20 questions · 73 marks · step-by-step working for every question. Marks are shown in square brackets. Every answer on this sheet has been checked by computer algebra.
Q1 · (1, 0) and (4, 3) [3]
Equate the two expressions for y: x2 − 4x + 3 = x − 1
x2 − 5x + 4 = 0 → (x − 1)(x − 4) = 0, so x = 1 or x = 4
x = 1 → y = 0; x = 4 → y = 3
(1, 0) and (4, 3)
Q2 · (2, 3) and (3, 2) [3]
From the first equation y = 5 − x
Substitute: x(5 − x) = 6 → x2 − 5x + 6 = 0
(x − 2)(x − 3) = 0, so x = 2 → y = 3 and x = 3 → y = 2
(2, 3) and (3, 2)
Q3 · (3, 4) and (4, 3) [3]
y = 7 − x, so x2 + (7 − x)2 = 25
x2 + 49 − 14x + x2 = 25 → 2x2 − 14x + 24 = 0
Divide by 2: x2 − 7x + 12 = 0 → (x − 3)(x − 4) = 0
(3, 4) and (4, 3)
Q4 · (4, 1) and (−1, −4) [3]
Substitute y = x − 3: x2 + (x − 3)2 = 17
x2 + x2 − 6x + 9 = 17 → 2x2 − 6x − 8 = 0
x2 − 3x − 4 = 0 → (x − 4)(x + 1) = 0
x = 4 → y = 1; x = −1 → y = −4 → (4, 1) and (−1, −4)
Q5 · (2, 4) and (4, 2) [3]
From the first equation y = 6 − x
Substitute: x(6 − x) = 8 → x2 − 6x + 8 = 0
(x − 2)(x − 4) = 0, so x = 2 → y = 4 and x = 4 → y = 2
(2, 4) and (4, 2)
Q6 · (0, −3) and (4, 5) [3]
Equate: x2 − 2x − 3 = 2x − 3
Collect everything on one side: x2 − 4x = 0 → x(x − 4) = 0
So x = 0 → y = −3, and x = 4 → y = 5
(0, −3) and (4, 5)
Q7 · (1, −1) and (3, 1) [3]
Equate: x2 − 3x + 1 = x − 2
x2 − 4x + 3 = 0 → (x − 1)(x − 3) = 0
x = 1 → y = −1; x = 3 → y = 1
(1, −1) and (3, 1)
Q8 · 5 and 7 [4]
x + y = 12 and x2 + y2 = 74
y = 12 − x, so x2 + (12 − x)2 = 74
x2 + 144 − 24x + x2 = 74 → 2x2 − 24x + 70 = 0
x2 − 12x + 35 = 0 → (x − 5)(x − 7) = 0, so the numbers are 5 and 7
Q9 · length 9 cm, width 4 cm [4]
2(l + w) = 26 → l + w = 13, and lw = 36
w = 13 − l, so l(13 − l) = 36
l2 − 13l + 36 = 0 → (l − 4)(l − 9) = 0
The length is the longer side, so l = 9 cm and w = 4 cm — check: area = 36 cm2 ✓
Q10 · 4 and 7 [4]
The integers are x and x + 3, so x2 + (x + 3)2 = 65
x2 + x2 + 6x + 9 = 65 → 2x2 + 6x − 56 = 0
Divide by 2: x2 + 3x − 28 = 0 → (x + 7)(x − 4) = 0
x = −7 is rejected (the integers are positive), so x = 4 and the integers are 4 and 7
Q11 · 5 and 9 [4]
The integers are x and x + 4, so x2 + (x + 4)2 = 106
x2 + x2 + 8x + 16 = 106 → 2x2 + 8x − 90 = 0
Divide by 2: x2 + 4x − 45 = 0 → (x + 9)(x − 5) = 0
x = −9 is rejected (the integers are positive), so the integers are 5 and 9
Q12 · width 7 cm, length 11 cm [4]
The length is w + 4, so w(w + 4) = 77
w2 + 4w − 77 = 0 → (w + 11)(w − 7) = 0
w = −11 is rejected (a length cannot be negative), so w = 7 cm
Length = 7 + 4 = 11 cm — check: 7 × 11 = 77 cm2 ✓
Q13 · 5 cm and 12 cm [4]
The shorter sides are x and x + 7, so by Pythagoras x2 + (x + 7)2 = 132
x2 + x2 + 14x + 49 = 169 → 2x2 + 14x − 120 = 0
Divide by 2: x2 + 7x − 60 = 0 → (x + 12)(x − 5) = 0
x = −12 is rejected (a length cannot be negative), so the sides are 5 cm and 12 cm — check: 25 + 144 = 169 ✓
Q14 · (3, 6) and (−3, −6) [4]
Substitute y = 2x: x2 + (2x)2 = 45
x2 + 4x2 = 45 → 5x2 = 45 → x2 = 9
x = 3 → y = 6; x = −3 → y = −6
(3, 6) and (−3, −6)
Q15 · (6, 5) and (−5, −6) [4]
x = y + 1, so (y + 1)2 + y2 = 61
y2 + 2y + 1 + y2 = 61 → 2y2 + 2y − 60 = 0
y2 + y − 30 = 0 → (y + 6)(y − 5) = 0
y = 5 → x = 6; y = −6 → x = −5 → (6, 5) and (−5, −6)
Q16 · (3, 5) and (−1, −3) [4]
Equate: x2 − 4 = 2x − 1
x2 − 2x − 3 = 0 → (x − 3)(x + 1) = 0
x = 3 → y = 2(3) − 1 = 5; x = −1 → y = −3
(3, 5) and (−1, −3)
Q17 · c = −4, touching at (3, 2) [4]
Equate: x2 − 4x + 5 = 2x + c
x2 − 6x + (5 − c) = 0 — exactly one solution means the discriminant is zero
(−6)2 − 4(1)(5 − c) = 0 → 36 − 20 + 4c = 0 → 4c = −16, so c = −4
The point is where x = 6 ÷ 2 = 3: y = 2(3) − 4 = 2, so the point is (3, 2)
Q18 · (1, 2) and (2, 1) [4]
Put the second equation over a common denominator: 1x + 1y = x + yxy
Since x + y = 3: 3xy = 32 → xy = 2
x and y are the roots of t2 − 3t + 2 = 0 → (t − 1)(t − 2) = 0, so t = 1 or 2
(1, 2) and (2, 1)
Q19 · k = 3, touching at (1, 6) [4]
Equate: x2 + x + 4 = 3x + k
x2 − 2x + (4 − k) = 0 — a tangent means exactly one solution, so the discriminant is zero
(−2)2 − 4(1)(4 − k) = 0 → 4 − 16 + 4k = 0 → 4k = 12, so k = 3
Contact is where x = 2 ÷ 2 = 1: y = 3(1) + 3 = 6, so the point is (1, 6)
Q20 · (3, 4) and (−4, −3) [4]
Substitute y = x + 1: x2 + (x + 1)2 = 25
x2 + x2 + 2x + 1 = 25 → 2x2 + 2x − 24 = 0
Divide by 2: x2 + x − 12 = 0 → (x + 4)(x − 3) = 0
x = 3 → y = 4; x = −4 → y = −3 → (3, 4) and (−4, −3)
