Sec 4 G1 Maths Solving quadratic equations worksheet with answers

A free Sec 4 G1 Maths worksheet on solving quadratic equations: 7 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 4 G1 Maths worksheet: Solving quadratic equations

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The diagram shows the graph of

y = \dfrac{4}{x}

and the line

y = 2x + 2

. Use the graph to solve the equation

\dfrac{4}{x} = 2x + 2

Answer[2]

The diagram shows the graph of

y = -x^{2} + 4

and the line

y = 0

. Use the graph to solve the equation

-x^{2} + 4 = 0

Answer[2]

Solve

x^{2} + 4x + 2 = 0

, giving your answers correct to 2 decimal places.

Answer[3]

The sum of a positive number and its square is

35

. Let the number be

x

. Find the number, correct to 2 decimal places.

Answer[3]

(a)A positive number is multiplied by the number that is

3

more than it. The result is

64

. Let the number be

x

. Find the larger of the two numbers, correct to 2 decimal places.

Answer (a)[3]

(b)A positive number is multiplied by the number that is

7

more than it. The result is

34

. Let the number be

x

. Find the number, correct to 2 decimal places.

Answer (b)[3]

Section B

(a)Solve

1 -x -5x^{2} = 0

, giving your answers correct to 2 decimal places.

Answer (a)[3]

(b)The length of a rectangle is

8

cm more than its width. Its area is

15

cm². Let the width be

x

cm. Find the length of the rectangle, correct to 2 decimal places.

Answer (b)[3]

(a)The graph of

y = -x^{2} + 4

has been drawn. Find the equation of the straight line that must be drawn on the same axes to solve the equation

-x^{2} - x + 2 = 0

Answer (a)[2]

(b)(i)Solve

2x^{2} -x = 8

, giving your answers correct to 2 decimal places.

Answer (b)(i)[3]

(b)(ii)Solve

-3 -7x -3x^{2} = 0

, giving your answers correct to 2 decimal places.

Answer (b)(ii)[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

x = -2

x = 1

[2]

Left-hand side:

y = \dfrac{4}{x}

, the curveRight-hand side:

y = 2x + 2

, the lineWhere the curve and the line meet, both sides have the same

y

-value for the same

x

, so the equation is true there.Number of points where they meet:

2

x = -2

x = 1

Question 2

Answer:

x = -2

x = 2

[2]

Left-hand side:

y = -x^{2} + 4

, the curveRight-hand side:

y = 0

, the horizontal lineWhere the curve and the line meet, both sides have the same

y

-value for the same

x

, so the equation is true there.Number of points where they meet:

2

x = -2

x = 2

Question 3

Answer:

x = -0.59

x = -3.41

(2 d.p.)[3]

a = 1,\ b = 4,\ c = 2

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-4 \pm \sqrt{(4)^2 - 4(1)(2)}}{2(1)}

x = \dfrac{-4 \pm \sqrt{8}}{2}

x = -0.5857\ldots

x = -3.4142\ldots

x = -0.59

x = -3.41

(2 d.p.)

Question 4

Answer: The number

= 5.44

(2 d.p.)[3]

x + x^2 = 35

x^{2} + x - 35 = 0

a = 1,\ b = 1,\ c = -35

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-35)}}{2(1)}

x = \dfrac{-1 \pm \sqrt{141}}{2}

x = 5.4371\ldots

x = -6.4371\ldots

x = 5.44

x = -6.44

(2 d.p.)

x = -6.44

is rejected because the number is positive. So

x = 5.44

.The number

= 5.44

(2 d.p.)

Question 5

(a)Answer:

= 9.64

(2 d.p.)[3]

x(x + 3) = 64

x^{2} + 3x - 64 = 0

a = 1,\ b = 3,\ c = -64

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-3 \pm \sqrt{(3)^2 - 4(1)(-64)}}{2(1)}

x = \dfrac{-3 \pm \sqrt{265}}{2}

x = 6.6394\ldots

x = -9.6394\ldots

x = 6.64

x = -9.64

(2 d.p.)

x = -9.64

is rejected because the number is positive. So

x = 6.64

.The larger number

= 6.6394\ldots + 3

= 9.64

(2 d.p.)

