Quiz

Algebra

Typical Exam Questions

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Quiz

Algebra Score: 0/10

1. Solve for x : ( x² - 4 ) ( x - 2 ) = 0

A. x = 4 or x = 2

B. x = ± 2 or x = 2

C. x = - 4 or x = 4

D. None of the above

2. Solve for x : ( x - 1 ) ( x + 8 ) = 10

A. x = 1 or x = -

\frac{5}{4}

B. x = 11 or x = 2

C. x = - 9 or x = 2

D. None of the above

3. Solve for x : 4x² + 11x + 4 = 0
(Leave your answer correct to TWO decimal places IF NECESSARY)

A. x = 0.43 or x = 2.32

B. x = 0 or x = 2

C. x = - 0.43 or x = - 2.32

D. None of the above

4. Solve for x : - x² ≤ 4 - 5x

A. x ≤ 1 or x ≥ 4

B. 1 ≤ x ≤ 4

C. 1 ≥ x ≥ 4

D. None of the above

5. Solve for x : 6x < 3x²

A. x < 0 or x > 2

B. 0 < x < 2

C. x < - 2 or x > 0

D. None of the above

6. Solve for x and y if y = x + 2 and x² + y² = 20

6. Solve for x and y if

y = x + 2 and x² + y² = 20

A. (x ; y) = ( - 2 ; 0 ) and (x ; y) = ( 4 ; 6 )

B. (x ; y) = ( 2 ; 4 ) and (x ; y) = ( 4 ; 6 )

C. (x ; y) = ( 2 ; 4 ) and (x ; y) = ( - 4 ; - 2 )

D. None of the above

7. Solve for x and y if 2x - y - 3 = 0 and x² + 5xy + y² = 15

7. Solve for x and y if

2x - y - 3 = 0 and x² + 5xy + y² = 15

A. (x ; y) = ( -

\frac{1}{5}

; -

\frac{17}{5}

) and (x ; y) = ( 2 ; 1 )

B. (x ; y) = (

\frac{1}{5}

;

\frac{17}{5}

) and (x ; y) = ( 2 ; 1 )

C. (x ; y) = ( - 5 ; -

\frac{5}{17}

) and (x ; y) = ( 2 ; 1 )

D. None of the above

8. Solve for x and y if $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if

$\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

A. (x ; y) = (

\frac{1}{3}

; -

\frac{1}{3}

) and (x ; y) = ( - 1 ; - 3 )

B. (x ; y) = ( -

\frac{1}{3}

;

\frac{1}{3}

) and (x ; y) = ( - 1 ; - 3 )

C. (x ; y) = ( 3 ; - 3 ) and (x ; y) = ( - 1 ; - 3 )

D. None of the above

9. Given: x = $\frac{± \sqrt[]{b 2 – 9}}{- 2}$

Determine the value(s) of b for which x is a Real Number.

A. b ≥ - 3 or b ≤ 3

B. - 3 ≤ b ≤ 3

C. b ≤ - 3 or b ≥ 3

D. None of the above

10. The roots of a quadratic equation are given by: x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

10. The roots of a quadratic equation are given by:

x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

For how many values of k > 0 will the roots be Rational ?

