Algebra Practice Questions — JAMB UTME

Question 1Question →

A binary operation $\otimes$ on the set of real numbers $\mathbb{R}$ is defined by $a \otimes b = a^2 + b^2 - ab$ . If $x \otimes 3 = 19$ and $x > 0$ , find the value of $x$ .

Show answer & explanation

Answer: 5

Answer

The value of

x

is

5

.

Applying the operation rule gives

x^2 + 3^2 - 3x = 19

, which simplifies to

x^2 - 3x - 10 = 0

. Factoring this equation yields

(x - 5)(x + 2) = 0

. Since

x

is constrained to be positive (

x > 0

), the unique valid answer is

5

.

Step-by-Step Solution

1

Apply the definition of the binary operation to

x \otimes 3

x^2 + 3^2 - 3(x) = x^2 - 3x + 9

Substitute

a = x

and

b = 3

into

a \otimes b = a^2 + b^2 - ab

.

2

Equate the result to 19 and rearrange into standard quadratic form

x^2 - 3x - 10 = 0

Subtract 19 from both sides to set the quadratic equation to zero.

3

Factor the quadratic equation and solve for

x

(x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2

Find two numbers that multiply to

-10

and add up to

-3

.

4

Apply the restriction

x > 0

x = 5

Reject the negative solution

x = -2

because

x

must be strictly positive.

Key Concept

Evaluation of Binary Operations and Solving Quadratic Equations

Question 2Question →

The perimeter of a rectangular playfield is $28\text{ m}$ and its area is $40\text{ m}^2$ . What is the positive difference, in metres, between its length and width?

Show answer & explanation

Answer: 6

Answer

The positive difference between the length and width of the playfield is 6 metres.

Formulating the system gives

x + y = 14

and

xy = 40

. Substituting

y = 14 - x

yields the quadratic equation

x^2 - 14x + 40 = 0

, which factors into

(x - 10)(x - 4) = 0

. The dimensions are

10\text{ m}

and

4\text{ m}

, giving a positive difference of

10 - 4 = 6\text{ m}

.

Step-by-Step Solution

1

Set up linear and quadratic equations for perimeter and area

x + y = 14

and

xy = 40

Perimeter formula is

2(x + y) = 28

which simplifies to

x + y = 14

, and area formula is

xy = 40

.

2

Substitute

y = 14 - x

into the quadratic area equation

x(14 - x) = 40 \implies x^2 - 14x + 40 = 0

Substitution reduces the simultaneous system to a single quadratic equation in terms of

x

.

3

Solve the quadratic equation by factoring

(x - 10)(x - 4) = 0 \implies x = 10 \text{ or } x = 4

The roots of the equation give the dimensions of the rectangle.

4

Calculate the positive difference between the two dimensions

10 - 4 = 6

Subtract the smaller dimension from the larger dimension.

Key Concept

Solving word problems involving simultaneous linear and quadratic equations

Estimated Time:1m 30s

Question 3Question →

A quantity $P$ varies partially as $x$ and partially as the square of $y$ . When $x = 2$ and $y = 3$ , $P = 24$ , and when $x = 5$ and $y = 1$ , $P = 17$ . What is the value of $P$ when $x = 4$ and $y = 3$ ?

30

35

26

21

Show answer & explanation

Answer: 30

Answer

The value of

P

is 30.

The relationship follows the partial variation formula

P = k_1 x + k_2 y^2

. Substituting the given conditions gives

2k_1 + 9k_2 = 24

and

5k_1 + k_2 = 17

. Solving these simultaneous equations yields

k_1 = 3

and

k_2 = 2

. Evaluating

P = 3(4) + 2(3^2)

produces

12 + 18 = 30

.

Step-by-Step Solution

1

Set up the general formula for partial variation.

P = k_1 x + k_2 y^2

, where

k_1

and

k_2

are constants.

Partial variation combines terms linearly with separate variation constants.

2

Substitute the given pairs of values to form simultaneous linear equations.

