs are not the same, so we change it to a uniform base. This will help us apply the laws to simplify the expression. We will change it to base 2, although we could have chosen to change it to base ½ log₂x______ log₂(1/2) + log₂(¹_

x) + 6 = 0 log₂x_ log₂(2⁻¹) + log₂(¹_

x) + 6 = 0 log₂x_ −log₂2 + log₂(¹_

x) = − 6 log₂x_ − 1 + log₂(¹_

x) = − 6 −log₂x + log₂(¹_

x) = − 6 log₂(x)⁻¹+ log₂(¹_

x) = − 6 log₂ 1/x × 1/x = − 6 log₂( 1_ x²) = − 6 1_ x²= 2⁻⁶1_ x²= 1/64 x²= 64 x = ±√64 = 8 or − 8 Truth set ={x : x = − 8, 8 } Compound Interest The amount of money a lender or a financial institution receives for lending out money is called interest. Interest is usually calculated as a percentage of the amount borrowed. The person giving the money is the lender. The person receiving the loan is called the borrower.

Simple interest (SI) = P × T × r Where:

P = The principal or the amount borrowed or the amount invested T = Time the borrower will keep the money until he/she pays.

r = rate of interest in percentage (per annum or yearly) The amount of money the borrower will pay the lender = Principal + Interest Compound Interest is the interest calculated on the principal and the interest accumulated over the period. The compound interest formula is given below:

Compound interest = Amount − Principal Amount (A)= P (1 + r/n)ⁿᵗWhere A = Amount, P = Principal, r = interest rate, n = number of times interest is compounded per year, t = time (in years) How did we arrive at this formula? The following activity will explain.

Activity 8.1: Compound Interest

In small groups, work through the following steps to discover the formula for compound interest.

Step 1: Calculate the amount at the end of the first year (A₁) using the simple interest formula.

Step 2: Calculate the amount at the end of the second year (A₂) using the amount obtained in step 1 as the principal.

Step 3: Calculate the amount at the end of the third year (A₃) using the amount obtained in step 2 as the principal.

Step 4: Repeat the process for 2 or 3 more terms. This will generate A₄, A₅, …….

Step 5: Find the tᵗʰterm of the sequence A₁, A₂, A₃, A₄, A₅, ………., Aₜ Compare your answer with the one below:

Step 1: Interest at the end of the first year = P × 1 × r = Pr Amount = Principal + Interest Amount at the end of the first year ( A₁) = P + Pr = P(1 + r)

Step 2: Interest at the end of the second year= P(1 + r) × 1 × r = P(1 + r)r Amount at the end of the second year (A₂) = Amount at the beginning of the first year + interest at the end of the second year = P(1 + r) + P(1 + r)r = P(1 + r)(1 + r) A₂ = P(1 + r )²Step 3: A₃ = P(1 + r )²+ P(1 + r )²× r A₃ = P(1 + r )²(1 + r) = P(1 + r )³Step 4: This follows that:

A₄ = P(1 + r )⁴A₅ = P(1 + r )⁵Step 5: This will generate the sequence: P(1 + r), P(1 + r )², P(1 + r )³, P(1 + r )⁴, …….

This is a geometric sequence with a = P(1 + r) and common ratio (R) = P(1 + r )²_______ P(1 + r) = (1 + r) The tᵗʰterm, Uₜ = Aₜ = a Rᵗ⁻¹Aₜ = [P(1 + r)](1 + r )ᵗ⁻¹Aₙ = P(1 + r )ᵗ⁻¹⁺¹Aₜ = P(1 + r )ᵗgives the amount at the end of the tᵗʰyear If the rate of interest is compounded n times per annum, then we the number of compounding periods = n × t = nt and the interest rate = r __ Aₜ = P (1 + r/n)ⁿᵗOther than the first year, interest compounded annually is greater than that of simple interest.

Most transactions in the banking and financial sector use compound interest.

Other applications include population growth, bacteria growth, inflation and depreciation.

What part do logarithms play in these formulas, A = P (1 + r/n)ⁿᵗand A = P(1 + r )ᵗ?

They prove essential when we are solving for t or r .