A. 2

B. 3

C. 4

D. None of the above

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Quiz Results

Your score is 0 out of 10

Solutions

Algebra Question 1 of 10

1. Solve for x : ( x² - 4 ) ( x - 2 ) = 0

A. x = 4 or x = 2

B. x = ± 2 or x = 2

C. x = - 4 or x = 4

D. None of the above

Correct Answer
Option B

Detailed Solution

Solution Details

Given ( x² - 4 ) ( x - 2 ) = 0

• ( x² - 4 ) ( x - 2 ) = 0

• ( x² - 4 ) = 0 or ( x - 2 ) = 0

∴ x = ± 2 or x = 2

For more info please see

Algebra

Algebra Question 2 of 10

2. Solve for x : ( x - 1 ) ( x + 8 ) = 10

A. x = 1 or x = -

\frac{5}{4}

B. x = 11 or x = 2

C. x = - 9 or x = 2

D. None of the above

Correct Answer
Option C

Detailed Solution

Solution Details

Given ( x - 1 ) ( x + 8 ) = 10

• ( x - 1 ) ( x + 8 ) = 10

• x² + 7x - 8 = 10

• x² + 7x - 18 = 0

• ( x + 9 ) ( x – 2 ) = 0

• ( x + 9 ) = 0 or ( x – 2 ) = 0

∴ x = – 9 or x = 2

For more info please see

Algebra

Algebra Question 3 of 10

3. Solve for x : 4x² + 11x + 4 = 0
(Leave your answer correct to TWO decimal places IF NECESSARY)

A. x = 0.43 or x = 2.32

B. x = 0 or x = 2

C. x = - 0.43 or x = - 2.32

D. None of the above

Correct Answer
Option C

Detailed Solution

Solution Details

Given 4x² + 11x + 4 = 0

• 4x² + 11x + 4 = 0

• x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$ The equation cannot be factorised so we use the quadratic formula

• x = $\frac{– 11 ± \sqrt[]{( 11 ) 2 – 4 ( 4 ) ( 4 )}}{2 ( 4 )}$ = $\frac{– 11 ± \sqrt[]{57}}{8}$

• x = $\frac{– 11 – \sqrt[]{57}}{8}$ or x = $\frac{– 11 + \sqrt[]{57}}{8}$

∴ x = – 0.43 or x = – 2.32

For more info please see

Algebra

Given 4x² + 11x + 4 = 0

• 4x² + 11x + 4 = 0

• x = $\frac{– b ± \sqrt[]{b 2 – 4ac}}{2a}$

The equation cannot be factorised so we use the quadratic formula

• x = $\frac{– 11 ± \sqrt[]{( 11 ) 2 – 4 ( 4 ) ( 4 )}}{2 ( 4 )}$

= $\frac{– 11 ± \sqrt[]{57}}{8}$

• x = $\frac{– 11 – \sqrt[]{57}}{8}$ or

x = $\frac{– 11 + \sqrt[]{57}}{8}$

∴ x = – 0.43 or x = – 2.32

For more info please see

Algebra

Algebra Question 4 of 10

4. Solve for x : - x² ≤ 4 - 5x

A. x ≤ 1 or x ≥ 4

B. 1 ≤ x ≤ 4

C. 1 ≥ x ≥ 4

D. None of the above

Correct Answer
Option A

Detailed Solution

Solution Details

Given - x² ≤ 4 - 5x

• - x² ≤ 4 - 5x

• x² - 5x + 4 ≥ 0

• ( x - 1 ) ( x – 4 ) ≥ 0

• x² - 5x + 4 = 0 at x = 1 and x = 4

+0-0+

||||

|14|

We have to find values of x where x² - 5x + 4 ≥ 0

∴ x ≤ 1 or x ≥ 4

For more info please see

Algebra

Algebra Question 5 of 10

5. Solve for x : 6x < 3x²

A. x < 0 or x > 2

B. 0 < x < 2

C. x < - 2 or x > 0

D. None of the above

Correct Answer
Option A

Detailed Solution

Solution Details

Given 6x < 3x²

• 6x < 3x²

• 3x² - 6x > 0

• 3x ( x – 2 ) > 0

• 3x² - 6x = 0 at x = 0 and x = 2

+0-0+

||||

|02|

We have to find values of x where 3x² - 6x > 0

∴ x < 0 or x > 2

For more info please see

Algebra

Algebra Question 6 of 10

6. Solve for x and y if y = x + 2 and x² + y² = 20

6. Solve for x and y if

y = x + 2 and x² + y² = 20

A. (x ; y) = ( - 2 ; 0 ) and (x ; y) = ( 4 ; 6 )

B. (x ; y) = ( 2 ; 4 ) and (x ; y) = ( 4 ; 6 )

C. (x ; y) = ( 2 ; 4 ) and (x ; y) = ( - 4 ; - 2 )

D. None of the above

Correct Answer
Option C

Detailed Solution

Solution Details

Given y = x + 2 and x² + y² = 20

• y = x + 2 ................................... 1

x² + y² = 20 ................................... 2

• Substitute 1 into 2

x² + ( x + 2 )² = 20

x² + x² + 4x + 4 = 20

2x² + 4x - 16 = 0

x² + 2x - 8 = 0

( x + 4 ) ( x - 2 ) = 0

( x + 4 ) = 0 or ( x - 2 ) = 0

∴ x = - 4 or x = 2

• Substitute the values of x into 1 to get the values of y

y = x + 2

For x = - 4

y = - 4 + 2 = - 2

For x = 2

y = 2 + 2 = 4

∴ ( x ; y ) = ( - 4 ; - 2 ) and ( x ; y ) = ( 2 ; 4 )