Equation (1):

2k_1 + 9k_2 = 24

; Equation (2):

5k_1 + k_2 = 17

.

Plugging in

(x=2, y=3, P=24)

and

(x=5, y=1, P=17)

creates a system of equations in terms of

k_1

and

k_2

.

3

Solve the simultaneous linear equations for

k_1

and

k_2

.

From Equation (2),

k_2 = 17 - 5k_1

. Substitute into Equation (1):

2k_1 + 9(17 - 5k_1) = 24 \implies -43k_1 = -129 \implies k_1 = 3

. Then

k_2 = 17 - 5(3) = 2

.

Finding the specific values of the variation constants is required to complete the formula.

4

Calculate

P

for

x = 4

and

y = 3

using the complete formula

P = 3x + 2y^2

.

P = 3(4) + 2(3^2) = 12 + 2(9) = 12 + 18 = 30

.

Evaluating the relationship with the target parameters produces the final answer.

Key Concept

Partial Variation and Simultaneous Linear Equations

Estimated Time:2m 0s

Question 4Question →

Given the $3 \times 3$ matrix $A = \begin{pmatrix} x & 2 & 1 \\ 3 & 1 & 2 \\ 1 & x & 0 \end{pmatrix}$ , find the positive value of $x$ for which $\det(A) = -2$ .

Show answer & explanation

Answer: 2.5

Answer

The positive value of

x

is 2.5.

Expanding the determinant of matrix

A

along the third row gives

1(4 - 1) - x(2x - 3) = -2x^2 + 3x + 3

. Setting this equal to

-2

yields

2x^2 - 3x - 5 = 0

. Factoring gives

(2x - 5)(x + 1) = 0

, yielding solutions

x = 2.5

and

x = -1

. Taking the positive value gives

x = 2.5

.

Step-by-Step Solution

1

Calculate the determinant of matrix

A

in terms of

x

\det(A) = -2x^2 + 3x + 3

Expanding along the third row simplifies computation because of the zero entry.

2

Set the determinant expression equal to

-2

and rearrange terms

2x^2 - 3x - 5 = 0

Setting

-2x^2 + 3x + 3 = -2

forms a standard quadratic equation.

3

Factorize the quadratic equation to find the roots

x = 2.5

or

x = -1

Factoring

(2x - 5)(x + 1) = 0

yields two real solutions.

4

Filter for the positive value requested in the stem

x = 2.5

The question specifically requires the positive value of

x

.

Key Concept

Determinant of a 3x3 Matrix and Quadratic Equation Solving

Estimated Time:2m 30s

Question 5Question →

Solve the simultaneous equations $y = 2x + 1$ and $y = x^2 - 2$ . Which of the following represents the complete set of solution pairs $(x, y)$ ?

(3, 7)

and

(-1, -1)

(-3, -5)

and

(1, 3)

(3, 5)

and

(-1, -3)

(3, 7)

and

(1, 3)

Show answer & explanation

Answer: $(3, 7)$ and $(-1, -1)$

Answer

The complete set of solution pairs

(x, y)

is

(3, 7)

and

(-1, -1)

.

Equating

2x + 1 = x^2 - 2

yields

x^2 - 2x - 3 = 0

. Factoring gives

(x - 3)(x + 1) = 0

, leading to

x = 3

or

x = -1

. Substituting these

x

-values into

y = 2x + 1

gives

y = 7

for

x = 3

, and

y = -1

for

x = -1

. Thus, the solution pairs are

(3, 7)

and

(-1, -1)

.

Step-by-Step Solution

1

Equate the linear expression for

y

to the quadratic expression for

y

.

2x + 1 = x^2 - 2

Since both expressions equal

y

, setting them equal eliminates

y

.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 2x - 3 = 0

Subtract

2x

and

1

from both sides.

3

Factor the quadratic equation to find the values of

x

.

(x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1

Determine two numbers that multiply to

-3

and add to

-2

.

