Sec 2 Maths Quadratic equations and graphs worksheet with answers

A free Sec 2 Maths worksheet on quadratic equations and graphs: 5 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 2 Maths worksheet: Quadratic equations and graphs

Total: 30 marksSuggested time: 45 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)Solve

5x^{2} +2x = 7

Answer (a)[2]

(b)Solve

\dfrac{1}{x - 2} + \dfrac{6}{x + 3} = 2

Answer (b)[3]

(a)Ali cycles

221

km from town A to town B at a steady speed of

x

km/h. On the way back, Ali cycles

8

km/h faster and takes 2 hours less. Find the time, in hours, taken for the journey back.

Answer (a)[3]

(b)A car travels

306

km at a steady speed of

x

km/h. If its speed had been

2

km/h less, the journey would have taken 30 minutes longer. Find the time, in hours, taken for the journey.

Answer (b)[3]

Section B

(a)The diagram shows the graph of

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

. Use the graph to find the values of

x

when

y = -7

Answer (a)[1]

(b)The sides of a right-angled triangle are

(x - 16)

cm,

(x - 8)

cm and

x

cm. Find the area of the triangle.

Answer (b)[3]

(a)The diagram shows the graph of

y = -x^{2} + 3x

. Use the graph to find the value of

y

when

x = 0

Answer (a)[1]

(b)(i)Complete the table of values for

y = x^{2} - x + 2

Answer (b)(i)[3]

(b)(ii)Complete the table of values for

y = -x^{2} - 3x + 1

Answer (b)(ii)[3]

(a)(i)Find the equation of the line of symmetry of the curve

y = -(x - 1)(x + 5)

Answer (a)(i)[2]

(a)(ii)Find the coordinates of the turning point of the curve

y = -(x + 4)^2 - 6

Answer (a)(ii)[2]

(b)(i)Solve

2x^{2} - 11x + 5 = 0

Answer (b)(i)[2]

(b)(ii)Solve

x^{2} = 3x

Answer (b)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

x = -\dfrac{7}{5}

x = 1

[2]

5x^{2} + 2x - 7 = 0

(5x + 7)(x - 1) = 0

5x + 7 = 0

x - 1 = 0

x = -\dfrac{7}{5}

x = 1

(b)Answer:

x = -\dfrac{1}{2}

x = 3

[3]

1(x + 3) + 6(x - 2) = 2(x - 2)(x + 3)

7x - 9 = 2x^{2} + 2x - 12

2x^{2} - 5x - 3 = 0

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

2x + 1 = 0

x - 3 = 0

x = -\dfrac{1}{2}

x = 3

Neither solution makes a denominator zero, so both are valid.

Question 2

(a)Answer:

= 6.5

hours[3]

x

km/h:

\dfrac{221}{x}

hoursAt

(x + 8)

km/h:

\dfrac{221}{x + 8}

hoursslower time − faster time = time difference

\dfrac{221}{x} - \dfrac{221}{x + 8} = 2

221(x + 8) - 221x = 2x(x + 8)

2x^{2} + 16x - 1768 = 0

x^{2} + 8x - 884 = 0

(x - 26)(x + 34) = 0

x - 26 = 0

x + 34 = 0

x = 26

x = -34

x = -34

is rejected because a speed cannot be negative. So

x = 26

.Time

= \dfrac{221}{26 + 8}

= 6.5

hours

(b)Answer:

= 8.5

hours[3]

x

km/h:

\dfrac{306}{x}

hoursAt

(x - 2)

km/h:

\dfrac{306}{x - 2}

hoursslower time − faster time = time difference

\dfrac{306}{x - 2} - \dfrac{306}{x} = \dfrac{1}{2}

612x - 612(x - 2) = x(x - 2)

x^{2} - 2x - 1224 = 0

(x - 36)(x + 34) = 0

x - 36 = 0

x + 34 = 0

x = 36

x = -34

x = -34

is rejected because a speed cannot be negative. So

x = 36

.Time

= \dfrac{306}{36}

= 8.5

hours

Question 3

(a)Answer:

x = -5

x = 1

[1]

The line

y = -7

meets the curve at two points.

x = -5

x = 1

(b)Answer:

