Revision Library

revisionlibrary.co.uk

HIGHER

8.1D Quadratic Simultaneous Equations: Fluency and Exam Drill

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Solve the simultaneous equations

y = x²
y = 2x + 3

(Total for Question 1 is 3 marks)

Solve the simultaneous equations

y = -x² + 5
y = x - 1

(Total for Question 2 is 3 marks)

Solve the simultaneous equations

y = x² - 4
x + y = 6

(Total for Question 3 is 4 marks)

For which value(s) of k does the line y = kx + 1 meet the parabola y = x² + 1 at exactly one point?

(Total for Question 4 is 2 marks)

Solve the simultaneous equations

y = x² - x
y = -x² + 3x - 2

(Total for Question 5 is 3 marks)

Solve the simultaneous equations

y = 2x² - 3
x - y = 1

(Total for Question 6 is 4 marks)

Show that the simultaneous equations

y = x² - 6x + 8
y = 0

have solutions x = 2 and x = 4.

(Total for Question 7 is 3 marks)

Find the value of k for which the horizontal line y = k meets the parabola y = x² - 2x at x = 0 and x = 2.

(Total for Question 8 is 2 marks)

Solve the simultaneous equations

x² + y² = 25
y = x + 1

(Total for Question 9 is 4 marks)

Solve the simultaneous equations

y = x² + x
y = -x² + 5

(Total for Question 10 is 4 marks)

Solve the simultaneous equations

y = x² + 4x + 1
y = -x - 3

(Total for Question 11 is 3 marks)

For which value(s) of a does the line y = 2x + 3 meet the parabola y = x² + ax + 4 at exactly one point? Give all real values of a.

(Total for Question 12 is 4 marks)

Mark scheme · 8.1D Quadratic Simultaneous Equations: Fluency and Exam Drill

Question 1

M1 equate the two expressions: x² = 2x + 3
M1 rearrange and factorise: x² - 2x - 3 = (x - 3)(x + 1)
A1 both solution pairs (3, 9) and (-1, 1) cao
Answer: (x, y) = (3, 9) and (-1, 1). (So x = 3 or -1, with corresponding y values 9 and 1.)

Question 2

M1 equate the two expressions: -x² + 5 = x - 1
M1 rearrange and factorise: x² + x - 6 = (x + 3)(x - 2)
A1 both solution pairs (-3, -4) and (2, 1) cao
Answer: (x, y) = (-3, -4) and (2, 1).

Question 3

M1 substitute y from the first equation into the second: x + x² - 4 = 6
M1 rearrange to a quadratic and apply the quadratic formula: x² + x - 10 = 0
A1 x = (-1 + √41)/2 and x = (-1 - √41)/2 (exact values) cao
A1 corresponding y values y = (13 - √41)/2 and y = (13 + √41)/2 cao
Answer: x = (-1 ± √41)/2. Corresponding y = 6 - x so y = (13 -/+ √41)/2. (Two solution pairs.)

Question 4

M1 equate: x² + 1 = kx + 1 so x² - kx = 0
A1 for a repeated root need k = 0 so k = 0 cao
Answer: k = 0.

Question 5

M1 equate and rearrange: x² - x = -x² + 3x - 2 => 2x² - 4x + 2 = 0
M1 divide by 2 and factorise: x² - 2x + 1 = (x - 1)²
A1 single solution (x, y) = (1, 0) cao
Answer: (x, y) = (1, 0).

Question 6

M1 substitute y = x - 1 into 2x² - 3 = x - 1 to form 2x² - x - 2 = 0
M1 apply quadratic formula: x = [1 ± √1 + 16]/4 = [1 ± √17]/4
A1 x = (1 + √17)/4 and x = (1 - √17)/4 (exact) cao
A1 corresponding y = x - 1 so y = (1 + √17)/4 - 1 and y = (1 - √17)/4 - 1 cao
Answer: x = (1 ± √17)/4. Corresponding y = x - 1 = (1 ± √17)/4 - 1.

Question 7

M1 substitute y = 0 into y = x² - 6x + 8 to get x² - 6x + 8 = 0 cso
M1 factorise to (x - 2)(x - 4) = 0 cso
A1 conclude x = 2 and x = 4 cao cso
Answer: x = 2 and x = 4 (so the solution pairs are (2, 0) and (4, 0)).

Question 8

M1 compute k from either x = 0 or x = 2: k = 0² - 0 = 0 or k = 2² - 4 = 0
A1 k = 0 cao
Answer: k = 0.

Question 9

M1 substitute y = x + 1 into x² + y² = 25 to get x² + (x + 1)² = 25
M1 simplify to 2x² + 2x - 24 = 0 and divide by 2 to get x² + x - 12 = 0
A1 factorise to (x + 4)(x - 3) so x = -4 or x = 3
A1 corresponding pairs (x, y) = (-4, -3) and (3, 4) cao
Answer: (x, y) = (-4, -3) and (3, 4).

Question 10

M1 equate and rearrange: x² + x = -x² + 5 => 2x² + x - 5 = 0
M1 use quadratic formula: x = [-1 ± √1 + 40]/4 = [-1 ± √41]/4
A1 x = (-1 + √41)/4 and x = (-1 - √41)/4 (exact) cao
A1 corresponding y = x² + x evaluated at these x values cao
Answer: x = (-1 ± √41)/4. Corresponding y values y = x² + x evaluated at these x.

Question 11

M1 equate and rearrange: x² + 4x + 1 = -x - 3 => x² + 5x + 4 = 0
M1 factorise: (x + 1)(x + 4) = 0
A1 solution pairs (x, y) = (-1, -2) and (-4, 1) cao
Answer: (x, y) = (-1, -2) and (-4, 1).

Question 12

M1 equate and form quadratic: x² + ax + 4 = 2x + 3 => x² + (a - 2)x + 1 = 0
M1 require discriminant = 0 for exactly one point: (a - 2)² - 4*1*1 = 0
A1 solve (a - 2)² = 4 so a - 2 = 2 or a - 2 = -2
A1 give a = 4 or a = 0 cao
Answer: a = 4 or a = 0.

Quadratic Simultaneous Equations: Fluency and Exam Drill - Worksheets, Questions and Revision