Worked solutions
15 Analysis and approaches:
Quadratics
These are worked solutions to the colour-coded problem-solving questions from the exercises in
the Student’s Book. This excludes the drill questions.
Exercise 15A
38 a 𝑦-intercept 𝑐
b 𝑦
𝑥
24
8 𝑥
Roots at 𝑎
39 a Roots at
3
3 and 𝑏
2 and 3 so 𝑝
b 𝑦-intercept 𝑎𝑝𝑞
8
2, 𝑞
18 so 𝑎
3
3
c
𝑦
3 𝑥
3𝑥
2 𝑥 3
3𝑥 18
40 a
𝑥
5𝑥
1
5
2
5
2
𝑥
𝑥
1
5
2
21
4
b Line of symmetry is 𝑥
41 a 𝑦-intercept is at 0,23
b
𝑦
12𝑥 23
2𝑥
6𝑥
23
2 𝑥
9
23
2 𝑥 3
5
2 𝑥 3
c Vertex is at
3, 5 .
42 a Vertex is at ℎ, 𝑘
So ℎ
2, 𝑘
2,7
7
b 𝑦-intercept 13
𝑎ℎ
𝑘
4𝑎
7 so 𝑎
5
c
𝑦
7
5 𝑥 2
20𝑥 13
5𝑥
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6𝑥
3𝑥
2
3 𝑥
2𝑥
3 𝑥 1
3 𝑥 1
2
1
5
Worked solutions
43 a
2
b Minimum value is 5
44 a
10𝑥
5𝑥
3
5 𝑥
2𝑥
5 𝑥 1
5 𝑥 1
3
1
2
3
b Positive quadratic with minimum at 1, 2
Range is f 𝑥 ⩾
2
Exercise 15B
29 a 𝑥
𝑥
12
𝑥
4 𝑥
So 𝑥
4
b
𝑥
𝑥
3
4 𝑥
3
0
0 or 𝑥
3
0
4 or 3
30 a
𝑥
6𝑥
2
𝑥
𝑥
3
11
𝑥
3
3
9 2
11
b
𝑥
31 a 𝑥
10𝑥
𝑥
0
3
√11
5
25
b
𝑥
25
𝑥
5
7
5
5
√32
4√2
32
𝑥
𝑥
6𝑥 5
1 𝑥 5
0
0
Positive quadratic is less than zero between the roots
1
𝑥
5
33
𝑥
𝑥
𝑏 𝑥
𝑏
𝑏
0
0
Positive quadratic is greater than zero outside the roots
𝑥⩽
𝑏 or 𝑥 ⩾ 𝑏
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4𝑥
7 𝑥
𝑥
𝑥
𝑥
21
3
7 or 𝑥
35 𝐴
Worked solutions
34
0
0
3
𝑥 and 𝑃
Require 𝑥
4𝑥
4𝑥 and (in context) 𝑥
0
Since 𝑥 is positive, dividing by 𝑥 will not change the solution set.
𝑥
4.
Tip: Ordinarily, cancelling an equation by an expression in x can only be done if you also allow
the possibility that the expression equals zero (so cancelling a factor of 𝑥 𝑎 on both sides of
an equation should only be done if you note the possible solution 𝑥 𝑎). In this case, by stating
𝑥 0 we can then cancel the factor x and avoid solving the quadratic inequality altogether.
36
3𝑥
3 𝑥
3
2
3
𝑥
4𝑥
4
𝑥
3
4
9
3 𝑥
1
0
1
0
1
0
7
3
2
2
3
𝑥
√7
3
37
𝑥
𝑥
𝑝
2𝑥𝑝
𝑥 𝑝
𝑞 𝑥 𝑝
𝑝
𝑞
𝑞
𝑞
0
0
Positive quadratic is greater than zero outside the roots
𝑥
𝑝
𝑞 or 𝑥
𝑝
𝑞
38 a The two numbers are 𝑥 and 𝑥
3.
If their product is 40 then 𝑥 𝑥
3
𝑥
b
𝑥
𝑥
3𝑥
8 𝑥
40
5
40
0
0
8 (and the larger number is 5) or 𝑥
39 a If the width is 𝑥 then the length is 𝑥
𝑥
b
𝑥
𝑥
5𝑥
Area
24
8 𝑥
𝑥 𝑥
0
3
5
5 (and the larger number is 8).