For more info please see

Algebra

Given y = x + 2 and x² + y² = 20

• y = x + 2 ................................... 1

x² + y² = 20 ................................... 2

• Substitute 1 into 2

x² + ( x + 2 )² = 20

x² + x² + 4x + 4 = 20

2x² + 4x - 16 = 0

x² + 2x - 8 = 0

( x + 4 ) ( x - 2 ) = 0

( x + 4 ) = 0 or ( x - 2 ) = 0

∴ x = - 4 or x = 2

• Substitute the values of x into 1 to get the values of y

y = x + 2

For x = - 4

y = - 4 + 2 = - 2

For x = 2

y = 2 + 2 = 4

∴ ( x ; y ) = ( - 4 ; - 2 ) and

( x ; y ) = ( 2 ; 4 )

For more info please see

Algebra

Algebra Question 7 of 10

7. Solve for x and y if 2x - y - 3 = 0 and x² + 5xy + y² = 15

7. Solve for x and y if

2x - y - 3 = 0 and x² + 5xy + y² = 15

A. (x ; y) = ( -

\frac{1}{5}

; -

\frac{17}{5}

) and (x ; y) = ( 2 ; 1 )

B. (x ; y) = (

\frac{1}{5}

;

\frac{17}{5}

) and (x ; y) = ( 2 ; 1 )

C. (x ; y) = ( - 5 ; -

\frac{5}{17}

) and (x ; y) = ( 2 ; 1 )

D. None of the above

Correct Answer
Option A

Detailed Solution

Solution Details

Given 2x - y - 3 = 0 and x² + 5xy + y² = 15

• 2x - y - 3 = 0 ................................... 1

x² + 5xy + y² = 15 ................................... 2

• From 1 we have

y = 2x - 3 ................................... 3

• Substitute 3 into 2

x² + 5x ( 2x - 3 ) + ( 2x - 3 )² = 15

x² + 10x² - 15x + 4x² - 12x + 9 = 15

15x² - 27x - 6 = 0

( 5x + 1 ) ( 3x - 6 ) = 0

( 5x + 1 ) = 0 or ( 3x - 6 ) = 0

∴ x = - $\frac{1}{5}$ or x = 2

• Substitute the values of x into 3 to get the values of y

y = 2x - 3

For x = - $\frac{1}{5}$

y = 2 ( - $\frac{1}{5}$ ) - 3 = - $\frac{17}{5}$

For x = 2

y = 2 ( 2 ) - 3 = 1

∴ ( x ; y ) = ( - $\frac{1}{5}$ ; - $\frac{17}{5}$ ) and ( x ; y ) = ( 2 ; 1 )

For more info please see

Algebra

Given 2x - y - 3 = 0 and x² + 5xy + y² = 15

• 2x - y - 3 = 0 ........................ 1

x² + 5xy + y² = 15 ........................ 2

• From 1 we have

y = 2x - 3 ........................ 3

• Substitute 3 into 2

x² + 5x ( 2x - 3 ) + ( 2x - 3 )² = 15

x² + 10x² - 15x + 4x² - 12x + 9 = 15

15x² - 27x - 6 = 0

( 5x + 1 ) ( 3x - 6 ) = 0

( 5x + 1 ) = 0 or ( 3x - 6 ) = 0

∴ x = - $\frac{1}{5}$ or x = 2

• Substitute the values of x into 3 to get the values of y

y = 2x - 3

For x = - $\frac{1}{5}$

y = 2 ( - $\frac{1}{5}$ ) - 3 = - $\frac{17}{5}$

For x = 2

y = 2 ( 2 ) - 3 = 1

∴ ( x ; y ) = ( - $\frac{1}{5}$ ; - $\frac{17}{5}$ ) and

( x ; y ) = ( 2 ; 1 )

