Completing the Square - Worksheets, Questions and Revision

19 original exam-style questions - 9 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 10)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

HIGHER

8.4 Completing the Square

EDEXCEL 1MA1 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write each expression in the form (x + a)² + b.

(a)x² + 6x + 5(2)

(b)x² - 8x + 1(2)

(c)x² + 4x(2)

(Total for Question 1 is 6 marks)

Answer the following multiple choice questions on completing the square.

(a)x² - 12x + 40 can be written in the form (x - 6)² + k. What is the value of k?(1)

A) 4
B) -4
C) 40
D) -32

(b)x² - 14x + 51 can be written in the form (x - 7)² + k. What is the value of k?(1)

A) 2
B) -2
C) 51
D) -49

(Total for Question 2 is 2 marks)

Given that x² + px + q = (x + 6)² - 11 for all values of x, find the values of p and q.

(Total for Question 3 is 2 marks)

The expression x² - 10x + 30 is to be written in completed square form.

(a)Write x² - 10x + 30 in the form (x - a)² + b.(2)

(b)Hence state the minimum value of x² - 10x + 30, and the value of x for which this minimum occurs.(2)

(Total for Question 4 is 4 marks)

Solve x² + 6x + 2 = 0 by completing the square. Give your answers as exact surds, simplified where possible.

(Total for Question 5 is 3 marks)

Consider the expression x² + 8x + 20.

(a)Show that x² + 8x + 20 can be written as (x + 4)² + 4.(2)

(b)Hence explain why x² + 8x + 20 is positive for all values of x.(1)

(Total for Question 6 is 3 marks)

The expression 2x² + 12x + 7 is to be written in completed square form.

(a)Write 2x² + 12x + 7 in the form 2(x + a)² + b.(3)

(b)Hence state the minimum value of 2x² + 12x + 7, and the value of x for which this minimum occurs.(2)

(Total for Question 7 is 5 marks)

The curve C has equation y = x² - 4x + 1.

(a)Write x² - 4x + 1 in the form (x - a)² + b.(2)

(b)Hence write down the coordinates of the minimum turning point of C.(2)

(c)Hence find the exact solutions of x² - 4x + 1 = 0.(2)

(Total for Question 8 is 6 marks)

The function f is defined by f(x) = x² - 3x + 7.

(a)Express f(x) in the form (x - a)² + b, where a and b are exact values.(3)

(b)Hence state the minimum value of f(x).(1)

(Total for Question 9 is 4 marks)

The height, h metres, of a ball t seconds after being thrown is modelled by h = -5t² + 20t + 1.

(a)Write -5t² + 20t + 1 in the form -5(t - p)² + q.(4)

(b)Hence, or otherwise, find the maximum height of the ball and the time at which it occurs.(2)

(Total for Question 10 is 6 marks)

The curve C has equation y = 2x² - 8x + 3.

(a)Express 2x² - 8x + 3 in the form 2(x - a)² + b.(3)

(b)Write down the coordinates of the minimum point of C.(1)

(c)On the grid provided, sketch the graph of C, showing clearly the coordinates of the minimum point and the y-intercept.(2)

(Total for Question 11 is 6 marks)

Consider the equation 4x² + 16x + 15 = 0.

(a)Show that 4x² + 16x + 15 can be written as (2x + 4)² - 1.(3)

(b)Hence solve 4x² + 16x + 15 = 0, giving your answers as exact fractions.(2)

(Total for Question 12 is 5 marks)

A gardener has 24 metres of fencing. She uses it to enclose a rectangular vegetable patch on three sides, using an existing straight wall as the fourth side. The width of the patch, perpendicular to the wall, is x metres.

Diagram NOT accurately drawn

(a)Show that the area, A square metres, of the patch is given by A = 24x - 2x².(2)

(b)Express A in the form q - 2(x - p)², by completing the square.(3)

(c)Hence find the maximum possible area of the patch and the width, x, for which this occurs.(2)

(Total for Question 13 is 7 marks)

A rectangular lawn has a perimeter of 50 metres. The width of the lawn is x metres.

Diagram NOT accurately drawn

(a)Show that the area, A square metres, of the lawn is given by A = 25x - x².(2)

(b)Express A in the form q - (x - p)², by completing the square.(3)

(c)Hence find the maximum possible area of the lawn, and describe the shape of the lawn when this maximum occurs.(2)

(Total for Question 14 is 7 marks)

Solve 3x² - 6x - 2 = 0 by completing the square. Give your answer in the form 1 ± (√15 / 3), fully simplified.

