Quadratic Simultaneous Equations Worksheet with Answers

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8.1 Quadratic Simultaneous Equations

EDEXCEL 1MA1 A19 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

1

Solve the simultaneous equations y = x + 3 and y = x² - 3

(Total for Question 1 is 4 marks)

2

Solve the simultaneous equations x² + y² = 25 and y = x - 1

(Total for Question 2 is 4 marks)

3

The equations y = x² - 1 and y = 2x + 2 are solved simultaneously. Which one of these points is a solution to both equations?

A) (2, 6)
B) (3, 8)
C) (-1, 2)
D) (0, -1)

(Total for Question 3 is 1 mark)

4

A curve has equation y = x² - x. A line has equation y = 2x + 4.

(a)Show that solving the simultaneous equations for the curve and the line leads to x² - 3x - 4 = 0(2)

(b)Hence solve the simultaneous equations.(3)

(Total for Question 4 is 5 marks)

5

Solve the simultaneous equations x - y = 3 and y = x² - 4x + 1

(Total for Question 5 is 5 marks)

6

A circle C has equation x² + y² = 25. A line L has equation y = 2x - 5. Find the coordinates of the points where L intersects C.

(Total for Question 6 is 5 marks)

7

Fatima is designing a rectangular allotment plot of length x metres and width y metres. The perimeter of the plot is 22 metres and the area of the plot is 24 square metres. Find the dimensions of the plot.

(Total for Question 7 is 5 marks)

8

Solve the simultaneous equations 2x + y = 8 and x² + y² = 20

(Total for Question 8 is 6 marks)

9

Show that the line y = x + 6 does not intersect the curve y = x² + 2x + 10

(Total for Question 9 is 4 marks)

10

Solve the simultaneous equations x = y² - 3y + 1 and x = 3y - 4

(Total for Question 10 is 5 marks)

11

A line has equation y = 2x - 1. A curve has equation y = x² - 4x + 8. The line meets the curve at exactly one point.

(a)Show that solving the simultaneous equations leads to x² - 6x + 9 = 0(2)

(b)Hence find the coordinates of the point where the line meets the curve.(3)

(Total for Question 11 is 5 marks)

12

Solve the simultaneous equations y = x² + x - 5 and y = 4x - 2, giving your answers to 3 significant figures where appropriate.

(Total for Question 12 is 5 marks)

13

A line and a curve are given by y = mx + c and y = ax² + bx + d. Explain how the discriminant of the quadratic equation formed by solving these simultaneously can be used to determine the number of points of intersection of the line and the curve.

(Total for Question 13 is 3 marks)

14

Solve the simultaneous equations y = x + 2 and y = x² - 3x - 1, giving your answers in surd form.

(Total for Question 14 is 5 marks)

15

The line y = x + k, where k is a constant, is a tangent to the curve y = x² + 5x + 7.

(a)Find the value of k.(4)

(b)Find the coordinates of the point where the line touches the curve.(2)

(Total for Question 15 is 6 marks)

16

The line y = x + k, where k is a constant, is a tangent to the circle x² + y² = 8. Find the two possible values of k.

(Total for Question 16 is 5 marks)

17

Priya is thinking of two numbers, x and y. The difference between the numbers is 3, so that x - y = 3. The sum of the squares of the numbers is 89, so that x² + y² = 89. Find the two possible pairs of values of x and y.

(Total for Question 17 is 5 marks)

18

A circle has equation x² + y² = 50. A line has equation y = x + 2. The line intersects the circle at two points, A and B. Calculate the length of AB, giving your answer in surd form.

(Total for Question 18 is 6 marks)

19

A curve C has equation y = x² - 6x + 11. A line L has equation y = 2x - k, where k is a constant. Given that L intersects C at two distinct points, find the range of possible values of k.

(Total for Question 19 is 5 marks)

20

Nadia is designing a right-angled triangular metal bracket. The two shorter sides have lengths x cm and (x+7) cm. The hypotenuse has length y cm, where y = 2x + 1.