For more info please see

Algebra

Algebra Question 8 of 10

8. Solve for x and y if $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if

$\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

A. (x ; y) = (

\frac{1}{3}

; -

\frac{1}{3}

) and (x ; y) = ( - 1 ; - 3 )

B. (x ; y) = ( -

\frac{1}{3}

;

\frac{1}{3}

) and (x ; y) = ( - 1 ; - 3 )

C. (x ; y) = ( 3 ; - 3 ) and (x ; y) = ( - 1 ; - 3 )

D. None of the above

Correct Answer
Option A

Detailed Solution

Solution Details

Given $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

• $\frac{2x}{1 + y}$ = 1 ................................... 1

( 3x - y ) ( x + y ) = 0 ................................... 2

• From 1 we have

y = 2x - 1 ................................... 3

• Substitute 3 into 2

( 3x - 2x + 1 ) ( x + 2x - 1 ) = 0

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

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

∴ x = - 1 or x = $\frac{1}{3}$

• Substitute the values of x into 3 to get the values of y

y = 2x - 1

For x = - 1

y = 2 ( - 1 ) - 1 = - 3

For x = $\frac{1}{3}$

y = 2 ( $\frac{1}{3}$ ) - 1 = - $\frac{1}{3}$

∴ ( x ; y ) = ( $\frac{1}{3}$ ; - $\frac{1}{3}$ ) and ( x ; y ) = ( - 1 ; - 3 )

For more info please see

Algebra

Given $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

• $\frac{2x}{1 + y}$ = 1 ........................ 1

( 3x - y ) ( x + y ) = 0 ........................ 2

• From 1 we have

y = 2x - 1 ........................ 3

• Substitute 3 into 2

( 3x - 2x + 1 ) ( x + 2x - 1 ) = 0

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

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

∴ x = - 1 or x = $\frac{1}{3}$

• Substitute the values of x into 3 to get the values of y

y = 2x - 1

For x = - 1

y = 2 ( - 1 ) - 1 = - 3

For x = $\frac{1}{3}$

y = 2 ( $\frac{1}{3}$ ) - 1 = - $\frac{1}{3}$

∴ ( x ; y ) = ( $\frac{1}{3}$ ; - $\frac{1}{3}$ ) and

( x ; y ) = ( - 1 ; - 3 )

For more info please see

Algebra

Algebra Question 9 of 10

9. Given: x = $\frac{± \sqrt[]{b 2 – 9}}{- 2}$

Determine the value(s) of b for which x is a Real Number.

A. b ≥ - 3 or b ≤ 3

B. - 3 ≤ b ≤ 3

C. b ≤ - 3 or b ≥ 3

D. None of the above

Correct Answer
Option C

Detailed Solution

Solution Details

Given x = $\frac{± \sqrt[]{b 2 – 9}}{- 2}$

• x = $\frac{± \sqrt[]{b 2 – 9}}{- 2}$

• b ² – 9 ≥ 0

• ( b - 3 ) ( b + 3 ) ≥ 0

+0-0+

||||

|-33|

∴ b ≤ - 3 or b ≥ 3

For more info please see

Algebra

Algebra Question 10 of 10

10. The roots of a quadratic equation are given by: x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

10. The roots of a quadratic equation are given by:

x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

For how many values of k > 0 will the roots be Rational ?

A. 2

B. 3

C. 4

D. None of the above

Correct Answer
Option C

Detailed Solution

Solution Details

Given x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

• x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

• The roots of a quadratic equation are Rational if Δ = 0, or if Δ is a Perfect Square

∴ 13 – 2k = 0

∴ k = $\frac{13}{2}$

• And there are three numbers that are less than 13 which are Perfect Squares. They are 1, 4 and 9.