Example 8.4

Ama borrowed GH¢ 10 000 from Bobo Bank for 3 years at an interest of 12% compounded annually. Find the compound interest and amount he has to pay at the end of 3 years.

Solution

Amount borrowed = Principal = 10 000 Number of years = 3 Interest rate = 12/100 = 0.12 Using the formula, Amount = P(1 + r)ⁿ, Amount owed at the end of the third year = 10 000 × (1 + 0.12)³= 10 000 × 1.12³= 10 000 × 1.404928 = 14049.28 Therefore, the amount at the end of the third year= GH¢ 14 049.28 Compound interest owed = 14 049.28 − 10 000 = GH¢ 4 049.28

Example 8.5

A bank decides to offer a business owner a GH¢ 750 000 loan at a compound interest of 8% per year.

Find the total amount, the bank will receive when the loan is repaid after 4 years.

How much interest will the business owner pay?

Solution

Principal = 750 000 Interest rate = 8/100 = 0.08 Time = 4 years.

Amount at the end of 4 years = 750 000 × (1 + 0.08)⁴= 750,000 × 1.08⁴= 750,000 × 1.36048896 = GH¢ 1 020 366.72 Interest paid by the business owner = 1,020,366.72 − 750,000 = GH¢ 270 366.72

Example 8.6

Sweety invested $2 400 into a fund that pays 5 percent interest each year, compounded semi-annually. Find the value of the investment after 4 years.

Solution

Principal = $2400 Interest rate = 5/100 = 0.05 interest is compounded semi-annually Time = 4 Since interest is compounded more than once in a year, we will use the formula Amount = P (1 + r/n)ⁿᵗn = 2 since interest is compounded semi-annually (twice a year) Amount = 2 400 × (1 + 0.05/2 ) 2×4 = 2 400 × (1 + 0.025)⁸= 2 400 × 1.025⁸= 2 924.1669 Therefore, the value of the investment after 4 years= GH¢ 2 924.17

Example 8.7

Baba invested €50 000 into a fund that pays 10% interest each year, compounded once every two years.

a. Find the total amount of his investment after 6 years.

b. After how many years will his investment double?

Solution

a. Principal = €50 000 Interest rate = 10/100 = 0.1 Time = 6 years Since interest is compounded less than once a year, we will use the formula Amount = P (1 + r/n)ⁿᵗn = 1/2 = 0.5 since interest is compounded once every two years (1 in every 2 = 1:2 = 1/2) Intuitively, once every two years translates to three times in six years.

Amount = 50 000 × (1 + 0.1/0.5) 0.5 × 6 = 50 000 × (1 + 0.2)³= 50 000 × 1.2³= 50 000 × 1.728 = 86 400 Therefore, the value of the investment after 6 years= GH¢ 86 400 .

b. Doubling his investment means that his investment will amount to:

50 000 + 50 000 = 100 000 Let the time it will take to double his investment be T .

This means that we seek to find T such that:

100 000 = 50 000 × (1 + 0.1/0.5)0.5 × T First divide bot sides by 50 000 100 000_ 50 000 = 50 000/50 000 × (1 + 0.2)^(0.5)× T 2 = 1.2^(0.5T)Take log to base 10 on both sides of the equation:

log₁₀2 = log₁₀1.2^(0.5T) log₁₀2 = 0.5T log₁₀1.2 log₁₀2_ log₁₀1.2 = 0.5T 3.80178 = 0.5T 3.80178_ 0.5 = 0.5T/0.5 T = 7.604 T = 7.604 years = 7 years, 0.604 × 12 = 7.24 months This means that it will take Baba about 7.6 years or 7 years and 7.24 months to double his investment.

Let us look back on our answer to check if it is correct.

Amount = 50 000 × (1 + 0.1/0.5) 0.5 × 7.604 = 50 000 × 1.2^(3.802)= 50 000 × 2 = 100 000 This confirms that our answer is correct.

Example 8.8

Calculate the annual rate of compound interest that will allow a loan of $30 000 to amount to $43 923 in four years.

Solution

We have:

Principal = $30 000.00 Amount = $43 923.00 Time = 4 years.