4

Substitute each

x

-value back into the linear equation

y = 2x + 1

to find the corresponding

y

-value.

For

x = 3

,

y = 2(3) + 1 = 7

. For

x = -1

,

y = 2(-1) + 1 = -1

.

Calculate the exact coordinate pairs

(x, y)

that satisfy both equations.

Key Concept

Solving simultaneous linear and quadratic equations using algebraic substitution.

Question 6Question →

Find the number of integer values of $x$ that satisfy the inequality $3x^2 - 10x - 8 \leq 0$ .

Show answer & explanation

Answer: 5

Answer

The number of integer values of

x

satisfying the inequality is 5.

Factoring

3x^2 - 10x - 8 \leq 0

gives

(3x + 2)(x - 4) \leq 0

. The region where the quadratic expression is non-positive lies between the roots

x = -\frac{2}{3}

and

x = 4

, yielding

-\frac{2}{3} \leq x \leq 4

. The integers falling within this closed interval are

0, 1, 2, 3,

and

4

, giving a total of 5 integer solutions.

Step-by-Step Solution

1

Factor the quadratic expression

(3x + 2)(x - 4) \leq 0

Factoring allows us to find the critical boundary values of the inequality.

2

Find the critical values (roots)

x = -\frac{2}{3}

and

x = 4

Setting each linear factor to zero determines where the expression changes sign.

3

Determine the solution set interval

-\frac{2}{3} \leq x \leq 4

Since the coefficient of

x^2

is positive, the quadratic curve is convex (U-shaped), so the expression is less than or equal to zero between the roots.

4

List and count the integer solutions

Integers:

0, 1, 2, 3, 4

(Total = 5)

The smallest integer greater than or equal to

-\frac{2}{3}

is

0

, and the largest integer less than or equal to

4

is

4

.

Key Concept

Solving quadratic inequalities and identifying integer solutions within a continuous range.

Question 7Question →

A binary operation $\star$ on the set of real numbers is defined by $a \star b = 3a + 2b - 1$ . What is the value of $x$ such that $4 \star x = 25$ ?

6

7

11

14

Show answer & explanation

Answer: 7

Answer

The value of

x

is

7

.

Applying the binary operation definition

a \star b = 3a + 2b - 1

with

a = 4

and

b = x

gives

3(4) + 2x - 1 = 25

. Simplifying gives

2x + 11 = 25

, leading to

2x = 14

and

x = 7

.

Step-by-Step Solution

1

Substitute

a = 4

and

b = x

into the definition of the binary operation

a \star b = 3a + 2b - 1

.

4 \star x = 3(4) + 2x - 1

Applying the given rule for the binary operation.

2

Simplify the expression on the left-hand side.

4 \star x = 12 + 2x - 1 = 2x + 11

Performing basic arithmetic multiplication and addition of constants.

3

Set the simplified expression equal to

25

and solve for

x

.

2x + 11 = 25 \implies 2x = 14 \implies x = 7

Subtracting

11

from both sides and dividing by

2

.

Key Concept

Evaluation of non-commutative binary operations and solving algebraic equations involving operational rules.

Question 8Question →

If $(x_1, y_1)$ and $(x_2, y_2)$ are the real solution pairs to the simultaneous equations $x - 2y = 1$ and $x^2 - xy + y^2 = 7$ , what is the value of $x_1 x_2 + y_1 y_2$ ?

-9

-11

-7

-1

Show answer & explanation

Answer: $-11$

Answer

The value of

x_1 x_2 + y_1 y_2

is

-11

.

Substituting

x = 2y + 1

into

x^2 - xy + y^2 = 7

gives

3y^2 + 3y - 6 = 0

, which reduces to

y^2 + y - 2 = 0

. The roots are

y = 1

and

y = -2

. Substituting back into the linear expression yields corresponding values

x = 3

and

x = -3

. The two solution pairs are

(3, 1)

and

(-3, -2)

. Calculating

x_1 x_2 + y_1 y_2 = (3)(-3) + (1)(-2)

yields

-11

.