∴ 13 – 2k = 1 or 13 – 2k = 4 or 13 – 2k = 9

∴ k = 6 or k = $\frac{9}{2}$ or k = $\frac{4}{2}$

∴ The roots are Rational if k ∈ ( $\frac{4}{2}$ ; $\frac{9}{2}$ ; $\frac{13}{2}$ ; 6 )

∴ There are 4 values k > 0 for which the roots are Rational

For more info please see

Algebra

Given x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

• x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

• The roots of a quadratic equation are Rational if Δ = 0, or if Δ is a Perfect Square

∴ 13 – 2k = 0

∴ k = $\frac{13}{2}$

• And there are three numbers that are less than 13 which are Perfect Squares. They are 1, 4 and 9.

∴ 13 – 2k = 1 or 13 – 2k = 4 or 13 – 2k = 9

∴ k = 6 or k = $\frac{9}{2}$ or k = $\frac{4}{2}$

∴ The roots are Rational if

k ∈ ( $\frac{4}{2}$ ; $\frac{9}{2}$ ; $\frac{13}{2}$ ; 6 )

∴ There are 4 values k > 0 for which the roots are Rational

For more info please see

Algebra

Quiz

Algebra

Typical Exam Questions

Quiz Quidelines

Quiz

1. Solve for x : ( x 2 - 4 ) ( x - 2 ) = 0

2. Solve for x : ( x - 1 ) ( x + 8 ) = 10

3. Solve for x : 4x 2 + 11x + 4 = 0(Leave your answer correct to TWO decimal places IF NECESSARY)

4. Solve for x : - x 2 ≤ 4 - 5x

5. Solve for x : 6x < 3x 2

6. Solve for x and y if y = x + 2 and x 2 + y 2 = 20

6. Solve for x and y if y = x + 2 and x 2 + y 2 = 20

6. Solve for x and y if y = x + 2 and x 2 + y 2 = 20

7. Solve for x and y if 2x - y - 3 = 0 and x 2 + 5xy + y 2 = 15

7. Solve for x and y if 2x - y - 3 = 0 and x 2 + 5xy + y 2 = 15

7. Solve for x and y if 2x - y - 3 = 0 and x 2 + 5xy + y 2 = 15

8. Solve for x and y if 2x 1 + y = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if 2x 1 + y = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if 2x 1 + y = 1 and ( 3x - y ) ( x + y ) = 0

9. Given: x = ± b 2 – 9 - 2

Determine the value(s) of b for which x is a Real Number.

10. The roots of a quadratic equation are given by: x = - 2 ± 13 – 2k 3

10. The roots of a quadratic equation are given by: x = - 2 ± 13 – 2k 3

10. The roots of a quadratic equation are given by: x = - 2 ± 13 – 2k 3

For how many values of k > 0 will the roots be Rational ?

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Quiz Results

Solutions

1. Solve for x : ( x 2 - 4 ) ( x - 2 ) = 0

Correct Answer Option B

Detailed Solution

Solution Details

Given ( x 2 - 4 ) ( x - 2 ) = 0

• ( x 2 - 4 ) ( x - 2 ) = 0 • ( x 2 - 4 ) = 0 or ( x - 2 ) = 0 ∴ x = ± 2 or x = 2

2. Solve for x : ( x - 1 ) ( x + 8 ) = 10

Correct Answer Option C

Detailed Solution

Solution Details

Given ( x - 1 ) ( x + 8 ) = 10

• ( x - 1 ) ( x + 8 ) = 10 • x 2 + 7x - 8 = 10 • x 2 + 7x - 18 = 0 • ( x + 9 ) ( x – 2 ) = 0 • ( x + 9 ) = 0 or ( x – 2 ) = 0 ∴ x = – 9 or x = 2

3. Solve for x : 4x 2 + 11x + 4 = 0(Leave your answer correct to TWO decimal places IF NECESSARY)

Correct Answer Option C

Detailed Solution

Solution Details

Given 4x 2 + 11x + 4 = 0

• 4x 2 + 11x + 4 = 0 • x = – b ± b 2 – 4ac 2a The equation cannot be factorised so we use the quadratic formula • x = – 11 ± ( 11 ) 2 – 4 ( 4 ) ( 4 ) 2 ( 4 ) = – 11 ± 57 8 • x = – 11 – 57 8 or x = – 11 + 57 8 ∴ x = – 0.43 or x = – 2.32

