Quadratic Formula Worksheet with Answers

Revision Library

revisionlibrary.co.uk

HIGHER

7.4 Quadratic Formula

EDEXCEL 1MA1 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

1

Quadratic equations can be written in the form ax² + bx + c = 0.

(a)Write down the values of a, b and c in the equation 4x² - 7x + 2 = 0.(1)

(b)Rearrange the equation 5x² = 3x + 8 into the form ax² + bx + c = 0, and state the values of a, b and c.(2)

(Total for Question 1 is 3 marks)

2

The equation 2x² - 3x - 4 = 0 is to be solved using the quadratic formula.
Which of the following is the correct value of the discriminant, b² - 4ac?

A) 41
B) -23
C) 9
D) 25

(Total for Question 2 is 1 mark)

3

Solve x² + 3x - 5 = 0. Give your solutions correct to 2 decimal places.

(Total for Question 3 is 3 marks)

4

Solve x² - 4x - 1 = 0. Give your solutions in surd form, simplified as far as possible.

(Total for Question 4 is 2 marks)

5

Solve 2x² + 5x - 1 = 0. Give your solutions correct to 2 decimal places.

(Total for Question 5 is 3 marks)

6

Solve (x + 2)(x - 1) = 9. Give your solutions correct to 2 decimal places.

(Total for Question 6 is 4 marks)

7

Solve (2x - 1)(x + 3) = 6. Give your solutions correct to 3 significant figures.

(Total for Question 7 is 4 marks)

8

Solve 5x² - 8x - 1 = 0. Give your solutions correct to 1 decimal place.

(Total for Question 8 is 3 marks)

9

For each equation, choose an efficient method (factorising or the quadratic formula) and solve it, showing your working clearly.

(a)x² - 5x + 6 = 0(2)

(b)x² - 5x + 2 = 0. Give your solutions correct to 2 decimal places.(3)

(Total for Question 9 is 5 marks)

10

Consider the equation 3x² - 2x + 5 = 0.

(a)Calculate the value of the discriminant, b² - 4ac.(1)

(b)Hence state, giving a reason, how many real solutions the equation has.(1)

(Total for Question 10 is 2 marks)

11

A rectangular garden has length (x + 4) metres and width x metres. The area of the garden is 35 m².

Diagram NOT accurately drawn

(a)Show that x² + 4x - 35 = 0.(2)

(b)Hence solve the equation to find the width of the garden. Give your answer correct to 2 decimal places, and explain why the other solution of the equation is not a valid width.(4)

(Total for Question 11 is 6 marks)

12

A ball is thrown in the air. Its height, h metres, after t seconds is given by h = 20t - 5t² + 1.
Find the time(s) at which the ball is at a height of 15 metres, giving your answer(s) correct to 2 decimal places.

(Total for Question 12 is 4 marks)

13

The equation x² + kx + 16 = 0 has equal roots, where k > 0. Find the value of k.

(Total for Question 13 is 3 marks)

14

A student attempts to solve 3x² + 4x - 2 = 0 using the quadratic formula. Their working is shown below:
x = (-4 ± √4² - 4x3x(-2)) / (2x3)
x = (-4 ± √16 - 24) / 6
Identify the error made by the student, and find the correct solutions to the equation, giving your answers correct to 2 decimal places.

(Total for Question 14 is 4 marks)

15

Solve x + 3/x = 7. Give your solutions correct to 2 decimal places.

(Total for Question 15 is 4 marks)

16

Without solving them fully, compare the equations x² + 4x + 7 = 0 and x² + 4x - 7 = 0.

(a)State, giving a reason, how many real solutions x² + 4x + 7 = 0 has.(1)

(b)State, giving a reason, how many real solutions x² + 4x - 7 = 0 has.(1)

(Total for Question 16 is 2 marks)

17

A right-angled triangle has a shorter side of length x cm, a second shorter side of length (x + 3) cm, and a hypotenuse of length (x + 9) cm.

Diagram NOT accurately drawn

(a)Show that x² - 12x - 72 = 0.(3)

(b)Hence find the length of the hypotenuse, giving your answer correct to 3 significant figures.(4)

(Total for Question 17 is 7 marks)

18

Show that the equation 2x² - 3x + 5 = 0 has no real solutions.

(Total for Question 18 is 3 marks)

19

The equation x² + px + 12 = 0 has two distinct real roots, where p is a positive integer. Find the smallest possible value of p.

(Total for Question 19 is 3 marks)

20

Starting from the general quadratic equation ax² + bx + c = 0 (a not equal to 0), show, by completing the square, that
x = (-b ± √b² - 4ac) / (2a)
You must show each step of your working.

(Total for Question 20 is 5 marks)

Mark scheme · 7.4 Quadratic Formula

Question 1

(a) B1 a = 4, b = -7, c = 2, all correct
(a) Answer: a = 4, b = -7, c = 2
(b) M1 5x² - 3x - 8 = 0 oe (correct rearrangement)
(b) A1 a = 5, b = -3, c = -8, all correct
(b) Answer: 5x² - 3x - 8 = 0; a = 5, b = -3, c = -8

Question 2

B1 A (41) cao
Answer: A) 41

Question 3

M1 correct substitution into the formula with a = 1, b = 3, c = -5, e.g. x = (-3 ± √3² - 4(1)(-5))/(2(1))
A1 x = 1.19 (awrt 1.19)
A1 x = -4.19 (awrt -4.19)
Answer: x = 1.19 or x = -4.19

Question 4

M1 correct substitution with a = 1, b = -4, c = -1, leading to √20 or 2sqrt(5) seen
A1 x = 2 + √5 or x = 2 - √5, both, simplified oe
Answer: x = 2 + √5 or x = 2 - √5