(b)Answer: The number

= 3.30

(2 d.p.)[3]

x(x + 7) = 34

x^{2} + 7x - 34 = 0

a = 1,\ b = 7,\ c = -34

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(1)(-34)}}{2(1)}

x = \dfrac{-7 \pm \sqrt{185}}{2}

x = 3.3007\ldots

x = -10.3007\ldots

x = 3.30

x = -10.30

(2 d.p.)

x = -10.30

is rejected because the number is positive. So

x = 3.30

.The number

= 3.30

(2 d.p.)

Question 6

(a)Answer:

x = 0.36

x = -0.56

(2 d.p.)[3]

5x^{2} + x - 1 = 0

a = 5,\ b = 1,\ c = -1

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-1 \pm \sqrt{(1)^2 - 4(5)(-1)}}{2(5)}

x = \dfrac{-1 \pm \sqrt{21}}{10}

x = 0.3582\ldots

x = -0.5582\ldots

x = 0.36

x = -0.56

(2 d.p.)

(b)Answer:

= 9.57

cm (2 d.p.)[3]

length × width = area

x(x + 8) = 15

x^{2} + 8x - 15 = 0

a = 1,\ b = 8,\ c = -15

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-8 \pm \sqrt{(8)^2 - 4(1)(-15)}}{2(1)}

x = \dfrac{-8 \pm \sqrt{124}}{2}

x = 1.5677\ldots

x = -9.5677\ldots

x = 1.57

x = -9.57

(2 d.p.)

x = -9.57

is rejected because a length cannot be negative. So

x = 1.57

.Length

= 1.5677\ldots + 8

= 9.57

cm (2 d.p.)

Question 7

(a)Answer: Gradient

1

y

-intercept

2

.[2]

-x^{2} - x + 2 = 0

-x^{2} + 4 = x + 2

The solutions are where the LHS graph

y = -x^{2} + 4

meets the RHS graph, so draw

y = x + 2

.Then read the

x

-coordinates where the curve and the line meet.Gradient

1

y

-intercept

2

(b)(i)Answer:

x = 2.27

x = -1.77

(2 d.p.)[3]

2x^{2} - x - 8 = 0

a = 2,\ b = -1,\ c = -8

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-8)}}{2(2)}

x = \dfrac{1 \pm \sqrt{65}}{4}

x = 2.2655\ldots

x = -1.7655\ldots

x = 2.27

x = -1.77

(2 d.p.)

(b)(ii)Answer:

x = -0.57

x = -1.77

(2 d.p.)[3]

3x^{2} + 7x + 3 = 0

a = 3,\ b = 7,\ c = 3

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)}

x = \dfrac{-7 \pm \sqrt{13}}{6}

x = -0.5657\ldots

x = -1.7675\ldots

x = -0.57

x = -1.77

(2 d.p.)

Sec 4 G1 Maths Solving quadratic equations worksheet with answers

In the print window, choose Save as PDF to download it.

Or practise it online with step-by-step help when you get one wrong.

Section B

(a)Solve

1 -x -5x^{2} = 0

, giving your answers correct to 2 decimal places.

Answer (a)[3]

(b)The length of a rectangle is

8

cm more than its width. Its area is

15

cm². Let the width be

x

cm. Find the length of the rectangle, correct to 2 decimal places.

Answer (b)[3]

Answers and worked solutions

Question 1

Answer:

x = -2

x = 1

[2]

Left-hand side:

y = \dfrac{4}{x}

, the curveRight-hand side:

y = 2x + 2

, the lineWhere the curve and the line meet, both sides have the same

y

-value for the same

x

, so the equation is true there.Number of points where they meet:

2

x = -2

x = 1

Question 2

Answer:

x = -2

x = 2

[2]

Left-hand side:

y = -x^{2} + 4

, the curveRight-hand side:

y = 0

, the horizontal lineWhere the curve and the line meet, both sides have the same

y

-value for the same

x

, so the equation is true there.Number of points where they meet:

2

x = -2

x = 2

Question 3

Answer:

x = -0.59

x = -3.41

(2 d.p.)[3]

a = 1,\ b = 4,\ c = 2

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-4 \pm \sqrt{(4)^2 - 4(1)(2)}}{2(1)}

x = \dfrac{-4 \pm \sqrt{8}}{2}

x = -0.5857\ldots

x = -3.4142\ldots

x = -0.59

x = -3.41

(2 d.p.)