Given 4x 2 + 11x + 4 = 0

• 4x 2 + 11x + 4 = 0 • x = – b ± b 2 – 4ac 2a The equation cannot be factorised so we use the quadratic formula • x = – 11 ± ( 11 ) 2 – 4 ( 4 ) ( 4 ) 2 ( 4 ) = – 11 ± 57 8 • x = – 11 – 57 8 or x = – 11 + 57 8 ∴ x = – 0.43 or x = – 2.32

4. Solve for x : - x 2 ≤ 4 - 5x

Correct Answer Option A

Detailed Solution

Solution Details

Given - x 2 ≤ 4 - 5x

• - x 2 ≤ 4 - 5x • x 2 - 5x + 4 ≥ 0 • ( x - 1 ) ( x – 4 ) ≥ 0 • x 2 - 5x + 4 = 0 at x = 1 and x = 4 +0-0+ |||| |14| We have to find values of x where x 2 - 5x + 4 ≥ 0 ∴ x ≤ 1 or x ≥ 4

5. Solve for x : 6x < 3x 2

Correct Answer Option A

Detailed Solution

Solution Details

Given 6x < 3x 2

• 6x < 3x 2 • 3x 2 - 6x > 0 • 3x ( x – 2 ) > 0 • 3x 2 - 6x = 0 at x = 0 and x = 2 +0-0+ |||| |02| We have to find values of x where 3x 2 - 6x > 0 ∴ x < 0 or x > 2

6. Solve for x and y if y = x + 2 and x 2 + y 2 = 20

6. Solve for x and y if y = x + 2 and x 2 + y 2 = 20

6. Solve for x and y if y = x + 2 and x 2 + y 2 = 20

Correct Answer Option C

Detailed Solution

Solution Details

Given y = x + 2 and x 2 + y 2 = 20

• y = x + 2 ................................... 1 x 2 + y 2 = 20 ................................... 2

• Substitute 1 into 2 x 2 + ( x + 2 ) 2 = 20 x 2 + x 2 + 4x + 4 = 20 2x 2 + 4x - 16 = 0 x 2 + 2x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0 ( x + 4 ) = 0 or ( x - 2 ) = 0 ∴ x = - 4 or x = 2

• Substitute the values of x into 1 to get the values of y y = x + 2 For x = - 4 y = - 4 + 2 = - 2 For x = 2 y = 2 + 2 = 4

∴ ( x ; y ) = ( - 4 ; - 2 ) and ( x ; y ) = ( 2 ; 4 )

Given y = x + 2 and x 2 + y 2 = 20

• y = x + 2 ................................... 1 x 2 + y 2 = 20 ................................... 2

• Substitute 1 into 2 x 2 + ( x + 2 ) 2 = 20 x 2 + x 2 + 4x + 4 = 20 2x 2 + 4x - 16 = 0 x 2 + 2x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0 ( x + 4 ) = 0 or ( x - 2 ) = 0 ∴ x = - 4 or x = 2

• Substitute the values of x into 1 to get the values of y y = x + 2 For x = - 4 y = - 4 + 2 = - 2 For x = 2 y = 2 + 2 = 4

∴ ( x ; y ) = ( - 4 ; - 2 ) and ( x ; y ) = ( 2 ; 4 )