Step-by-Step Solution

1

Express

x

in terms of

y

from the linear equation.

x = 2y + 1

Isolating variable

x

allows direct substitution into the quadratic equation.

2

Substitute

x = 2y + 1

into the quadratic equation

x^2 - xy + y^2 = 7

.

(2y + 1)^2 - (2y + 1)y + y^2 = 7 \implies 3y^2 + 3y - 6 = 0

This simplifies the system to a single quadratic equation in

y

.

3

Solve the quadratic equation

3y^2 + 3y - 6 = 0

.

y^2 + y - 2 = 0 \implies (y + 2)(y - 1) = 0

, giving

y_1 = 1

and

y_2 = -2

Factoring determines the two possible

y

-coordinates.

4

Determine corresponding

x

-coordinates for each

y

-value.

For

y_1 = 1

,

x_1 = 2(1) + 1 = 3 \implies (3, 1)

. For

y_2 = -2

,

x_2 = 2(-2) + 1 = -3 \implies (-3, -2)

.

Substituting each

y

into

x = 2y + 1

yields the complete solution pairs.

5

Compute

x_1 x_2 + y_1 y_2

.

(3)(-3) + (1)(-2) = -9 - 2 = -11

Evaluates the targeted algebraic expression.

Key Concept

Solving simultaneous linear and quadratic equations using algebraic substitution.

Question 9Question →

If the matrix $A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}$ , which matrix represents the inverse $A^{-1}$ ?

\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

\begin{pmatrix} 5 & 2 \\ 7 & 3 \end{pmatrix}

\begin{pmatrix} -5 & 2 \\ 7 & -3 \end{pmatrix}

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}$

Answer

\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

The correct inverse matrix is computed by evaluating the determinant

\det(A) = 3(5) - 2(7) = 1

and constructing the adjugate matrix by swapping the diagonal elements

3

and

5

while changing the signs of

2

and

7

, resulting in

\begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

.

Step-by-Step Solution

1

Calculate the determinant of matrix

A

\det(A) = (3)(5) - (2)(7) = 15 - 14 = 1

The inverse requires dividing the adjugate matrix by the determinant of

A

.

2

Find the adjugate of matrix

A

\text{adj}(A) = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

For a 2x2 matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, swap the main diagonal elements (

a

and

d

) and change the signs of the off-diagonal elements (

b

and

c

).

3

Compute

A^{-1} = \frac{1}{\det(A)} \text{adj}(A)

A^{-1} = \frac{1}{1} \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix}

Multiply the adjugate matrix by the reciprocal of the determinant.

Key Concept

2x2 Matrix Inversion

Estimated Time:1m 0s

Question 10Question →

Solve for $x$ in the logarithmic equation $\log_2 (x + 2) + \log_2 (x - 4) = 4$ .

4

5

6

9

Show answer & explanation

Answer: 6

Answer

The value of

x

is 6.

Using the product rule for logarithms,

\log_2 (x + 2) + \log_2 (x - 4) = \log_2 [(x + 2)(x - 4)] = 4

. Converting to exponential form yields

(x + 2)(x - 4) = 2^4 = 16

. Expanding gives

x^2 - 2x - 8 = 16

, which simplifies to

x^2 - 2x - 24 = 0

. Factoring gives

(x - 6)(x + 4) = 0

. Since the logarithmic arguments must be positive (

x > 4

), the negative root

-4

is discarded, leaving

x = 6

.

Step-by-Step Solution

1

Apply the product rule of logarithms:

\log_b M + \log_b N = \log_b (M \cdot N)

\log_2 [(x + 2)(x - 4)] = 4

Logarithms with the same base being added combine by multiplying their arguments.

2

Convert the logarithmic equation to its exponential form

(x + 2)(x - 4) = 2^4 = 16

If

\log_b Y = X

, then

Y = b^X

.