= 384

cm²[3]

The hypotenuse is the longest side,

x

.hypotenuse² = sum of the squares of the other two sides

(x - 16)^2 + (x - 8)^2 = x^2

x^{2} - 32x + 256 + x^{2} - 16x + 64 = x^2

x^{2} - 48x + 320 = 0

(x - 40)(x - 8) = 0

x - 40 = 0

x - 8 = 0

x = 40

x = 8

x = 8

is rejected because it makes a side negative. So

x = 40

.Area

= \dfrac{1}{2} \times (40 - 16) \times (40 - 8)

= 384

cm²

Question 4

(a)Answer:

y = 0

[1]

From

x = 0

, go up to the curve, then across to the

y

-axis.

y = 0

(b)(i)Answer: The

y

-values are the same for

x = -2

and

x = 3

, and so on: they are symmetrical about

x = 0.5

.[3]

When

x = -4

y = (-4)^2 - (-4) + 2

= 16 + 4 + 2

= 22

When

x = -1

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

= 1 + 1 + 2

= 4

When

x = 0

y = (0)^2 - (0) + 2

= 0 + 2

= 2

When

x = 3

y = (3)^2 - (3) + 2

= 9 - 3 + 2

= 8

The

y

-values are the same for

x = -2

and

x = 3

, and so on: they are symmetrical about

x = 0.5

(b)(ii)Answer: The

y

-values are the same for

x = -3

and

x = 0

, and so on: they are symmetrical about

x = -1.5

.[3]

When

x = -3

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

= -9 + 9 + 1

= 1

When

x = 0

y = -(0)^2 - 3(0) + 1

= 0 + 1

= 1

When

x = 1

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

= -1 - 3 + 1

= -3

When

x = 2

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

= -4 - 6 + 1

= -9

The

y

-values are the same for

x = -3

and

x = 0

, and so on: they are symmetrical about

x = -1.5

Question 5

(a)(i)Answer:

x = -2

[2]

On the

x

-axis,

y = 0

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

x = 1

x = -5

x = \dfrac{1 + (-5)}{2}

x = -2

(a)(ii)Answer: Turning point:

(-4, -6)

[2]

The square has a minus in front, so it is a maximum point.A square is never negative:

(x + 4)^2 = 0

when

x = -4

, and with the minus in front y is greatest there.Then

y = -(0)^2 - 6

y = -6

Turning point:

(-4, -6)

(b)(i)Answer:

x = \dfrac{1}{2}

x = 5

[2]

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

2x - 1 = 0

x - 5 = 0

x = \dfrac{1}{2}

x = 5

(b)(ii)Answer:

x = 0

x = 3

[2]

x^{2} - 3x = 0

Dividing both sides by

x

would lose a solution, so don't.

x(x - 3) = 0

x = 0

x - 3 = 0

x = 0

x = 3

Practice questions by skill

Sec 2 Maths Quadratic equations and graphs worksheet with answers

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

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

Section B

(a)The diagram shows the graph of

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

. Use the graph to find the values of

x

when

y = -7

Answer (a)[1]

(b)The sides of a right-angled triangle are

(x - 16)

cm,

(x - 8)

cm and

x

cm. Find the area of the triangle.

Answer (b)[3]

Answers and worked solutions

Question 1

(a)Answer:

x = -\dfrac{7}{5}

x = 1

[2]

5x^{2} + 2x - 7 = 0

(5x + 7)(x - 1) = 0

5x + 7 = 0

x - 1 = 0

x = -\dfrac{7}{5}

x = 1

(b)Answer:

x = -\dfrac{1}{2}

x = 3

[3]

1(x + 3) + 6(x - 2) = 2(x - 2)(x + 3)

7x - 9 = 2x^{2} + 2x - 12

2x^{2} - 5x - 3 = 0

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

2x + 1 = 0

x - 3 = 0

x = -\dfrac{1}{2}

x = 3

Neither solution makes a denominator zero, so both are valid.