7. Solve for x and y if 2x - y - 3 = 0 and x 2 + 5xy + y 2 = 15

7. Solve for x and y if 2x - y - 3 = 0 and x 2 + 5xy + y 2 = 15

7. Solve for x and y if 2x - y - 3 = 0 and x 2 + 5xy + y 2 = 15

Correct Answer Option A

1. Solve for x : ( x² - 4 ) ( x - 2 ) = 0

3. Solve for x : 4x² + 11x + 4 = 0
(Leave your answer correct to TWO decimal places IF NECESSARY)

4. Solve for x : - x² ≤ 4 - 5x

5. Solve for x : 6x < 3x²

6. Solve for x and y if y = x + 2 and x² + y² = 20

6. Solve for x and y if y = x + 2 and x² + y² = 20

6. Solve for x and y if

y = x + 2 and x² + y² = 20

7. Solve for x and y if 2x - y - 3 = 0 and x² + 5xy + y² = 15

7. Solve for x and y if 2x - y - 3 = 0 and x² + 5xy + y² = 15

7. Solve for x and y if

2x - y - 3 = 0 and x² + 5xy + y² = 15

8. Solve for x and y if $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if

$\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

9. Given: x = $\frac{± \sqrt[]{b 2 – 9}}{- 2}$

10. The roots of a quadratic equation are given by: x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

10. The roots of a quadratic equation are given by: x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

10. The roots of a quadratic equation are given by:

x = $\frac{- 2 ± \sqrt[]{13 – 2k}}{3}$

1. Solve for x : ( x² - 4 ) ( x - 2 ) = 0

Correct Answer
Option B

Given ( x² - 4 ) ( x - 2 ) = 0

• ( x² - 4 ) ( x - 2 ) = 0

• ( x² - 4 ) = 0 or ( x - 2 ) = 0

∴ x = ± 2 or x = 2

Correct Answer
Option C

• ( x - 1 ) ( x + 8 ) = 10

• x² + 7x - 8 = 10

• x² + 7x - 18 = 0

• ( x + 9 ) ( x – 2 ) = 0

• ( x + 9 ) = 0 or ( x – 2 ) = 0

∴ x = – 9 or x = 2

3. Solve for x : 4x² + 11x + 4 = 0
(Leave your answer correct to TWO decimal places IF NECESSARY)

Correct Answer
Option C

Given 4x² + 11x + 4 = 0

Given 4x² + 11x + 4 = 0

4. Solve for x : - x² ≤ 4 - 5x

Correct Answer
Option A

Given - x² ≤ 4 - 5x

• - x² ≤ 4 - 5x

• x² - 5x + 4 ≥ 0

• ( x - 1 ) ( x – 4 ) ≥ 0

• x² - 5x + 4 = 0 at x = 1 and x = 4

+0-0+

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|14|

We have to find values of x where x² - 5x + 4 ≥ 0

∴ x ≤ 1 or x ≥ 4

5. Solve for x : 6x < 3x²

Correct Answer
Option A

Given 6x < 3x²

• 6x < 3x²

• 3x² - 6x > 0

• 3x ( x – 2 ) > 0

• 3x² - 6x = 0 at x = 0 and x = 2

+0-0+

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|02|

We have to find values of x where 3x² - 6x > 0

∴ x < 0 or x > 2

6. Solve for x and y if y = x + 2 and x² + y² = 20

6. Solve for x and y if y = x + 2 and x² + y² = 20

6. Solve for x and y if

y = x + 2 and x² + y² = 20

Correct Answer
Option C

Given y = x + 2 and x² + y² = 20

• y = x + 2 ................................... 1

x² + y² = 20 ................................... 2

• Substitute 1 into 2

x² + ( x + 2 )² = 20

x² + x² + 4x + 4 = 20

2x² + 4x - 16 = 0

x² + 2x - 8 = 0

( x + 4 ) ( x - 2 ) = 0

( x + 4 ) = 0 or ( x - 2 ) = 0

∴ x = - 4 or x = 2

• Substitute the values of x into 1 to get the values of y

y = x + 2

For x = - 4

y = - 4 + 2 = - 2

For x = 2

y = 2 + 2 = 4

Given y = x + 2 and x² + y² = 20

• y = x + 2 ................................... 1

x² + y² = 20 ................................... 2

• Substitute 1 into 2

x² + ( x + 2 )² = 20

x² + x² + 4x + 4 = 20

2x² + 4x - 16 = 0

x² + 2x - 8 = 0

( x + 4 ) ( x - 2 ) = 0

( x + 4 ) = 0 or ( x - 2 ) = 0

∴ x = - 4 or x = 2

• Substitute the values of x into 1 to get the values of y

y = x + 2

For x = - 4

y = - 4 + 2 = - 2

For x = 2

y = 2 + 2 = 4

∴ ( x ; y ) = ( - 4 ; - 2 ) and

( x ; y ) = ( 2 ; 4 )