(a)Using Pythagoras' theorem, show that x² - 5x - 24 = 0(3)

(b)Hence find the length of the hypotenuse of the bracket.(3)

(Total for Question 20 is 6 marks)

Mark scheme · 8.1 Quadratic Simultaneous Equations

Question 1

M1 substitutes to eliminate y, e.g. x + 3 = x² - 3 oe
M1 rearranges to a 3-term quadratic = 0, e.g. x² - x - 6 = 0 oe
A1 x = 3 and x = -2 (both, cao)
A1 y = 6 and y = 1 (both, cao, ft from their x values)
Answer: x = 3, y = 6 and x = -2, y = 1

Question 2

M1 substitutes y = x - 1 into x² + y² = 25
M1 expands and simplifies to a 3-term quadratic = 0, e.g. x² - x - 12 = 0 oe
A1 x = 4 and x = -3 (both, cao)
A1 y = 3 and y = -4 (both, cao, ft from their x values)
Answer: x = 4, y = 3 and x = -3, y = -4

Question 3

B1 B) (3, 8) selected
Answer: B) (3, 8)

Question 4

(a) M1 equates x² - x = 2x + 4 oe
(a) A1 rearranges correctly to x² - 3x - 4 = 0 (cso, answer given)
(a) Answer: x² - 3x - 4 = 0 (shown)
(b) M1 factorises or solves x² - 3x - 4 = 0, e.g. (x-4)(x+1) = 0
(b) A1 x = 4 and x = -1 (both, cao)
(b) A1 y = 12 and y = 2 (both, cao, ft from their x values)
(b) Answer: x = 4, y = 12 and x = -1, y = 2

Question 5

M1 rearranges x - y = 3 to y = x - 3 oe
M1 substitutes to give x - 3 = x² - 4x + 1 oe
dM1 rearranges and solves the resulting 3-term quadratic = 0, e.g. (x-4)(x-1) = 0
A1 x = 4 and x = 1 (both, cao)
A1 y = 1 and y = -2 (both, cao, ft from their x values)
Answer: x = 4, y = 1 and x = 1, y = -2

Question 6

M1 substitutes y = 2x - 5 into x² + y² = 25
M1 expands and simplifies to a 3-term quadratic = 0, e.g. 5x² - 20x = 0 oe
dM1 factorises or solves for x, e.g. 5x(x-4) = 0
A1 x = 0 and x = 4 (both, cao)
A1 y = -5 and y = 3 (both, cao, ft from their x values)
Answer: (0, -5) and (4, 3)

Question 7

M1 forms x + y = 11 oe from the perimeter
M1 forms xy = 24 oe from the area and substitutes y = 11 - x
dM1 rearranges and solves the resulting 3-term quadratic = 0, e.g. (x-8)(x-3) = 0
A1 x = 8 and x = 3 (or equivalent y values), cao
A1 states dimensions of the plot are 3 m by 8 m, with units
Answer: The plot measures 3 m by 8 m

Question 8

M1 rearranges 2x + y = 8 to y = 8 - 2x oe
M1 substitutes into x² + y² = 20 and expands
M1 simplifies to a 3-term quadratic = 0, e.g. 5x² - 32x + 44 = 0 oe
dM1 solves using the quadratic formula or factorising
A1 x = 2 and x = 4.4 (oe 22/5), both, cao
A1 y = 4 and y = -0.8 (oe -4/5), both, cao, ft from their x values
Answer: x = 2, y = 4 and x = 4.4, y = -0.8

Question 9

M1 equates x + 6 = x² + 2x + 10 oe
M1 rearranges to a 3-term quadratic = 0, e.g. x² + x + 4 = 0 oe
M1 calculates the discriminant, e.g. 1² - 4(1)(4) = -15
A1 correct conclusion: since the discriminant is negative there are no real solutions, so the line does not intersect the curve (cso)
Answer: Discriminant = -15 < 0, so no real solutions exist and the line does not intersect the curve (shown)

Question 10

M1 equates y² - 3y + 1 = 3y - 4 oe
M1 rearranges to a 3-term quadratic = 0, e.g. y² - 6y + 5 = 0 oe
dM1 factorises or solves for y, e.g. (y-1)(y-5) = 0
A1 y = 1 and y = 5 (both, cao)
A1 x = -1 and x = 11 (both, cao, ft from their y values)
Answer: y = 1, x = -1 and y = 5, x = 11

Question 11

(a) M1 equates 2x - 1 = x² - 4x + 8 oe
(a) A1 rearranges correctly to x² - 6x + 9 = 0 (cso, answer given)
(a) Answer: x² - 6x + 9 = 0 (shown)
(b) M1 recognises x² - 6x + 9 = (x-3)² = 0, a repeated root
(b) A1 x = 3 (cao)
(b) A1 y = 5 (cao, ft from their x value)
(b) Answer: (3, 5)