We have to find r = rate of interest Substitute these values into A = P(1 + r)ᵗ43 923 = 30 000(1 + r)⁴Divide through by 30 000 43 923_ 30 000 = 30 000 _ 30 000 (1 + r)4 1.4641 = (1 + r )⁴ Find the 4ᵗʰroot on both sides of the equation or raise each side of the equation to the exponent 1/4 4 √1.4641 = ⁴√(1 + r )⁴or (1.4641)¹__ 4 = ((1 + r )⁴)¹__ 4 1.1 = 1 + r 1.1 − 1 = r 0.1 = r r = 0.1 = 10/100 = 10% Therefore, the annual rate of compound interest would need to be 10% to make the amount to be repaid raise to $43 923.

Population Growth and Decay

The growth and decay of certain populations follow an exponential equation. The following examples will show us how the exponential equation is applied when dealing with a population problem.

Example 8.9

Currently, the population of cats in a country is 12 million. This figure is expected to increase at a constant rate of 4% each year. Estimate the population of cats in the country in 5 years.

Solution

We can use the compound interest formula to answer this question. Since the principal (here population) compounds once every year, we will use:

A = P (1 + r)ᵗ, where:

P = current population of cats A = Future population of cats T = Time = 5 years Population in 5 years = 12 000 000 × (1 + 4/100)⁵= 12 000 000 × 1.04⁵= 14 599 834.8288 This means that, in 5 years, there will be approximately 14.6 million cats in the country.

Example 8.10

The population of elephants in a country is currently 640 000 and is predicted to decrease at a rate of 6% per year.

a. What will the population be in 4 years? Give your answer to 3 significant figures.

b. How long will it take for the population to decrease to 390 120?

c. At this rate of decrease, Hannah thinks there will be no more than 300 000 elephants in a decade. Is she correct? Justify your response.

Solution

We can use the compound interest formula to answer this question. Since the principal (population of elephants) decreases every year, we will use:

A = P (1 − r)ᵗ, where:

A = population of elephants in 4 years P = current population of elephants r = 6 % = 0.06 = the rate of decrease t = 4 years

a. Population of elephants in 4 years= 640 000 × (1 − 6 _ 100)⁴b. = 640 000 × 0.94⁴= 499 679.3344 = 500 000 (3 sf) The population of elephants in 4 years is about 500 000

c. We seek to find t such that A = 390,120. This will give us the equation 390120 = 640000 × (1 − 6/100)ᵗ390120_ 640000 = 640000/640000 × 0.94ᵗ0.6095625 = 0.94ᵗlog0.6095625 = log 0.94ᵗlog0.6095625 = tlog0.94 log0.6095625_ log0.94 = t ∴ t = 8 So, it will take about 8 years for the population to decrease to 390120.

In 10 years (a decade), the population of elephants = 640 000 × (1 − 6/100)¹⁰= 640 000 × 0.94¹⁰= 344 713.673 This figure is greater than 300 000. So, Hannah is not correct.

Example 8.11

A body eliminates the concentration of drugs in the blood at a constant rate per hour. The initial concentration of the drug in the blood is 8.46mg/ml. After 5 hours, the concentration is 5mg/ml.

a. Find the rate at which the body eliminates the drug.

b. The concentration is considered negligible if it is 0.125 mg/ml. When will the concentration be negligible?

Solution

a. Using A = P (1 − r)ᵗ, where:

A = Concentration of the drug after 5 hours P = initial concentration of the drug = 8.46 mg/ml r is the rate at which the body eliminates the drug from the blood t = 5 hours We seek to find r, such that:

5 = 8.46 × (1 − r)⁵5_ 8.46 = 8.46/8.46 × (1 − r)⁵5_ 8.46 = (1 − r )⁵Take the fifth root on both sides of the equation 5 √5_ 8.46 = ⁵√(1 − r )⁵or ( 5_ 8.46) 1/5 = ((1 − r)⁵)¹__ 5 (0.591)¹__ 5 = 1 − r 0.9 = 1 − r r = 1 − 0.9 r = 0.1 = 10/100 = 10% Therefore, the rate at which the body eliminates the drug is 10%

b. We have to find t such that 0.125 = 8.46 × (1 − 0.1 )ᵗ0.125/8.46 = 8.46/8.46 × 0.9ᵗ0.014775 = 0.9ᵗlog0.014775 = t × log0.9 log0.014775_________ log0.9 = t × log0.9_____ log0.9 t = 40 It will be 40 hours before the concentration will be negligible.