3

Expand and rearrange into a standard quadratic equation

x^2 - 2x - 8 = 16 \implies x^2 - 2x - 24 = 0

Expanding

(x+2)(x-4)

yields

x^2 - 2x - 8

, and subtracting 16 sets the equation to zero.

4

Factor the quadratic equation to find potential solutions

(x - 6)(x + 4) = 0 \implies x = 6 \text{ or } x = -4

The roots of

x^2 - 2x - 24 = 0

are

x = 6

and

x = -4

.

5

Check the domain restrictions for logarithmic functions

x = 6

The arguments of the original logarithms require

x + 2 > 0 \implies x > -2

and

x - 4 > 0 \implies x > 4

. Therefore,

x = -4

is extraneous and

x = 6

is the only valid solution.

Key Concept

Solving logarithmic equations using product rule and domain constraints

Question 11Question →

If $\log_4 x + \log_2 x = 6$ , find the value of $x$ .

Show answer & explanation

Answer: 16

Answer

The value of

x

is 16.

Using the change of base identity

\log_b a = \frac{\log_c a}{\log_c b}

, we convert

\log_4 x

into

\frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

. Substituting this into the equation yields

\frac{3}{2} \log_2 x = 6

, which simplifies to

\log_2 x = 4

, leading directly to

x = 2^4 = 16

.

Step-by-Step Solution

1

Apply the change of base formula to

\log_4 x

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

Converting all terms to a common base (base 2) simplifies addition of logarithmic terms.

2

Substitute the expression back into the original equation and collect like terms

\frac{1}{2}\log_2 x + \log_2 x = \frac{3}{2}\log_2 x = 6

Adding the coefficients of

\log_2 x

gives

\frac{3}{2}

.

3

Isolate

\log_2 x

\log_2 x = 6 \cdot \frac{2}{3} = 4

Multiplying both sides by

\frac{2}{3}

isolates the logarithmic term.

4

Convert from logarithmic form to exponential form

x = 2^4 = 16

If

\log_b a = c

, then

a = b^c

.

Key Concept

Change of base rule for logarithms

Question 12Question →

What is the positive integer solution to the logarithmic equation $\log_2 x - 3\log_x 2 = 2$ ?

8

6

3

12

Show answer & explanation

Answer: $8$

Answer

The positive integer solution is

8

.

Using the change of base identity

\log_x 2 = \frac{1}{\log_2 x}

, we rewrite the equation as

u - \frac{3}{u} = 2

where

u = \log_2 x

. Rearranging yields the quadratic

u^2 - 2u - 3 = 0

, which factors as

(u - 3)(u + 1) = 0

. This gives

u = 3

or

u = -1

. Converting back to

x = 2^u

, we get

x = 2^3 = 8

or

x = 2^{-1} = \frac{1}{2}

. Since the question asks for the positive integer solution, the correct answer is 8.

Step-by-Step Solution

1

Apply the change of base rule to express

\log_x 2

in terms of base 2.

\log_2 x - \frac{3}{\log_2 x} = 2

The change of base identity states that

\log_b a = \frac{1}{\log_a b}

.

2

Substitute

u = \log_2 x

into the equation and clear the fraction.

u - \frac{3}{u} = 2 \implies u^2 - 2u - 3 = 0

Multiplying through by

u

(where

u \neq 0

) transforms the equation into standard quadratic form.

3

Factor the quadratic equation to solve for

u

.

(u - 3)(u + 1) = 0 \implies u = 3 \text{ or } u = -1

Finding the roots of the quadratic equation in terms of

u

.

4

Convert back to

x

using

x = 2^u

and select the positive integer solution.

For

u = 3

:

x = 2^3 = 8

. For

u = -1

:

x = 2^{-1} = \frac{1}{2}

.

The question asks specifically for the positive integer solution, which is 8.

Key Concept

Logarithmic Change of Base and Quadratic Reduction

Question 13Question →

Find the value of $x$ that satisfies the exponential equation $27^{x - 1} = \frac{9^{x + 1}}{3^{x - 5}}$ .