5
24
0
8 (reject due to context) or 𝑥
The perimeter is 2𝑥
2 𝑥
5
3
22 cm
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𝑥
𝑥
2
𝑥
𝑥 6
𝑥
3 𝑥 2
0
0
4
Worked solutions
40 Substituting: 𝑥
3 or 2
2, 0 and 3, 5
Coordinates of intersection are
41
6𝑥 9
𝑥 3
𝑥
0
0
A square is only non-positive when it is zero. The inequality has solution 𝑥
3.
42 a
10𝑡 5𝑡
5𝑡 2 𝑡
0
0
The ball is at ground level at 𝑡
0 (when it is hit) and 𝑡
2 (when it lands).
The ball lands 2 seconds after being hit.
b
2𝑡
5 𝑡
5 𝑡 1
5 𝑡 1
ℎ
1
5
Maximum height is 5 m at 𝑡
1 second
c
5 𝑡
ℎ
5
1
5 𝑡
1
𝑡
1
Roots of 𝑡
1
1
4
5
4
are 𝑡
1
1
√
Positive quadratic is less than zero between the roots
The ball is
so for
√
1 m off the ground for 1
seconds
√
𝑡
√
1
,
√
1.79 seconds.
43 Substituting:
𝑥
𝑥 3
6𝑥 9 16
2𝑥
3𝑥
7
2 𝑥
3
9
2 𝑥
7
2
4
23
3
2 𝑥
2
2
𝑥
Coordinates are therefore
16
0
0
0
0
3
2
√23
2
√
,
√
,
√
,
√
.
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6
3𝑥
𝑥 1
3𝑥 4 𝑥 1
3𝑥
𝑥 10
3𝑥 5 𝑥 2
45 4
𝑥 and 𝑥
9, so
2 or 𝑥
2 and
𝑥
4
6
0
0
5
or
3
𝑥
Worked solutions
44
2
3
𝑥
3
The coincident regions of the two requirements are
46 𝑥
3𝑥 ⩽ 0 and 𝑥
𝑥 𝑥
5𝑥
3 ⩽ 0 and 𝑥
4
3
𝑥
2 or 2
𝑥
3
0
1 𝑥
4
0
Positive quadratics are less than zero between the roots.
0 ⩽ 𝑥 ⩽ 3 and 1
𝑥
4
The coincident region of the two requirements is 1
𝑥⩽3
47
𝑦
𝑦
3𝑥𝑦
𝑥 𝑦
𝑥
𝑥
𝑥
𝑥
2𝑥
2𝑥
𝑦
0
0
𝑥 or 2𝑥
48
2𝑘𝑥
𝑘
𝑘 𝑥
𝑘
𝑥
𝑘
𝑘
𝑘
1
𝑥
0
0
0
𝑘 or 𝑘
1
49
𝑥
Since
𝑦
𝑥
2𝑦 1
𝑦 1
0
0 and 𝑦
1
𝑥
0, the only possible solution is at 𝑥
0, 𝑦
1.