Question 4

Answer: The number

= 5.44

(2 d.p.)[3]

x + x^2 = 35

x^{2} + x - 35 = 0

a = 1,\ b = 1,\ c = -35

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-1 \pm \sqrt{(1)^2 - 4(1)(-35)}}{2(1)}

x = \dfrac{-1 \pm \sqrt{141}}{2}

x = 5.4371\ldots

x = -6.4371\ldots

x = 5.44

x = -6.44

(2 d.p.)

x = -6.44

is rejected because the number is positive. So

x = 5.44

.The number

= 5.44

(2 d.p.)

Question 5

(a)Answer:

= 9.64

(2 d.p.)[3]

x(x + 3) = 64

x^{2} + 3x - 64 = 0

a = 1,\ b = 3,\ c = -64

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-3 \pm \sqrt{(3)^2 - 4(1)(-64)}}{2(1)}

x = \dfrac{-3 \pm \sqrt{265}}{2}

x = 6.6394\ldots

x = -9.6394\ldots

x = 6.64

x = -9.64

(2 d.p.)

x = -9.64

is rejected because the number is positive. So

x = 6.64

.The larger number

= 6.6394\ldots + 3

= 9.64

(2 d.p.)

(b)Answer: The number

= 3.30

(2 d.p.)[3]

x(x + 7) = 34

x^{2} + 7x - 34 = 0

a = 1,\ b = 7,\ c = -34

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(1)(-34)}}{2(1)}

x = \dfrac{-7 \pm \sqrt{185}}{2}

x = 3.3007\ldots

x = -10.3007\ldots

x = 3.30

x = -10.30

(2 d.p.)

x = -10.30

is rejected because the number is positive. So

x = 3.30

.The number

= 3.30

(2 d.p.)

Question 6

(a)Answer:

x = 0.36

x = -0.56

(2 d.p.)[3]

5x^{2} + x - 1 = 0

a = 5,\ b = 1,\ c = -1

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-1 \pm \sqrt{(1)^2 - 4(5)(-1)}}{2(5)}

x = \dfrac{-1 \pm \sqrt{21}}{10}

x = 0.3582\ldots

x = -0.5582\ldots

x = 0.36

x = -0.56

(2 d.p.)

(b)Answer:

= 9.57

cm (2 d.p.)[3]

length × width = area

x(x + 8) = 15

x^{2} + 8x - 15 = 0

a = 1,\ b = 8,\ c = -15

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-8 \pm \sqrt{(8)^2 - 4(1)(-15)}}{2(1)}

x = \dfrac{-8 \pm \sqrt{124}}{2}

x = 1.5677\ldots

x = -9.5677\ldots

x = 1.57

x = -9.57

(2 d.p.)

x = -9.57

is rejected because a length cannot be negative. So

x = 1.57

.Length

= 1.5677\ldots + 8

= 9.57

cm (2 d.p.)

Question 7

(a)Answer: Gradient

1

y

-intercept

2

.[2]

-x^{2} - x + 2 = 0

-x^{2} + 4 = x + 2

The solutions are where the LHS graph

y = -x^{2} + 4

meets the RHS graph, so draw

y = x + 2

.Then read the

x

-coordinates where the curve and the line meet.Gradient

1

y

-intercept

2

(b)(i)Answer:

x = 2.27

x = -1.77

(2 d.p.)[3]

2x^{2} - x - 8 = 0

a = 2,\ b = -1,\ c = -8

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-8)}}{2(2)}

x = \dfrac{1 \pm \sqrt{65}}{4}

x = 2.2655\ldots

x = -1.7655\ldots

x = 2.27

x = -1.77

(2 d.p.)

(b)(ii)Answer:

x = -0.57

x = -1.77

(2 d.p.)[3]

3x^{2} + 7x + 3 = 0

a = 3,\ b = 7,\ c = 3

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)}

x = \dfrac{-7 \pm \sqrt{13}}{6}

x = -0.5657\ldots

x = -1.7675\ldots

x = -0.57

x = -1.77

(2 d.p.)