Show answer & explanation

Answer: 5

Answer

The value of

x

is 5.

Rewriting the terms with a common base of 3 transforms the equation into

3^{3x-3} = 3^{x+7}

. Equating exponents gives

3x - 3 = x + 7

, which yields

x = 5

.

Step-by-Step Solution

1

Convert all terms to base 3

27^{x-1} = 3^{3x-3}

and

9^{x+1} = 3^{2x+2}

Laws of indices require identical bases to manipulate exponents.

2

Apply division rule of indices to the right-hand side

\frac{3^{2x+2}}{3^{x-5}} = 3^{x+7}

When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

3

Equate the exponents and solve for

x

3x - 3 = x + 7 \implies 2x = 10 \implies x = 5

If

a^f(x) = a^g(x)

for

a > 0

and

a \neq 1

, then

f(x) = g(x)

.

Key Concept

Solving exponential equations using base reduction and exponent laws

Estimated Time:1m 30s

Question 14Question →

What is the square root of the surd expression $14 + 6\sqrt{5}$ ?

3 + \sqrt{5}

3 - \sqrt{5}

\sqrt{14} + \sqrt{5}

9 + \sqrt{5}

Show answer & explanation

Answer: $3 + \sqrt{5}$

Answer

3 + \sqrt{5}

Expanding the square of

3 + \sqrt{5}

yields

(3)^2 + 2(3)(\sqrt{5}) + (\sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}

, which accurately equals the original expression under the radical.

Step-by-Step Solution

1

Set up the general form for the square root of a binomial surd

Let

\sqrt{14 + 6\sqrt{5}} = \sqrt{a} + \sqrt{b}

The square root of a compound surd expression takes the form of a sum of radical terms

2

Square both sides of the equation

14 + 6\sqrt{5} = a + b + 2\sqrt{ab}

Eliminate the outer radical to equate real and surd parts

3

Equate the rational parts and the surd parts

a + b = 14

and

2\sqrt{ab} = 6\sqrt{5} \implies \sqrt{ab} = 3\sqrt{5} = \sqrt{45} \implies ab = 45

Match integer terms together and radical terms together

4

Solve for values of

a

and

b

Two positive numbers with sum

14

and product

45

are

9

and

5

, so

a = 9

and

b = 5

Determine the factors satisfying both equations

5

Substitute

a

and

b

into the radical expression

\sqrt{14 + 6\sqrt{5}} = \sqrt{9} + \sqrt{5} = 3 + \sqrt{5}

Simplify

\sqrt{9}

to

3

to obtain the final simplified expression

Key Concept

Finding the square root of a surd expression by equating rational and radical parts

Question 15Question →

If $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} = x\sqrt{15}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 2

Answer

The value of

x

is 2.

Rationalising both fractions yields

4 + \sqrt{15}

and

4 - \sqrt{15}

. Subtracting the second from the first gives

(4 + \sqrt{15}) - (4 - \sqrt{15}) = 2\sqrt{15}

. Comparing

2\sqrt{15}

with

x\sqrt{15}

gives

x = 2

.

Step-by-Step Solution

1

Rationalise the denominator of the first fraction

\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{5 + 2\sqrt{15} + 3}{5 - 3} = 4 + \sqrt{15}

Multiplying numerator and denominator by the conjugate of the denominator removes the surd from the denominator.

2

Rationalise the denominator of the second fraction

\frac{(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{5 - 2\sqrt{15} + 3}{5 - 3} = 4 - \sqrt{15}

Multiply by the conjugate

(\sqrt{5}-\sqrt{3})

to simplify the second surd term.

3

Subtract the simplified expressions

(4 + \sqrt{15}) - (4 - \sqrt{15}) = 4 - 4 + \sqrt{15} + \sqrt{15} = 2\sqrt{15}

Distribute the negative sign and combine like surd terms.