Depreciation Most things we use do not last forever. This is because they are being used up little by little. The item will not continue to look new or perform the same way as time passes. This is even true with some natural things. The “using up” or “the reduction in value” or “reduction in performance” is called depreciation. If the item is expected to last for 10 years, this 10 years is called the item’s “useful life”.

The value of the item at the end of its useful life is called scrap value.

There are many methods of calculating depreciation. Here we will apply percentages to calculate depreciation.

Depreciation = Rate of depreciation × Value at the beginning of the year.

The item’s value at the beginning of the year is often called the original value.

If the rate of depreciation is r , then depreciation = r × original value The rate of depreciation for each year may vary.

If the rate of depreciation is same for every year, then Current Value or Scrap Value = P(1 − r)ᵗwhere:

P = Purchase price r = rate of depreciation t = time in years.

The current value of the item = Original value − Depreciation.

Depreciation = Purchase price – Current Value

Example 8.12

The value of a generator which originally cost GH¢7 500.00 depreciates by 10% of its original value each year. Find the value of the computer at the end of the third year.

Solution

Current Value = P(1 − r)ᵗP = GH¢ 7500.00 r = 10 % = 0.1 t = 3.

Current Value = 7 500(1 − 0.1)³= 7 500 × 0.9³= 7 500 × 0.729 = 5 467.5 The value of the generator after 3 years is GH¢ 5 467.50

Example 8.13

Find the scrap value of a machine costing GH¢ 50 000.00, having a useful life of fifteen years and a constant annual rate of depreciation of 12%

Solution

Scrap Value = 50 000 × (1 − 0.12 )¹⁵= 50 000 × −0.88¹⁵= 50 000 × 0.1469738539 = GH₵7 348.69 Logarithmic scales are used when calculating the loudness of sound.

Loudness of Sound

The loudness of sound is the intensity or the amount of energy in a sound wave. It shows how loud or soft a listener perceives a sound. It is measured in decibels (dB) The loudness of a sound, L, perceived by the human ear depends on the ratio of the intensity (I) of the sound, to the threshold (lₒ) of hearing for the average human ear. The intensity of sound

(I) is the average power of sound wave per unit area. The formula for calculating loudness of sound is given by:

L = 10Log( I/lₒ)

Example 8.14

Find the loudness of a sound that has an intensity 100 000 times the threshold of hearing for the average human ear.

Solution

Let the threshold of hearing be lₒ In ten si ty, I = 100 000lₒ L oudn ess, L = 10Log( I/lₒ) L = 10Log( 100 000 lₒ_______ lₒ ) = 10 log (100 000) = 10 × 5 = 50dB

1. Equation of the form Y = aXⁿ2. Any relation of a non-linear form Y = a Xⁿ, where x and y are variables and a , n are constants can be reduced to a linear equation of the form y = mx +

c. How do we achieve this? The following activity will explain:

Activity 8.2: Reducing the Exponential function to a linear function In small groups, work through the following steps to see how this is done.

Part A

Step 1: Take the logarithm on both sides of Y = a XⁿStep 2: Apply the addition rule to expand Log(a Xⁿ) into two terms

Step 3: Apply the power rule to rewrite Log Xⁿas nLogX

Step 4: Compare the resulting Log equation to the general equation of a straight line.

Step 5: Based on your comparison, find the values of the unknown constants.

Compare your answer to the one below:

Step 1: Take the logarithm on both sides of Y = a XⁿLogY = Log(a Xⁿ)

Step 2: Apply the addition rule to expand Log(a Xⁿ) into two terms Log Y = Loga + Log XⁿStep 3: Apply the power rule to rewrite Log Xⁿas nLogX LogY = Loga + nLogX

Step 4: Compare the resulting Log equation to the general equation of a straight line.