Exercise 15C
10 a Repeated root when Δ
8
4 3 𝑘
12𝑘
𝑘
0
0
64
16
3
b 𝑥
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4 5 20
𝑘
𝑘
𝑘
0
0
400
20
12 Distinct real roots when Δ
3
4 𝑘 2
9 8𝑘
13 No real roots when Δ
0
4 3 2𝑘
25 24𝑘
0
0
25
24
𝑘
14 a Repeated root when Δ
𝑘
4 2 𝑘
8𝑘
𝑘
0
0
0
9
8
𝑘
5
Worked solutions
11 Repeated root when Δ
2
16
0
0
0
b
𝑘
0
4
4
𝑘
15 a Repeated root when Δ
𝑘
3
𝑘
4 𝑘
1
10𝑘 9
0
0
0
b
𝑘
𝑘
1 𝑘
9
0
1 or 9
16 Distinct real roots when Δ
𝑘
4 2 2
16
𝑘
𝑘
4 or 𝑘
0
0
0
4
17 At least one real root when Δ ⩾ 0
𝑎
4 𝑎 3 ⩾0
12𝑎 ⩾ 0
𝑎
𝑎 𝑎 12 ⩾ 0
Positive quadratic is greater than zero outside the roots
𝑎 ⩽ 0 or 𝑎 ⩾ 12
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4 1 9
𝑎
𝑎
𝑎
0
0
36
6
19 No real roots when Δ
3
0
4 2 𝑐
9 8𝑐
0
0
9
8
𝑐
20 No real roots when Δ
4 1 2𝑏
8𝑏
𝑏
𝑏 𝑏 8
𝑏
Worked solutions
18 Repeated root when Δ
0
0
0
0
Positive quadratic is less than zero between the roots
0
𝑏
8
21 Distinct real roots when Δ
𝑎
4 3 4
𝑎 1
1
𝑎
1
𝑎
0
48
√48 or 𝑎
1
1
4√3 or 𝑎
22 No real roots when Δ
√48
1
4√3
0
4 𝑘 𝑘
𝑘
3 𝑘
2𝑘
3 3𝑘 3 𝑘
3
0
0
0
0
Negative quadratic is less than zero outside the roots
𝑘
23 𝑥
3 or 𝑘
1
0 is not a solution, so we can multiply by 𝑥.
3𝑥
𝑥
𝑎
0
Real roots when Δ ⩾ 0
3
4 1 𝑎 ⩾0
9
𝑎⩽
4
24 For the graph to be always positive, the quadratic must be positive with no real roots.
No real roots when Δ
𝑎
𝑎
4 1 4
16
𝑎
4 𝑎 4
0
0
0
0
Positive quadratic is less than zero between the roots
4
𝑎
4
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Negative quadratic: 𝑎
0
No real roots when Δ
0
4
4 𝑎 𝑎
12𝑎
4𝑎
3𝑎
4 𝑎
𝑎 1 𝑎
3
16
4
4
Worked solutions
25 For the graph to lie entirely below the 𝑥-axis, the quadratic must be negative with no real
roots.
1
0
0
0
0
Positive quadratic is less than zero between the roots
1
𝑎
4
2
Taking these two conditions together 1
𝑎
0
26
Tip: This could be answered using calculus and gradients, but it is elegant to solve by
considering the number of intersection points.
If the line is tangent to the curve then there will be a single (repeated) point of intersection.
𝑥
7
𝑥
𝑥
𝑘
𝑘
𝑥
0
7
Single repeated root when Δ
1
4 1 7
𝑘
7
0
0
1
4
27
4
𝑘
𝑘
2𝑥
27 Intersection occurs when 𝑥
5𝑥
4𝑘𝑥
𝑘
5
𝑘
5
0
Real roots when Δ ⩾ 0. When Δ 0 the line would be tangent to the circle (non properly
intersecting), so restrict to Δ 0 for intersections.
4𝑘
4 5 𝑘
4𝑘
5
100
𝑘
5
0
0
25
𝑘 5
28 Distinct real roots when Δ
Δ
𝑎
0
4 3
2
2
𝑎
2
24
This must be greater than zero for all real 𝑎, and hence there must always be two real roots
to the original quadratic.
𝑏𝑥 9 0 to have no solutions, the quadratic 𝑦 𝑥
𝑏𝑥 9 must lie entirely
29 For 𝑥
above the 𝑥-axis; that is, it has no real roots. Let the discriminant of this quadratic be Δ .
No real roots when Δ
Δ
𝑏
Therefore
0
4 1 9
6
𝑏
𝑏
36
0
6
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Δ
9
𝑥
9𝑥
81
4𝑏
4 1 𝑏
𝑏 be Δ
Worked solutions
Let the discriminant of 𝑦
Given the values of 𝑏 found above,
57
Δ
105
Therefore Δ
0 and hence the second quadratic always has two distinct real roots.
Mixed Practice
1 a 𝑥
7𝑥
18
𝑥
9 𝑥
2
9,0 and 2,0
b
2
2𝑥
9
3
4𝑥
3 2𝑥
0
0
𝑥
2
3
,0 .
Coordinates of the roots are
3 a
8𝑥
2𝑥
3
2 𝑥
4𝑥
2 𝑥 2
2 𝑥 2
3
4
5
3
b Vertex is 2, 5 .