7. Solve for x and y if 2x - y - 3 = 0 and x² + 5xy + y² = 15

7. Solve for x and y if 2x - y - 3 = 0 and x² + 5xy + y² = 15

7. Solve for x and y if

2x - y - 3 = 0 and x² + 5xy + y² = 15

Correct Answer
Option A

Given 2x - y - 3 = 0 and x² + 5xy + y² = 15

• 2x - y - 3 = 0 ................................... 1

x² + 5xy + y² = 15 ................................... 2

• From 1 we have

y = 2x - 3 ................................... 3

• Substitute 3 into 2

x² + 5x ( 2x - 3 ) + ( 2x - 3 )² = 15

x² + 10x² - 15x + 4x² - 12x + 9 = 15

15x² - 27x - 6 = 0

( 5x + 1 ) ( 3x - 6 ) = 0

( 5x + 1 ) = 0 or ( 3x - 6 ) = 0

∴ x = - $\frac{1}{5}$ or x = 2

• Substitute the values of x into 3 to get the values of y

y = 2x - 3

For x = - $\frac{1}{5}$

y = 2 ( - $\frac{1}{5}$ ) - 3 = - $\frac{17}{5}$

For x = 2

y = 2 ( 2 ) - 3 = 1

∴ ( x ; y ) = ( - $\frac{1}{5}$ ; - $\frac{17}{5}$ ) and ( x ; y ) = ( 2 ; 1 )

Given 2x - y - 3 = 0 and x² + 5xy + y² = 15

• 2x - y - 3 = 0 ........................ 1

x² + 5xy + y² = 15 ........................ 2

• From 1 we have

y = 2x - 3 ........................ 3

• Substitute 3 into 2

x² + 5x ( 2x - 3 ) + ( 2x - 3 )² = 15

x² + 10x² - 15x + 4x² - 12x + 9 = 15

15x² - 27x - 6 = 0

( 5x + 1 ) ( 3x - 6 ) = 0

( 5x + 1 ) = 0 or ( 3x - 6 ) = 0

∴ x = - $\frac{1}{5}$ or x = 2

• Substitute the values of x into 3 to get the values of y

y = 2x - 3

For x = - $\frac{1}{5}$

y = 2 ( - $\frac{1}{5}$ ) - 3 = - $\frac{17}{5}$

For x = 2

y = 2 ( 2 ) - 3 = 1

∴ ( x ; y ) = ( - $\frac{1}{5}$ ; - $\frac{17}{5}$ ) and

( x ; y ) = ( 2 ; 1 )

8. Solve for x and y if $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

8. Solve for x and y if

$\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

Correct Answer
Option A

Given $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

• $\frac{2x}{1 + y}$ = 1 ................................... 1

( 3x - y ) ( x + y ) = 0 ................................... 2

• From 1 we have

y = 2x - 1 ................................... 3

• Substitute 3 into 2

( 3x - 2x + 1 ) ( x + 2x - 1 ) = 0

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

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

∴ x = - 1 or x = $\frac{1}{3}$

• Substitute the values of x into 3 to get the values of y

y = 2x - 1

For x = - 1

y = 2 ( - 1 ) - 1 = - 3

For x = $\frac{1}{3}$

y = 2 ( $\frac{1}{3}$ ) - 1 = - $\frac{1}{3}$

∴ ( x ; y ) = ( $\frac{1}{3}$ ; - $\frac{1}{3}$ ) and ( x ; y ) = ( - 1 ; - 3 )

Given $\frac{2x}{1 + y}$ = 1 and ( 3x - y ) ( x + y ) = 0

• $\frac{2x}{1 + y}$ = 1 ........................ 1

( 3x - y ) ( x + y ) = 0 ........................ 2

• From 1 we have

y = 2x - 1 ........................ 3

• Substitute 3 into 2

( 3x - 2x + 1 ) ( x + 2x - 1 ) = 0

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

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

∴ x = - 1 or x = $\frac{1}{3}$

• Substitute the values of x into 3 to get the values of y

y = 2x - 1

For x = - 1

y = 2 ( - 1 ) - 1 = - 3

For x = $\frac{1}{3}$

y = 2 ( $\frac{1}{3}$ ) - 1 = - $\frac{1}{3}$

∴ ( x ; y ) = ( $\frac{1}{3}$ ; - $\frac{1}{3}$ ) and

( x ; y ) = ( - 1 ; - 3 )