LogY = nLogX + Loga y = m x + c

Step 5: Based on your comparison, find the values of the unknown constants.

Comparing terms, we have y = LogY , m = n and c = Loga .

This implies that a = 10ᶜThis shows that to reduce an exponential function of the form Y = a Xⁿto linear, plot LogY against LogX. The gradient of the resulting line = n .

Part B Graphs of the form Y = a b^(X), where a and b are constants, can also be reduced to linear form. We will use the same procedure to evaluate this one too:

Step 1: Take the logarithm on both sides of Y = a Xⁿlog₁₀Y = log₁₀a b^(X)

Step 2: Apply the addition rule to expand Log₁₀a b^(X)into two terms Log₁₀Y = Log₁₀a + Log₁₀b^(X)Step 3: Apply the power rule to rewrite Log₁₀b^(X)as X Log₁₀b Log₁₀Y = Log₁₀a + X Log₁₀b

Step 4: Compare the resulting Log equation to the general equation of a straight line.

Log₁₀Y = X Log₁₀b + Log₁₀a y = x m + c

Step 5: Based on your comparison, find the values of the unknown constants.

From the comparisons We have log₁₀Y = y, y coordinates of the resulting line X = x , x coordinates of the resulting line Log₁₀b = m, gradient of the resulting line This implies that b = 10ᵐLog₁₀a = C, y − cordinate of the y-intercept This implies that a = 10^(C)To draw a linear graph for Y = a b^(X), plot LogY against X

Example 8.15

Convert the following to linear functions.

a. y = 3 x^(1.5)b. y = 2/x

c. V = 5 P²Solution

a. y = 3 x^(1.5)Take logs on both sides of the equation Log y = Log(3 x^(1.5)) Log y = log 3 + log x^(1.5)Log y = Log 3 + 1.5 Log x Log y = 1.5 Log x + log 3

b. y = 2/x Log y = Log (²__ x) Log y = Log 2 – Log x Log y = -Log x + Log 2

c. V = 5 P²Log V = log (5P²) Log V = log 5 + Log P²Log V = 2logP + Log 5

Example 8.16

Convert the following to linear functions

a. M = 5 × 3ʳb. 100/5ˣSolution

a. M = 5 × 3ʳTake logs on both sides of the equation Log M = Log (5 × 3ʳ) Log M = Log 5 + Log 3ʳLog M = Log 5 + r Log 3 Log M = r Log 3 + Log5

b. y = 100/5ˣLog y = Log(100_ 5ˣ) Log y = Log 100− Log 5ˣLog y = 2 − xLog5 Logy = − Log5 + 2

Example 8.17

Convert P = 10t³to a linear function and graph it.

Solution

P = 10 t³Log P = log (10t³) Log P = log 10 + Log t³Log P = 3Log t + Log10 Log P = 3Log t +1 y=3x+1 This is the same as the line y = 3x + 1, where y=Log P and x =Log t

Figure 8.1: Graph of linear function y = 3x+1

Example 8.18

The table below gives some values of two related variables, x and y.

x 1 1.2 1.4 1.6 1.8 2 y 2 3.46 5.49 8.19 11.66 16 The relationship between y and x is of the form y = A xᵇwhere A and b are constants.

a. Draw a suitable linear graph for y = A xᵇ

b. Use your graph to find A and b, correct to the nearest whole number.

c. Use your graph to find, correct to the nearest whole number, the value of y when x = 1.5

Solution

a. To reduce y = A xᵇto a linear graph, take logs of both sides:

log₁₀y = log₁₀A xᵇlog₁₀y = log₁₀A + b log₁₀x log₁₀y = b log₁₀x+ log₁₀A ……………….. (1) Comparing (1) to the general equation of a line Y = mX + C, we have:

log₁₀y = Y, y coordinates of the resulting line log₁₀x = X , x coordinates of the resulting line b = m, gradient of the resulting line log₁₀A = C, y − cordinate of the y − intercept .

Using the values given, we construct a table for log₁₀y and log₁₀x log₁₀x 0 0.079 0.146 0.204 0.255 0.301 log₁₀y 0.30 0.54 0.74 0.91 1.07 1.20 The graph of log₁₀y = b log₁₀x+ log₁₀A is presented below

Figure 8.2: Graph of log₁₀y = b log₁₀x+ log₁₀A b.