Question 12

M1 equates x² + x - 5 = 4x - 2 oe
M1 rearranges to a 3-term quadratic = 0, e.g. x² - 3x - 3 = 0 oe
M1 applies the quadratic formula correctly with a = 1, b = -3, c = -3
A1 x = awrt 3.79 and x = awrt -0.79 (both), or exact form (3+-√21)/2
A1 y = awrt 13.2 and y = awrt -5.17 (both), ft from their x values, or exact form 4+-2sqrt(21)
Answer: x = (3+√21)/2 or (3-√21)/2 [awrt 3.79 or -0.79]; y = 4+2sqrt(21) or 4-2sqrt(21) [awrt 13.2 or -5.17]

Question 13

B1 states that substituting the linear equation into the quadratic gives a single quadratic equation of the form px² + qx + r = 0, whose discriminant is q² - 4pr
B1 states that if q² - 4pr > 0 there are two distinct points of intersection, and if q² - 4pr = 0 the line is a tangent to the curve (one point of intersection)
B1 states that if q² - 4pr < 0 there are no real solutions, so the line does not intersect the curve
Answer: Substitute to form one quadratic equation; if its discriminant is positive there are two intersection points, if zero the line is a tangent (one point), if negative there are none.

Question 14

M1 equates x + 2 = x² - 3x - 1 oe
M1 rearranges to a 3-term quadratic = 0, e.g. x² - 4x - 3 = 0 oe
M1 applies the quadratic formula correctly with a = 1, b = -4, c = -3
A1 x = 2 + √7 and x = 2 - √7 (both, exact surd form, cao)
A1 y = 4 + √7 and y = 4 - √7 (both, exact surd form, ft from their x values)
Answer: x = 2 + √7 or 2 - √7; y = 4 + √7 or 4 - √7

Question 15

(a) M1 equates x + k = x² + 5x + 7 oe
(a) M1 rearranges to x² + 4x + (7 - k) = 0 oe
(a) M1 sets the discriminant equal to zero, e.g. 16 - 4(7-k) = 0
(a) A1 k = 3 (cao)
(a) Answer: k = 3
(b) M1 substitutes k = 3 to solve x² + 4x + 4 = 0, giving x = -2 (repeated root)
(b) A1 y = 1 (cao, ft from their x value)
(b) Answer: (-2, 1)

Question 16

M1 substitutes y = x + k into x² + y² = 8 and expands
M1 rearranges to a 3-term quadratic in x = 0 in terms of k, e.g. 2x² + 2kx + (k²-8) = 0
M1 sets the discriminant equal to zero, e.g. (2k)² - 4(2)(k²-8) = 0
dM1 solves the resulting equation for k, e.g. k² = 16
A1 k = 4 and k = -4 (both, cao)
Answer: k = 4 or k = -4

Question 17

M1 rearranges x - y = 3 to x = y + 3 oe
M1 substitutes into x² + y² = 89 and expands
M1 simplifies to a 3-term quadratic = 0, e.g. y² + 3y - 40 = 0 oe
dM1 factorises or solves for y, e.g. (y+8)(y-5) = 0
A1 both pairs stated: x = 8, y = 5 and x = -5, y = -8 (cao)
Answer: x = 8, y = 5 or x = -5, y = -8

Question 18

M1 substitutes y = x + 2 into x² + y² = 50 and expands
M1 simplifies to a 3-term quadratic = 0, e.g. x² + 2x - 23 = 0 oe
M1 uses the quadratic formula to find x1 - x2 = √discriminant/a, e.g. 4sqrt(6)
M1 recognises that since the line has gradient 1, y1 - y2 = x1 - x2
A1 correctly applies AB = √(x1-x2)² + (y1-y2)²
A1 AB = 8sqrt(3) (cao, oe awrt 13.9)
Answer: AB = 8sqrt(3) (awrt 13.9)

Question 19

M1 equates x² - 6x + 11 = 2x - k oe
M1 rearranges to x² - 8x + (11+k) = 0 oe
M1 sets the discriminant greater than zero, e.g. 64 - 4(11+k) > 0
A1 simplifies correctly to 20 - 4k > 0 oe
A1 k < 5 (cao)
Answer: k < 5

Question 20

(a) M1 forms the Pythagoras equation x² + (x+7)² = y²
(a) M1 substitutes y = 2x+1 and expands both sides
(a) A1 simplifies correctly to x² - 5x - 24 = 0 (cso, answer given)
(a) Answer: x² - 5x - 24 = 0 (shown)
(b) M1 factorises or solves x²-5x-24 = 0, e.g. (x-8)(x+3) = 0
(b) A1 selects x = 8, rejecting x = -3 since a length cannot be negative
(b) A1 hypotenuse y = 17 cm (cao, ft from their x)
(b) Answer: Hypotenuse = 17 cm

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8.1 Quadratic Simultaneous Equations

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Question 3

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Question 6

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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

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Question 10

Question 11

Question 12

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