(Total for Question 15 is 4 marks)

Prove algebraically that x² - 6x + 11 ≥ 2 for all real values of x.

(Total for Question 16 is 3 marks)

Solve x² + 5x - 3 = 0 by completing the square. Give your answers in the form (a ± √b) / c, where a, b and c are integers.

(Total for Question 17 is 4 marks)

Prove that the equation x² + 6x + 10 = 0 has no real roots.

(Total for Question 18 is 3 marks)

A circle has equation x² + y² - 6x + 4y - 3 = 0.

(a)By completing the square twice, show that the equation can be written as (x - 3)² + (y + 2)² = 16.(3)

(b)Hence state the coordinates of the centre and the radius of the circle.(2)

(Total for Question 19 is 5 marks)

Mark scheme · 8.4 Completing the Square

Question 1

(a) M1 (x + 3)² seen, or correct method to complete the square
(a) A1 (x + 3)² - 4 cao
(a) Answer: (x + 3)² - 4
(b) M1 (x - 4)² seen, or correct method to complete the square
(b) A1 (x - 4)² - 15 cao
(b) Answer: (x - 4)² - 15
(c) M1 (x + 2)² seen
(c) A1 (x + 2)² - 4 cao
(c) Answer: (x + 2)² - 4

Question 2

(a) B1 A
(a) Answer: A
(b) B1 A
(b) Answer: A

Question 3

B1 p = 12 cao
B1 q = 25 cao
Answer: p = 12, q = 25

Question 4

(a) M1 (x - 5)² seen
(a) A1 (x - 5)² + 5 cao
(a) Answer: (x - 5)² + 5
(b) B1 ft minimum value = 5
(b) B1 ft occurs when x = 5
(b) Answer: Minimum value 5, at x = 5

Question 5

M1 (x + 3)² - 9 + 2 = 0 oe, correct completed square set to zero
M1 (x + 3)² = 7 leading to x + 3 = +-√7
A1 x = -3 + √7 and x = -3 - √7 oe, both exact
Answer: x = -3 + √7 or x = -3 - √7

Question 6

(a) M1 (x + 4)² = x² + 8x + 16 obtained
(a) A1 cso: 16 + 4 = 20, so x² + 8x + 20 = (x + 4)² + 4, answer printed
(a) Answer: (x + 4)² + 4, as shown
(b) B1 (x + 4)² ≥ 0 for all real x, so (x + 4)² + 4 ≥ 4, which is always greater than 0 oe
(b) Answer: (x+4)² ≥ 0, so the expression is always at least 4, hence always positive

Question 7

(a) M1 2(x² + 6x) + 7 oe, factor of 2 taken out correctly
(a) M1 2[(x + 3)² - 9] + 7 oe
(a) A1 2(x + 3)² - 11 cao
(a) Answer: 2(x + 3)² - 11
(b) B1 ft minimum value = -11
(b) B1 ft occurs when x = -3
(b) Answer: Minimum value -11, at x = -3

Question 8

(a) M1 (x - 2)² seen
(a) A1 (x - 2)² - 3 cao
(a) Answer: (x - 2)² - 3
(b) M1 ft identifies x = 2 from the completed square form
(b) A1 ft (2, -3) cao
(b) Answer: (2, -3)
(c) M1 ft sets (x - 2)² - 3 = 0 and rearranges to (x-2)² = 3
(c) A1 x = 2 + √3 and x = 2 - √3 oe, both exact
(c) Answer: x = 2 + √3 or x = 2 - √3

Question 9

(a) M1 (x - 1.5)² oe seen (allow x - 3/2)
(a) M1 correct subtraction of (1.5)² = 2.25 from 7
(a) A1 (x - 1.5)² + 4.75 oe (accept (x - 3/2)² + 19/4)
(a) Answer: (x - 1.5)² + 4.75
(b) B1 ft 4.75 oe (accept 19/4)
(b) Answer: 4.75

Question 10

(a) M1 -5(t² - 4t) + 1 oe, factor of -5 taken out correctly
(a) M1 -5[(t - 2)² - 4] + 1 oe
(a) M1 correct expansion: -5(t-2)² + 20 + 1
(a) A1 -5(t - 2)² + 21 cao
(a) Answer: -5(t - 2)² + 21
(b) B1 ft maximum height = 21 metres
(b) B1 ft occurs when t = 2 seconds
(b) Answer: Maximum height 21 m, at t = 2 seconds