Figure 8.3: Graph of log₁₀y = b log₁₀x+ log₁₀A From the graph, the y-coordinates intersect the line at 0.3, ⟹ log₁₀A = 0.3 A = 10^(0.3)= 1.995 ≈ 2 Also, the gradient of the line is equal to b, ⟹ b = change in log₁₀y____________ change log₁₀x = 0.9/0.3 = 3 Hence, A = 2 and b = 3 ⟹ y = 2 x³c. Note that the graph has log₁₀x values not x values. Hence, to read x = 1.5 on the graph, we have to convert it to a logarithm of the base 10.

Log₁₀1.5 = 0.18 Locate 0.18 on the log₁₀x axis. Draw a vertical line to intersect the green line, from this point of intersection, draw a horizontal line to log₁₀y axis.

Find the corresponding log₁₀y value of 0.18

Figure 8.4: graph of log₁₀y = b log₁₀x+ log₁₀A From the graph, when log₁₀x = 1.18, Log₁₀y = 0.84 ⟹ y = 10^(0.84)= 6.9183 ≈ 7 Therefore, when x = 1.5, y = 7

Example 8.19

The population of trees in a forest has been decreasing since 1970. A tree population census was conducted every 5 years to assess the decline. It is believed that the decline follows the relation P = a bᵗ. Where P = population of trees after 1970 and t = time (years) after 1970. The census data is shown below.

t 5 10 15 20 25 30 35 40 45 P 108470 98049 88628 80113 72416 65458 59169 53484 48345

a. Use the table to draw a suitable linear graph for the relation.

b. Use your graph to find a and b , correct your answers to two significant figures.

c. Use your graph to estimate the population of trees in 1997.

d. From your graph, in which year did the population of trees decrease to about 85 000

Solution

a. To draw a line graph for the relation P = a bᵗ, take logs on both sides of the equation.

b.

LogP = Log(a bᵗ) LogP = Loga + tLogb LogP = tLogb + Loga The equation is similar to y = mx + c , where y = LogP, t = x, m = Logb and c = Loga To obtain a linear equation, we have to plot LogP against t .

Let us first create a table of values for Log P and t.

t 5 10 15 20 25 30 35 40 45 Log P 5.04 4.99 4.95 4.90 4.86 4.82 4.77 4.73 4.68 Find below the graph of LogP against t

Figure 8.5: Graph of the relation P = a bᵗ,

b. From our relation, m = Logb and c = Loga

Figure 8.6: Graph of the relation P = a bᵗ, Since the line graph is a decreasing function, m = −change in LogP___________ change in t = −0.27/30 = − 0.009 From the relation, we have m = Logb − 0.009 = Logb b = 10^(−0.009)= 0.9795 = 0.98 (to 2 significant figures) From the relation c = Loga . Recall that C is the y-intercept of the graph.

From the graph c = 5.08 This implies that 5.08 = Loga a = 10^(5.08)= 120 226.443 a = 120 000 (to 2 significant figures) The relation is P = 120 000 × 0.98ᵗc. From 1970 to 1997 = 27 years. This means that we will use the graph to find the value of P when t = 27.

Figure 8.7: Graph of the relation P = a bᵗ, From the graph, when t = 27, LogP = 4.84 This implies that P = 10^(4.84)= 69 183.097 In 1997, the population of trees was about 69 183

d. Since the graph has LogP values, we will find Log 85 000 and use the graph to calculate its t value Log85 000 = 4.93 Log85 000 = 4.93

Figure 8.8: Graph of the relation P = a bᵗ, This will take us to 17 years after 1970, which is the same as the year = 1970 + 17 = 1987 Therefore, the population of trees reached 85 000 in 1987

Example 8.20

The value of an investment (V) in cocoa at the end of t years satisfies the relation:

V = P Rᵗwhere P and R are constants.

