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5.8F Solving Quadratics: Foundation Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 55 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the value of x.
x² = 49

(Total for Question 1 is 1 mark)

Work out the value of x.
(x + 3)(x - 2) = 0

(Total for Question 2 is 1 mark)

Work out the values of x.
x² - 5x = 0

(Total for Question 3 is 2 marks)

Solve the equation by factorising.
2x² + 3x = 0

(Total for Question 4 is 2 marks)

Work out the values of x.
x² - x - 6 = 0

(Total for Question 5 is 3 marks)

Solve the equation.
3x² - 12 = 0

(Total for Question 6 is 3 marks)

Work out x.
x² + 4x + 4 = 0

(Total for Question 7 is 3 marks)

A rectangle has length (x + 5) metres and width (x - 1) metres. The area is 40 m².
Write down an equation and solve for x, explaining why only one of your solutions is valid in this context.

(Total for Question 8 is 4 marks)

Solve the equation by factorising.
4x² - 9 = 0

(Total for Question 9 is 4 marks)

A small garden is a square of side (2x - 1) metres. Its area is 49 m².
Work out the values of x.

(Total for Question 10 is 5 marks)

Priya is making a rectangular flowerbed. The length is 3 metres more than the width. The area is 70 m².
Let the width be x metres.
(a) Write down an equation in x.
(b) Solve the equation and give the width of the flowerbed.

(a)(a) Write down an equation in x.(1)

(b)(b) Solve the equation and give the width of the flowerbed.(5)

(Total for Question 11 is 6 marks)

A rectangular shed has length 4x metres and width (x + 2) metres. The area of the shed is 96 m².
Work out the value of x and then find the actual length and width of the shed.

(Total for Question 12 is 6 marks)

Mark scheme · 5.8F Solving Quadratics: Foundation Tier Practice

Question 1

B1 x = 7 or x = -7 cao
Answer: x = 7 or x = -7

Question 2

B1 x = -3 or x = 2 cao
Answer: x = -3 or x = 2

Question 3

M1 factorise to x(x - 5) = 0 or equivalent
A1 x = 0 and x = 5 cao
Answer: x = 0 and x = 5

Question 4

M1 factorise to x(2x + 3) = 0 or equivalent
A1 x = 0 and x = -3/2 cao
Answer: x = 0 and x = -3/2

Question 5

M1 factorise to (x - 3)(x + 2) = 0 or equivalent
A1 x = 3 cao
A1 x = -2 cao
Answer: x = 3 and x = -2

Question 6

M1 rearrange to 3x² = 12 and divide to x² = 4 or equivalent
A1 x = 2 cao
A1 x = -2 cao
Answer: x = 2 and x = -2

Question 7

M1 recognise perfect square: (x + 2)² = 0 or factorise to (x + 2)(x + 2)
A1 x = -2 cao
isw second A1 is dependent so isw
Answer: x = -2

Question 8

M1 form equation (x + 5)(x - 1) = 40 or equivalent
M1 expand to x² + 4x - 5 = 40 then rearrange to x² + 4x - 45 = 0
A1 factorise to (x + 9)(x - 5) = 0 giving x = -9 or x = 5
A1 reject x = -9 (width x - 1 must be positive, so x > 1); x = 5 cao
Answer: x = 5 (length = 10 m, width = 4 m, area = 40 m²). x = -9 is rejected as it gives a negative width.

Question 9

M1 recognise difference of squares: (2x - 3)(2x + 3) = 0 or show correct factorisation
A1 2x - 3 = 0 gives x = 3/2 cao
A1 2x + 3 = 0 gives x = -3/2 cao
isw final A1 dependent on correct factorisation
Answer: x = 3/2 and x = -3/2

Question 10

M1 form equation (2x - 1)² = 49 or equivalent
M1 expand or take square roots: (2x - 1) = 7 or (2x - 1) = -7
A1 from 2x - 1 = 7, x = 4 cao
A1 from 2x - 1 = -7, x = -3 cao
C1 communicate both solutions with working shown
Answer: x = 4 or x = -3

Question 11

(a) M1 equation x(x + 3) = 70 or equivalent
(a) Answer: x(x + 3) = 70
(b) M1 expand to x² + 3x - 70 = 0 or rearrange to standard form
(b) M1 factorise to (x + 10)(x - 7) = 0 or equivalent
(b) M1 state both roots x = 7 and x = -10 or show solving each factor
(b) A1 x = 7 cao
(b) A1 width = 7 m stated as the positive practical solution
(b) Answer: x = 7 (width = 7 m)

Question 12

M1 form equation 4x(x + 2) = 96 or equivalent
M1 simplify to 4x² + 8x - 96 = 0 and divide by 4 to get x² + 2x - 24 = 0
M1 factorise to (x + 6)(x - 4) = 0 or equivalent
A1 x = 4 cao
A1 x = -6 cao
A1 state length = 16 m and width = 6 m for the positive solution
Answer: x = 4 (discard x = -6). Length = 4x = 16 m, width = x + 2 = 6 m. Working: 4x(x + 2) = 96 -> x² + 2x - 24 = 0 -> (x + 6)(x - 4) = 0.

Solving Quadratics: Foundation Tier Practice - Worksheets, Questions and Revision