Question 5

M1 correct substitution into the formula with a = 2, b = 5, c = -1
A1 x = 0.19 (awrt 0.19)
A1 x = -2.69 (awrt -2.69)
Answer: x = 0.19 or x = -2.69

Question 6

M1 expands and rearranges to x² + x - 11 = 0 oe
M1 correct substitution into the formula with a = 1, b = 1, c = -11 (ft)
A1 x = 2.85 (awrt 2.85)
A1 x = -3.85 (awrt -3.85)
Answer: x = 2.85 or x = -3.85

Question 7

M1 expands and rearranges to 2x² + 5x - 9 = 0 oe
M1 correct substitution into the formula with a = 2, b = 5, c = -9 (ft)
A1 x = 1.21 (awrt 1.21)
A1 x = -3.71 (awrt -3.71)
Answer: x = 1.21 or x = -3.71

Question 8

M1 correct substitution into the formula with a = 5, b = -8, c = -1
A1 x = 1.7 (awrt 1.7)
A1 x = -0.1 (awrt -0.1)
Answer: x = 1.7 or x = -0.1

Question 9

(a) M1 correct factorisation (x - 2)(x - 3) = 0
(a) A1 x = 2 and x = 3, both cao
(a) Answer: x = 2 or x = 3
(b) M1 recognises the equation does not factorise with integers and substitutes correctly into the formula with a = 1, b = -5, c = 2
(b) A1 x = 4.56 (awrt 4.56)
(b) A1 x = 0.44 (awrt 0.44)
(b) Answer: x = 4.56 or x = 0.44

Question 10

(a) B1 -56 cao
(a) Answer: -56
(b) B1 0 (no real solutions), since the discriminant is negative (ft from part a)
(b) Answer: No real solutions

Question 11

(a) M1 forms the equation x(x + 4) = 35 oe
(a) A1 x² + 4x - 35 = 0 cso (fully correct working shown, answer given)
(a) Answer: x² + 4x - 35 = 0 (shown)
(b) M1 correct substitution into the formula with a = 1, b = 4, c = -35 (ft)
(b) A1 x = 4.24 (awrt 4.24)
(b) B1 rejects x = -8.24, with reason that a length cannot be negative
(b) A1 width = 4.24 m cao (ft)
(b) Answer: Width = 4.24 m

Question 12

M1 forms the equation 5t² - 20t + 14 = 0 oe from 20t - 5t² + 1 = 15
M1 correct substitution into the formula with a = 5, b = -20, c = 14 (ft)
A1 t = 0.90 (awrt 0.90)
A1 t = 3.10 (awrt 3.10)
Answer: t = 0.90 s or t = 3.10 s

Question 13

M1 sets the discriminant equal to zero: k² - 4(1)(16) = 0 oe
M1 k² = 64
A1 k = 8 (rejecting k = -8 since k > 0)
Answer: k = 8

Question 14

B1 identifies the sign error, e.g. 4 x 3 x (-2) = -24 so 4² - 4ac = 16 - (-24) = 16 + 24, not 16 - 24
M1 correct discriminant of 40 used in the formula with a = 3, b = 4, c = -2
A1 x = 0.39 (awrt 0.39)
A1 x = -1.72 (awrt -1.72)
Answer: x = 0.39 or x = -1.72

Question 15

M1 multiplies through by x to obtain x² - 7x + 3 = 0 oe
M1 correct substitution into the formula with a = 1, b = -7, c = 3 (ft)
A1 x = 6.54 (awrt 6.54)
A1 x = 0.46 (awrt 0.46)
Answer: x = 6.54 or x = 0.46

Question 16

(a) B1 no real solutions, since b² - 4ac = 16 - 28 = -12, which is negative
(a) Answer: No real solutions
(b) B1 two real solutions, since b² - 4ac = 16 + 28 = 44, which is positive
(b) Answer: Two real solutions

Question 17

(a) M1 correct use of Pythagoras' theorem: x² + (x + 3)² = (x + 9)²
(a) M1 correct expansion: x² + x² + 6x + 9 = x² + 18x + 81
(a) A1 x² - 12x - 72 = 0 cso (fully correct working shown, answer given)
(a) Answer: x² - 12x - 72 = 0 (shown)
(b) M1 correct substitution into the formula with a = 1, b = -12, c = -72 (ft)
(b) A1 x = 16.4 (awrt 16.4), rejecting the negative root as x is a length
(b) M1 hypotenuse = x + 9 (ft)
(b) A1 25.4 cm (awrt 25.4)
(b) Answer: Hypotenuse = 25.4 cm

Question 18

M1 substitutes a = 2, b = -3, c = 5 into b² - 4ac
A1 b² - 4ac = -31
B1 cso: since -31 < 0, the discriminant is negative, so the equation has no real solutions
Answer: No real solutions (shown)

Question 19

M1 sets up the inequality p² - 4(1)(12) > 0 oe
M1 p² > 48 (or p > √48 = 6.928...)
A1 p = 7, with justification that p = 6 gives discriminant -12 < 0 while p = 7 gives discriminant 1 > 0
Answer: p = 7

Question 20

M1 divides through by a: x² + (b/a)x + c/a = 0 oe
M1 completes the square: (x + b/(2a))² - b²/(4a²) + c/a = 0 oe
M1 rearranges to (x + b/(2a))² = (b² - 4ac)/(4a²) oe, combining the right-hand side over a common denominator
M1 takes the square root of both sides: x + b/(2a) = ± √b² - 4ac/(2a)
A1 x = (-b ± √b² - 4ac)/(2a) cso (fully correct derivation, answer given)
Answer: x = (-b ± √b² - 4ac) / (2a) (shown)

Quadratic Formula - Worksheets, Questions and Revision

More on this topic

Next in this strand

7.4 Quadratic Formula

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20