The table below gives some values of V at the end of t years.

t 2 4 6 8 10 V($) 14 400 20 700 29 900 43 000 61 900

a. Draw a suitable linear graph for V = P Rᵗb. Use your graph to find:

i. P , correct to the nearest whole number

ii. R, correct to two significant figures.

iii. the Value (V ) of an investment at the end of 5 years, correct to two significant figures.

Solution

a. To reduce V = P Rᵗto a linear graph, take log₁₀ on both sides of the equation.

log₁₀V = log₁₀P Rᵗlog₁₀V = log₁₀P + t log₁₀R log₁₀V = t log₁₀R + log₁₀P ……………….. (1) Comparing (1) to the general equation of a line Y = mX + C, we have:

log₁₀V = Y, y coordinates of the resulting line t = X , x coordinates of the resulting line log₁₀R = m, gradient of the resulting line log₁₀P = C, y − coordinate of the y− intersept.

Using the values given, we construct a table for log₁₀A and x x 2 4 6 8 10 log₁₀V 4.16 4.32 4.48 4.63 4.79 The graph of log₁₀V = t log₁₀R+log₁₀P is presented below

Figure 8.9: Graph of log₁₀V = t log₁₀R+ log₁₀P

b. i. To find P, extend the line to intersect log₁₀V axis.

From the graph, the line intersects it at 4 P = 10⁴= 10 000

ii. log₁₀R = m (gradient of the line)

Figure 8.10: graph of log₁₀V = t log₁₀R+ log₁₀P m = 0.47/6 = 0.07833 ⟹ R = 10^(0.07833)= 1.19765 = 1.2 (correct to 2 significant figures) Hence, P = 10 000 and R = 1.2 ⟹ V = 10 000( 1.2)ᵗc. To find the corresponding value of t = 5 on the graph, locate 5 on the t - axis of your graph. Draw a vertical line to intersect your line. From this point of intersection, draw a horizontal line to log₁₀V axis and find its value.

Figure 8.11: graph of log₁₀V = t log₁₀R+ log₁₀P From the graph when t = 5, log₁₀V = 4.4 ⟹ V = 10^(4.4)= 25118.864315 = $25 000 (correct to 2 significant figures)

Example 8.21

Figure 8.12 below shows the graph of Log y against x

Figure. 8.12: Graph of y = 10^(2+1.2x)a. Find the equation of Logy in terms of x.

b. Find the equation of y in terms of x.

Solution

a. the equation of the graph is of the form: Logy = mx + c

b. Gradient of the line = 5− 2/2.5 − 0 = 3/2.5 = 1.2 The line intersects Log y axis at 2 this implies c = 2 Therefore, the equation of Logy in terms of x is:

Logy = 1.2x + 2

c. The equation on y in terms of x is of the form y = a bˣ, where a and b are constants.

Writing y = a bˣin linear form, we have Logy = xlogb+loga.

Comparing Logy = xlogb+loga to Logy = 1.2 x + 2 Log y = Logy. This implies that y = y x = x Log b = 1.2 Laking antilog on both sides, we have b =10^(1.2)And Loga = 2 Taking antilog on both sides, we have a = 10²= 100 Therefore, the equation of y in terms of x is:

y = 100 × (10^(1.2))ˣ, y = 10²× 10^(1.2x)y = 10^(2+1.2x)Example 8.22

Figure 8.13 shows the relationship between LogV and LogP .

Figure 8.13: Graph of V = 1000 P⁻²a. Find the equation of LogV in terms of LogP

b. Find the equation of V in terms of P

Solution

a. The line passes through the points (0,3) and (0.5, 2)

b. The gradient of the line is m = 2− 3/0.5 − 0 = − 1/0.5 = − 2 The equation of LogV in terms of LogP is:

LogV = − 2LogP + 3 The relation is of the form V = a Pⁿ, where a and n are constants.