4

Solve for the unknown coefficient x

2\sqrt{15} = x\sqrt{15} \implies x = 2

Divide both sides of the equation by

\sqrt{15}

to isolate

x

.

Key Concept

Binomial Surd Rationalisation and Simplification

Estimated Time:1m 30s

Question 16Question →

If $\sqrt{x + 4} - \sqrt{x - 1} = 1$ , what is the value of $x$ ?

5

17

3

10

Show answer & explanation

Answer: $5$

Answer

The value of

x

is

5

.

Isolating

\sqrt{x + 4}

gives

\sqrt{x + 4} = 1 + \sqrt{x - 1}

. Squaring both sides yields

x + 4 = 1 + 2\sqrt{x - 1} + x - 1

, which simplifies to

4 = 2\sqrt{x - 1}

. Dividing by 2 gives

2 = \sqrt{x - 1}

. Squaring both sides once more gives

4 = x - 1

, which leads to

x = 5

.

Step-by-Step Solution

1

Isolate one of the surd terms on the left side of the equation.

\sqrt{x + 4} = 1 + \sqrt{x - 1}

Rearranging terms prevents dealing with complex cross-products when squaring.

2

Square both sides of the equation.

x + 4 = 1 + 2\sqrt{x - 1} + (x - 1)

Squaring eliminates the outer radical on the left side.

3

Simplify both sides and isolate the remaining radical term.

4 = 2\sqrt{x - 1} \implies 2 = \sqrt{x - 1}

Subtracting

x

from both sides simplifies the algebraic expression.

4

Square both sides again to solve for

x

.

4 = x - 1 \implies x = 5

Squaring eliminates the remaining radical term.

Key Concept

Solving Surd Equations by Rational Elimination

Question 17Question →

Given the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , with subsets $P = \{x \in \mathcal{U} : x \text{ is a prime number}\}$ and $Q = \{x \in \mathcal{U} : x \text{ is an odd number}\}$ , what is the number of elements in the set $(P \cup Q)'$ ?

Show answer & explanation

Answer: 6; six; 6 elements

Answer

The number of elements in

(P \cup Q)'

is 6.

The set

P \cup Q

consists of all numbers from 1 to 15 that are either prime or odd:

\{1, 2, 3, 5, 7, 9, 11, 13, 15\}

. The complement

(P \cup Q)'

relative to

\mathcal{U}

contains all elements of

\mathcal{U}

that are neither prime nor odd, which are the even composite numbers:

\{4, 6, 8, 10, 12, 14\}

. Counting these elements gives 6.

Step-by-Step Solution

1

List all elements of the universal set

\mathcal{U}

, subset

P

, and subset

Q

.

\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\}

,

P = \{2, 3, 5, 7, 11, 13\}

, and

Q = \{1, 3, 5, 7, 9, 11, 13, 15\}

.

Listing the explicit elements allows precise execution of set union and complement operations.

2

Find the union

P \cup Q

.

P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15\}

.

The union combines all unique elements that belong to either set

P

, set

Q

, or both.

3

Determine the complement set

(P \cup Q)'

relative to

\mathcal{U}

and count its elements.

(P \cup Q)' = \{4, 6, 8, 10, 12, 14\}

, which contains 6 elements.

The complement set consists of all elements in

\mathcal{U}

that are not present in

P \cup Q

.

Key Concept

Complement of Set Union

Question 18Question →

When the polynomial $P(x) = 3x^3 - kx^2 + 4x - 7$ is divided by $x - 2$ , the remainder is $9$ . What is the value of $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of

k

is

4

.

According to the Remainder Theorem, dividing

P(x)

by

x - 2

leaves a remainder of

P(2)

. Evaluating

P(2) = 3(2)^3 - k(2)^2 + 4(2) - 7

gives

25 - 4k

. Setting

25 - 4k = 9

and solving yields

k = 4

.

Step-by-Step Solution

1

Apply the Remainder Theorem

The remainder when

P(x)

is divided by

x - 2

is

P(2)

.

