Sec 3 G2 Maths Quadratic equations worksheet with answers

A free Sec 3 G2 Maths worksheet on quadratic equations: 6 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.

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

Need the formulas first? Read the Quadratic equations notes. Or practise it online with step-by-step help when you get one wrong.

Try a question: Quadratic equations

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 3 G2 Maths worksheet: 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

A group of

x

pupils shared the cost of a $

18

gift equally. If

3

more pupils had joined the group, each pupil would have paid $

1

less. Find the number of pupils in the group.

Answer[3]

(a)Solve

x^{2} - 8x + 1 = 0

, giving your answers correct to 2 decimal places.

Answer (a)[3]

(b)Solve

x^{2} + 4x - 1 = 0

, giving your answers correct to 2 decimal places.

Answer (b)[3]

Section B

The sum of the squares of two consecutive positive integers is

181

. The smaller integer is

n

(a)Show that

n^2 + n - 90 = 0

[2]

(b)Find the two integers.

Answer (b)[2]

A ball is thrown upwards. Its height,

h

metres, above the ground

t

seconds after it is thrown is given by

h = 20t - 5t^2

(a)Find the height of the ball after

2

seconds.

Answer (a)m[1]

(b)Find the values of

t

when the ball is

15

m above the ground.

Answer (b)[3]

(c)Explain why there are two answers to part (b).

[1]

A photograph

20

cm long and

12

cm wide is placed in a frame of width

x

cm all the way round. The total area of the photograph and the frame is

384

cm².

A 20 cm by 12 cm photograph inside a frame of width x cm

(a)Show that

x^2 + 16x - 36 = 0

[3]

(b)Find the width of the frame.

Answer (b)cm[2]

Ravi cycles

30

km from his home to a park at an average speed of

x

km/h. On the way back, he cycles

5

km/h faster and the journey takes

1

hour less.

(a)Write down an expression, in terms of

x

, for the time, in hours, that Ravi takes to cycle to the park.

Answer (a)h[1]

(b)Show that

x^2 + 5x - 150 = 0

[3]

(c)Solve

x^2 + 5x - 150 = 0

, and hence find the time, in hours, that Ravi takes for the journey back.

Answer (c)h[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer: Number of pupils

= 6

[3]

Now:

\dfrac{18}{x}

dollarsWith

3

\dfrac{18}{x + 3}

dollarsbigger share − smaller share = difference

\dfrac{18}{x} - \dfrac{18}{x + 3} = 1

18(x + 3) - 18x = x(x + 3)

x^{2} + 3x - 54 = 0

(x - 6)(x + 9) = 0

x - 6 = 0

x + 9 = 0

x = 6

x = -9

x = -9

is rejected because the number of people or things cannot be negative. So

x = 6

.Number of pupils

= 6

Question 2

(a)Answer:

x = 7.87

x = 0.13

(2 d.p.)[3]

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

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

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

x = \dfrac{8 \pm \sqrt{60}}{2}

x = 7.8729\ldots

x = 0.1270\ldots

x = 7.87

x = 0.13

(2 d.p.)

(b)Answer:

x = 0.24

x = -4.24

(2 d.p.)[3]

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

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

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

x = \dfrac{-4 \pm \sqrt{20}}{2}

x = 0.2360\ldots

x = -4.2360\ldots

x = 0.24

x = -4.24

(2 d.p.)

Question 3

(a)[2]

n^2 + (n + 1)^2 = 181

2n^2 + 2n + 1 = 181

2n^2 + 2n - 180 = 0

n^2 + n - 90 = 0

(shown)

(b)Answer:

9

and

10

[2]

(n + 10)(n - 9) = 0

n = 9

(

n = -10

is rejected: the integers are positive)The integers are

9

and

10

Question 4

(a)Answer:

20

m[1]

h = 20(2) - 5(2)^2

\phantom{h} = 40 - 20

\phantom{h} = 20

(b)Answer:

t = 1

t = 3

[3]

20t - 5t^2 = 15

5t^2 - 20t + 15 = 0

t^2 - 4t + 3 = 0

(t - 1)(t - 3) = 0

t = 1

t = 3

(c)[1]

The ball is at

15

m on the way up, after

1

s, and again on the way down, after

3

Question 5

(a)[3]

The outside measures

(20 + 2x)

cm by

(12 + 2x)

cm.