Question 2

(a)Answer:

= 6.5

hours[3]

x

km/h:

\dfrac{221}{x}

hoursAt

(x + 8)

km/h:

\dfrac{221}{x + 8}

hoursslower time − faster time = time difference

\dfrac{221}{x} - \dfrac{221}{x + 8} = 2

221(x + 8) - 221x = 2x(x + 8)

2x^{2} + 16x - 1768 = 0

x^{2} + 8x - 884 = 0

(x - 26)(x + 34) = 0

x - 26 = 0

x + 34 = 0

x = 26

x = -34

x = -34

is rejected because a speed cannot be negative. So

x = 26

.Time

= \dfrac{221}{26 + 8}

= 6.5

hours

(b)Answer:

= 8.5

hours[3]

x

km/h:

\dfrac{306}{x}

hoursAt

(x - 2)

km/h:

\dfrac{306}{x - 2}

hoursslower time − faster time = time difference

\dfrac{306}{x - 2} - \dfrac{306}{x} = \dfrac{1}{2}

612x - 612(x - 2) = x(x - 2)

x^{2} - 2x - 1224 = 0

(x - 36)(x + 34) = 0

x - 36 = 0

x + 34 = 0

x = 36

x = -34

x = -34

is rejected because a speed cannot be negative. So

x = 36

.Time

= \dfrac{306}{36}

= 8.5

hours

Question 3

(a)Answer:

x = -5

x = 1

[1]

The line

y = -7

meets the curve at two points.

x = -5

x = 1

(b)Answer:

= 384

cm²[3]

The hypotenuse is the longest side,

x

.hypotenuse² = sum of the squares of the other two sides

(x - 16)^2 + (x - 8)^2 = x^2

x^{2} - 32x + 256 + x^{2} - 16x + 64 = x^2

x^{2} - 48x + 320 = 0

(x - 40)(x - 8) = 0

x - 40 = 0

x - 8 = 0

x = 40

x = 8

x = 8

is rejected because it makes a side negative. So

x = 40

.Area

= \dfrac{1}{2} \times (40 - 16) \times (40 - 8)

= 384

cm²

Question 4

(a)Answer:

y = 0

[1]

From

x = 0

, go up to the curve, then across to the

y

-axis.

y = 0

(b)(i)Answer: The

y

-values are the same for

x = -2

and

x = 3

, and so on: they are symmetrical about

x = 0.5

.[3]

When

x = -4

y = (-4)^2 - (-4) + 2

= 16 + 4 + 2

= 22

When

x = -1

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

= 1 + 1 + 2

= 4

When

x = 0

y = (0)^2 - (0) + 2

= 0 + 2

= 2

When

x = 3

y = (3)^2 - (3) + 2

= 9 - 3 + 2

= 8

The

y

-values are the same for

x = -2

and

x = 3

, and so on: they are symmetrical about

x = 0.5

(b)(ii)Answer: The

y

-values are the same for

x = -3

and

x = 0

, and so on: they are symmetrical about

x = -1.5

.[3]

When

x = -3

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

= -9 + 9 + 1

= 1

When

x = 0

y = -(0)^2 - 3(0) + 1

= 0 + 1

= 1

When

x = 1

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

= -1 - 3 + 1

= -3

When

x = 2

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

= -4 - 6 + 1

= -9

The

y

-values are the same for

x = -3

and

x = 0

, and so on: they are symmetrical about

x = -1.5

Question 5

(a)(i)Answer:

x = -2

[2]

On the

x

-axis,

y = 0

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

x = 1

x = -5

x = \dfrac{1 + (-5)}{2}

x = -2

(a)(ii)Answer: Turning point:

(-4, -6)

[2]

The square has a minus in front, so it is a maximum point.A square is never negative:

(x + 4)^2 = 0

when

x = -4

, and with the minus in front y is greatest there.Then

y = -(0)^2 - 6

y = -6

Turning point:

(-4, -6)

(b)(i)Answer:

x = \dfrac{1}{2}

x = 5

[2]

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

2x - 1 = 0

x - 5 = 0

x = \dfrac{1}{2}

x = 5

(b)(ii)Answer:

x = 0

x = 3

[2]

x^{2} - 3x = 0

Dividing both sides by

x

would lose a solution, so don't.

x(x - 3) = 0

x = 0

x - 3 = 0

x = 0

x = 3