By the Remainder Theorem, dividing a polynomial

P(x)

by

x - a

leaves a remainder equal to

P(a)

.

2

Substitute

x = 2

into

P(x)

P(2) = 3(2)^3 - k(2)^2 + 4(2) - 7 = 25 - 4k

Evaluating the polynomial at

x = 2

expresses the remainder in terms of

k

.

3

Equate

P(2)

to the given remainder and solve for

k

25 - 4k = 9 \implies 4k = 16 \implies k = 4

Setting the calculated expression equal to

9

forms a linear equation that yields

k = 4

.

Key Concept

Polynomial Remainder Theorem

Estimated Time:1m 30s

Question 19Question →

What is the remainder when the polynomial $P(x) = 2x^3 - 5x^2 + 7x - 3$ is divided by $(2x - 1)$ ?

-\frac{1}{2}

-8

1

-17

Show answer & explanation

Answer: $-\frac{1}{2}$

Answer

The remainder is

-\frac{1}{2}

.

By the Remainder Theorem, when

P(x)

is divided by

(ax - b)

, the remainder is

P\left(\frac{b}{a}\right)

. Setting

2x - 1 = 0

yields

x = \frac{1}{2}

. Evaluating

P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + 7\left(\frac{1}{2}\right) - 3 = \frac{1}{4} - \frac{5}{4} + \frac{7}{2} - 3 = -\frac{1}{2}

.

Step-by-Step Solution

1

Find the root of the linear divisor

2x - 1 = 0 \implies x = \frac{1}{2}

According to the Remainder Theorem, dividing

P(x)

by a linear divisor

(ax - b)

yields a remainder of

P\left(\frac{b}{a}\right)

.

2

Substitute

x = \frac{1}{2}

into

P(x) = 2x^3 - 5x^2 + 7x - 3

P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 5\left(\frac{1}{2}\right)^2 + 7\left(\frac{1}{2}\right) - 3

Evaluating the polynomial at the root of the divisor determines the remainder.

3

Calculate the arithmetic value

P\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) - 5\left(\frac{1}{4}\right) + \frac{7}{2} - 3 = \frac{1}{4} - \frac{5}{4} + \frac{7}{2} - 3 = -1 + \frac{7}{2} - 3 = -4 + 3.5 = -0.5 = -\frac{1}{2}

Simplifying the fractional terms gives the final remainder value.

Key Concept

Remainder Theorem for Linear Divisors

(ax - b)

Question 20Question →

In a geometric progression of positive terms, the sum of the first two terms is $12$ and the sum of the third and fourth terms is $48$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 128

Answer

The 6th term of the geometric progression is 128.

Dividing

ar^2(1+r) = 48

by

a(1+r) = 12

yields

r^2 = 4

, so

r = 2

for positive terms. Substituting

r = 2

into

a(1+r) = 12

gives

a = 4

. Using

T_n = a r^{n-1}

for

n=6

, we get

T_6 = 4 \times 2^5 = 128

.

Step-by-Step Solution

1

Set up algebraic equations for the given sums using first term

a

and common ratio

r

.

a(1+r) = 12

and

ar^2(1+r) = 48

The terms of a geometric progression are given by

T_n = a r^{n-1}

.

2

Divide the equation for the third and fourth terms by the equation for the first and second terms.

r^2 = 4 \implies r = 2

Dividing eliminates

a

and

(1+r)

, giving

r^2 = 4

. Since terms are positive,

r > 0

.

3

Substitute

r = 2

into

a(1+r) = 12

to solve for

a

.

a = 4

3a = 12

leads directly to

a = 4

.

4

Evaluate the 6th term using the formula

T_6 = a r^{5}

.

T_6 = 4 \times 2^5 = 128

Applying the general term formula

T_n = a r^{n-1}

with

n=6

.

Key Concept

Geometric Progression term relations and finding the common ratio from consecutive term pairs

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