(20 + 2x)(12 + 2x) = 384

240 + 40x + 24x + 4x^2 = 384

4x^2 + 64x - 144 = 0

x^2 + 16x - 36 = 0

(shown)

(b)Answer:

2

cm[2]

(x + 18)(x - 2) = 0

x = 2

(

x = -18

is rejected: a width can't be negative)The frame is

2

cm wide.

Question 6

(a)Answer:

\dfrac{30}{x}

hours[1]

Time

= \dfrac{\text{distance}}{\text{speed}}

= \dfrac{30}{x}

hours

(b)[3]

Time back

= \dfrac{30}{x + 5}

hours

\dfrac{30}{x} - \dfrac{30}{x + 5} = 1

30(x + 5) - 30x = x(x + 5)

150 = x^2 + 5x

x^2 + 5x - 150 = 0

(shown)

(c)Answer:

2

hours[3]

(x + 15)(x - 10) = 0

x = 10

(

x = -15

is rejected: a speed can't be negative)Time back

= \dfrac{30}{10 + 5}

= 2

hours

Sec 3 G2 Maths Quadratic equations worksheet with answers

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

Need the formulas first? Read the Quadratic equations notes. Or practise it online with step-by-step help when you get one wrong.

Answers and worked solutions

Question 1

Answer: Number of pupils

= 6

[3]

Now:

\dfrac{18}{x}

dollarsWith

3

\dfrac{18}{x + 3}

dollarsbigger share − smaller share = difference

\dfrac{18}{x} - \dfrac{18}{x + 3} = 1

18(x + 3) - 18x = x(x + 3)

x^{2} + 3x - 54 = 0

(x - 6)(x + 9) = 0

x - 6 = 0

x + 9 = 0

x = 6

x = -9

x = -9

is rejected because the number of people or things cannot be negative. So

x = 6

.Number of pupils

= 6

Question 2

(a)Answer:

x = 7.87

x = 0.13

(2 d.p.)[3]

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

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

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

x = \dfrac{8 \pm \sqrt{60}}{2}

x = 7.8729\ldots

x = 0.1270\ldots

x = 7.87

x = 0.13

(2 d.p.)

(b)Answer:

x = 0.24

x = -4.24

(2 d.p.)[3]

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

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

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

x = \dfrac{-4 \pm \sqrt{20}}{2}

x = 0.2360\ldots

x = -4.2360\ldots

x = 0.24

x = -4.24

(2 d.p.)

Question 3

(a)[2]

n^2 + (n + 1)^2 = 181

2n^2 + 2n + 1 = 181

2n^2 + 2n - 180 = 0

n^2 + n - 90 = 0

(shown)

(b)Answer:

9

and

10

[2]

(n + 10)(n - 9) = 0

n = 9

(

n = -10

is rejected: the integers are positive)The integers are

9

and

10

Question 4

(a)Answer:

20

m[1]

h = 20(2) - 5(2)^2

\phantom{h} = 40 - 20

\phantom{h} = 20

(b)Answer:

t = 1

t = 3

[3]

20t - 5t^2 = 15

5t^2 - 20t + 15 = 0

t^2 - 4t + 3 = 0

(t - 1)(t - 3) = 0

t = 1

t = 3

(c)[1]

The ball is at

15

m on the way up, after

1

s, and again on the way down, after

3

Question 5

(a)[3]

The outside measures

(20 + 2x)

cm by

(12 + 2x)

cm.

(20 + 2x)(12 + 2x) = 384

240 + 40x + 24x + 4x^2 = 384

4x^2 + 64x - 144 = 0

x^2 + 16x - 36 = 0

(shown)

(b)Answer:

2

cm[2]

(x + 18)(x - 2) = 0

x = 2

(

x = -18

is rejected: a width can't be negative)The frame is

2

cm wide.

Question 6

(a)Answer:

\dfrac{30}{x}

hours[1]

Time

= \dfrac{\text{distance}}{\text{speed}}

= \dfrac{30}{x}

hours

(b)[3]

Time back

= \dfrac{30}{x + 5}

hours

\dfrac{30}{x} - \dfrac{30}{x + 5} = 1

30(x + 5) - 30x = x(x + 5)

150 = x^2 + 5x

x^2 + 5x - 150 = 0

(shown)

(c)Answer:

2

hours[3]

(x + 15)(x - 10) = 0

x = 10

(

x = -15

is rejected: a speed can't be negative)Time back

= \dfrac{30}{10 + 5}

= 2

hours