Question 11

(a) M1 2(x² - 4x) + 3 oe
(a) M1 2[(x - 2)² - 4] + 3 oe
(a) A1 2(x - 2)² - 5 cao
(a) Answer: 2(x - 2)² - 5
(b) B1 ft (2, -5) cao
(b) Answer: (2, -5)
(c) B1 ft correct u-shape with minimum point plotted at (2, -5)
(c) B1 y-intercept plotted at (0, 3)
(c) Answer: U-shaped curve, minimum (2, -5), y-intercept (0, 3)

Question 12

(a) M1 (2x + 4)² = 4x² + 16x + 16 obtained
(a) M1 16 - 1 = 15 identified
(a) A1 cso: (2x+4)² - 1 = 4x² + 16x + 15, answer printed
(a) Answer: (2x + 4)² - 1, as shown
(b) M1 ft (2x + 4)² = 1 leading to 2x + 4 = ±1
(b) A1 x = -3/2 and x = -5/2 oe, both exact
(b) Answer: x = -3/2 or x = -5/2

Question 13

(a) M1 length parallel to the wall = 24 - 2x oe
(a) A1 cso: A = x(24 - 2x) = 24x - 2x², answer printed
(a) Answer: A = 24x - 2x², as shown
(b) M1 -2(x² - 12x) oe, factor of -2 taken out correctly
(b) M1 -2[(x - 6)² - 36] oe
(b) A1 72 - 2(x - 6)² cao
(b) Answer: 72 - 2(x - 6)²
(c) B1 ft maximum area = 72 square metres
(c) B1 ft occurs when x = 6
(c) Answer: Maximum area 72 m², when x = 6 m

Question 14

(a) M1 length = 25 - x oe, from 2(length + width) = 50
(a) A1 cso: A = x(25 - x) = 25x - x², answer printed
(a) Answer: A = 25x - x², as shown
(b) M1 -(x² - 25x) oe, factor of -1 taken out correctly
(b) M1 -[(x - 12.5)² - 156.25] oe
(b) A1 156.25 - (x - 12.5)² oe (accept 625/4 - (x - 12.5)²)
(b) Answer: 156.25 - (x - 12.5)²
(c) B1 ft maximum area = 156.25 square metres
(c) B1 ft occurs when x = 12.5, so the lawn is a square (length also 12.5 m)
(c) Answer: Maximum area 156.25 m², when the lawn is a 12.5 m by 12.5 m square

Question 15

M1 3(x² - 2x) - 2 = 0 oe, factor of 3 taken out correctly
M1 3[(x - 1)² - 1] - 2 = 0 oe leading to 3(x-1)² = 5
M1 (x - 1)² = 5/3 leading to x - 1 = +-√15/3 (rationalising √5/3)
A1 x = 1 + √15/3 and x = 1 - √15/3 oe, both exact and simplified
Answer: x = 1 + √15/3 or x = 1 - √15/3

Question 16

M1 completes the square: x² - 6x + 11 = (x - 3)² - 9 + 11
A1 (x - 3)² + 2 obtained
B1 dep on M1: since (x - 3)² ≥ 0 for all real x, (x - 3)² + 2 ≥ 2, so x² - 6x + 11 ≥ 2 for all real x, cso
Answer: x² - 6x + 11 = (x-3)² + 2 ≥ 2, since (x-3)² ≥ 0 for all real x

Question 17

M1 (x + 2.5)² - 6.25 - 3 = 0 oe, correct completed square set to zero
M1 (x + 2.5)² = 9.25 oe (accept 37/4)
M1 x + 2.5 = +-√37/2 oe, correct rearrangement
A1 x = (-5 + √37) / 2 and x = (-5 - √37) / 2 oe, both exact
Answer: x = (-5 + √37) / 2 or x = (-5 - √37) / 2

Question 18

M1 completes the square: x² + 6x + 10 = (x + 3)² - 9 + 10
A1 (x + 3)² + 1 = 0 leading to (x + 3)² = -1
B1 dep: since (x + 3)² ≥ 0 for all real x, it can never equal -1, so the equation has no real roots, cso
Answer: (x+3)² = -1 has no real solution, since a square cannot be negative; hence no real roots