When V = a Pⁿis written in linear form, it becomes LogV = nLogP + Loga Comparing LogV = nLogP + Loga to LogV = − 2logP + 3 We have:

n = − 2 and Loga =3 ⟹ a = 10³= 1000 Therefore, the relation of V in terms of P is:

V = 1000 P⁻²or V = 1000____

1. Explore plotting of logarithmic graphs using GeoGebra software.

a. Open the GeoGebra Application

b. In the input bar, type the equation of the logarithmic function (E.g., y = log(x))

c. Press Enter and GeoGebra will display the graph of y=log(x)

d. Modify the logarithmic function to explore transformations:

Input the following functions:

i. y = log (x)+2.

ii. y = log (x − 3).

iii. y = −log(x).

iv. y = 2log(x).

e. Observe how the graph changes with each transformation.

f. Discuss the changes with your classmates.

g. Save the GeoGebra file by clicking on the Save or Export option.

2. Talbert, J.F. & Heng, H.H., (2007). Additional Mathematics (Pure and Applied). Pages 370 – 379

3. Haese M., Haese S., Humphries M., Sangwin C., (2014). Cambridge Additional Mathematics. Pages 140, 141

1. Find the value of x that satisfies the equation:

a. log₁₀(2x − 1) = log₁₀x − log₁₀3

b. log₁₀(3x + 1) − log₁₀(1 − x) = 2 log₁₀2

c. Solve log₃(x + 4) − log₂(x − 4) = 2

d. log₂ ³√4x + 5 = ¹_ ₃[log₂x − log₂3]

2. Make y the subject of the relation 2 + logₐb + 3 logₐy = 2 logₐa²y

3. Find the truth set of the equation:

a. logₓ4x − log₄ₓx = 1 ¹_ ₂

b. log₄x + logₓ16 = 3

4. Two equations are defined by: log₂x + 2 log₄y = 4 and x − y = 6 ,

5. for x > y. Find x + y .

6. Ato borrowed GH¢ 12 000 from a Bank for 4 years at an interest of 15% compounded annually. Find the compound interest and amount he has to pay at the end of 4 years.

7. A bank decides to offer Fatima a GH¢ 800 000 loan at a compound interest of 10% per year. Find the total amount the bank will receive when the loan is repaid after 5 years.

8. How much interest will Fatima pay?

9. A housewife invested $30 000 into a fund that pays 12 percent interest each year, compounded quarterly. Find the value of the investment at the end of the fourth year.

10. Ato invested €70 000 into a fund that pays 10% interest each year, compounded once every two years.

a. Find the total amount of his investment after 10 years.

b. After how many years will his investment double?

11. Calculate the annual rate of compound interest that will allow a loan of $50 000 to amount to $73 205 in four years.

12. Currently, the population of a country is 72 million. This figure is expected to increase at a constant rate of 2.5% each year. Estimate the population of the country in 5 years.

13. Find the loudness of a sound that has an intensity 10 000 times the threshold of hearing for the average human ear.

14. Find the scrap value of a generator costing GH¢ 25 000.00, having a useful life span of fifteen years and a constant annual rate of depreciation of 10%.

15. Convert the following to linear functions:

a. y = 1/2 x⁴b. P = 10/r

c. V = 2 P³16. Convert the following to linear functions

a. M = 10 × 7ʳb. y = 1000/2ˣ17. Convert P = 100 t²to a linear function and graph it.

18. The table below gives some values of two related variables, x and y.

x 1 1.2 1.4 1.6 1.8 2 2.2 y 10 14.4 19.6 25.6 32.4 40 48.4 The relationship between y and x is of the form y = A xᵇwhere A and b are constants.

a. Draw a suitable linear graph for y = A xᵇb. Use your graph to find A and b, correct to the nearest whole number.

c. Use your graph to find, correct to the nearest whole number the value of y when x = 1.7

19. The value of an investment (V) at the end of t years satisfies the relation:

20. V = P Rᵗwhere P and R are constants. The table below gives some values of V at the end of t years.

t 1 2 3 4 5 6 7 V (GH¢) 22400.00 25088.00 28098.56 31470.39 35246.83 39476.45 44213.63

a. Draw a suitable line graph for V = P Rᵗb. Use your graph to find

i. P , correct to the nearest thousand

ii. R, correct to three significant figures.

21. The figure below shows the graph of Log y against x

22. a. Find the equation of Logy in terms of x

b. Find the equation of y in terms of x

23. The figure shows the relationship between LogV and LogP .

a. Find the equation of LogV in terms of LogP

b. Find the equation of V in terms of P