Exercise 2G
1 a i f(x) = x2 + 8x + 3 2 x2 + 6x + k = 0
b2 − 4ac = 82 − 4(1)(3) a = 1, b = 6 and c = k
= 64 − 12 For two real solutions, b2 − 4ac > 0
= 52 62 − 4 × 1 × k > 0
36 − 4k > 0
ii g(x) = 2x2 − 3x + 4 36 > 4k
b2 − 4ac = (−3)2 − 4(2)(4) 9>k
= 9 − 32 So k < 9
= −23
3 2x2 − 3x + t = 0
iii h(x) = −x2 + 7x − 3 a = 2, b = −3 and c = t
b2 − 4ac = 72 − 4(−1)(−3) For exactly one solution, b2 − 4ac = 0
= 49 − 12 (−3)2 − 4 × 2 × t = 0
= 37 9 − 8t = 0
9
So t = 8
iv j(x) = x2 − 8x + 16
b2 − 4ac = (−8)2 − 4(1)(16) 4 f(x) = sx2 + 8x + s
= 64 − 64 a = s, b = 8 and c = s
=0 For equal roots, b2 − 4ac = 0
82 − 4 × s × s = 0
v k(x) = 2x − 3x2 − 4 64 − 4s2 = 0
= − 3x2 + 2x − 4 64 = 4s2
b − 4ac = (2)2 − 4(−3)(−4)
2
16 = s2
= 4 − 48 So s = ±4
= −44 The positive solution is s = 4.
b i This graph has two distinct real roots 5 3x2 − 4x + k = 0
and has a maximum, so a < 0: h(x). a = 3, b = −4 and c = k
For no real solutions, b2 − 4ac < 0
ii This graph has two distinct real roots (−4)2 − 4 × 3 × k < 0
and has a minimum, so a > 0: f(x). 16 − 12k < 0
16 < 12k
iii This graph has no real roots and has a 4 < 3k
maximum, so a < 0: k(x).
So k > 34
iv This graph has one repeated root and has
a minimum, so a > 0: j(x). 6 a g(x) = x2 + 3px + (14p − 3)
a = 1, b = 3p and c = (14p – 3)
v This graph has no real roots and has a For two equal roots, b2 − 4ac = 0
minimum, so a > 0: g(x). (3p)2 − 4 × 1 × (14p − 3) = 0
9p2 − 56p + 12 = 0
(p − 6)(9p − 2) = 0
p = 6 or p = 92
p is an integer, so p = 6
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6 b When p = 6,
x2 + 3px + (14p − 3)
= x2 + 3(6)x + (14(6) − 3)
= x2 + 18x + 81
x2 + 18x + 81 = 0
(x + 9)(x + 9) = 0
So x = −9
7 a h(x) = 2x2 + (k + 4)x + k
a = 2, b = (k + 4) and c = k
b2 − 4ac = (k + 4)2 − 4 × 2 × k
= k2 + 8k + 16 − 8k = k2 + 16
b k2 ⩾ 0, therefore k2 + 16 is also > 0.
If b2 − 4ac > 0, then h(x) has two
distinct real roots.
Challenge
a For distinct real roots, b2 − 4ac > 0.
Therefore b2 > 4ac
If a > 0 and c > 0, or a < 0 and c < 0,
choose b such that b > 4ac .
If a > 0 and c < 0, or a < 0 and c > 0,
4ac < 0, therefore 4ac < b2 for all b.
b For equal roots, b2 − 4ac = 0.
Therefore b2 = 4ac.
If 4ac < 0, then there is no value for b to
satisfy b2 = 4ac as b2 is always positive.
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