4 a Positive quadratic with negative 𝑦-intercept: Graph 3.
b Positive quadratic with positive 𝑦-intercept: Graph 1.
c Negative quadratic: Graph 2.
5 a Vertex is at 3, 2 so 𝑎
3, 𝑏
2
b
𝑥
2
3
3
𝑥
√2
Coordinates of the roots are 3
6 a 𝑦
𝑥
5 𝑥
Roots are at
b 𝑦
𝑥
1
√2, 0
3
3, 0 and 5, 0
16
Line of symmetry is 𝑥
1
c Minimum value is 16, the 𝑦-coordinate of the vertex.
7 a
2𝑥
b Δ
𝑥
5
Δ
0
1
4 2
5
41
0 so there are two distinct real roots.
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4 𝑘 5
4
0
Worked solutions
8 No real roots when Δ
0
4
5
𝑘
9 a Axis of symmetry is vertical through the vertex: 𝑥
2
b
c By symmetry, if one root is 1 unit to the left of the axis, the other must be one unit to the
right, so 𝐷 has 𝑥-coordinate 3.
10 a
6𝑥
3𝑥
10
3 𝑥
2𝑥
3 𝑥 1
3 𝑥 1
10
1
10
7
b Positive quadratic with vertex at 1, 7 has range f 𝑥 ⩾ 7
11 a
4𝑥
𝑥
3
𝑥
4𝑥
𝑥 2
𝑥 2
3
4
7
2, 7 has range f 𝑥 ⩽ 7
b Negative quadratic with vertex at
12 Equal roots when Δ
𝑘
2
3
0
4 2 3
𝑘 2
𝑘
0
24
2
√24
2
2√6
𝑥
6𝑥
13 a Let the length be 𝑦.
Then 2𝑥
2𝑦
12 so 𝑦
Then the area is given by 𝐴
6
𝑥
𝑥𝑦
𝑥 6
𝑥
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𝐴
𝑥
9
6𝑥
𝑥 3
𝑥 3
Worked solutions
b
9
So the maximum area is 9 cm
14 At least one real root when Δ ⩾ 0
4 3 6 ⩾0
𝑘 ⩾ 72
6√2 ⩽ 𝑘 ⩽ 6√2
𝑘
15 a i Completing the square:
f 𝑥
2 𝑥
2
𝑥
2 𝑥
Vertex is at
𝑘
𝑘
𝑥
2
𝑘
4
4
𝑘
16
32
𝑘
4
4
𝑘
8
,
so
1.25
5
ii The 𝑦-coordinate is
b
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Worked solutions
16
a f 𝑥
10 𝑥
4 𝑥
6
b Expanding and then completing the square:
f 𝑥
10
10
10
10
𝑥
2𝑥
2𝑥
𝑥
𝑥 1
𝑥 1
24
240
1
240
250
c Expanding fully:
f 𝑥
10 𝑥
10 𝑥
10𝑥
2𝑥
2𝑥
20𝑥
24
240
240
17 a
Δ
10
𝑝
100
100
5𝑝
4
5𝑝
4 𝑝
20𝑝
4𝑝
𝑝
b Equal roots when Δ
4𝑝
𝑝
100
5
6𝑥
15
5
20𝑝
0
18 a
𝑥
𝑥
𝑥
3
3
9
6
15
b This function has minimum value 6, so its reciprocal has maximum value
Then
has maximum value
19 When the line is tangent, the intersection of line and circle will have only one root.
Substituting for intersection: 𝑥
5𝑥
4𝑐𝑥
𝑐
3
2𝑥
𝑐
3
0
0
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4𝑐
0
4 5 𝑐
4𝑐
3
60
𝑐
𝑐
20 Multiplying through by 𝑥
𝑎𝑥
2𝑥
1
21 𝑦
2𝑥
Δ
4 𝑎 1
4𝑎
𝑎
4𝑘𝑥
4𝑘
16𝑘
8𝑘
0
0
15
√15
0 and rearranging:
0
No real roots when Δ
2
Worked solutions
Single root when Δ
0
0
4
1
3𝑘 is a positive quadratic.
4 2 3𝑘
24𝑘
0 for all 𝑘
Since Δ 0, it follows that there are no real roots to the quadratic, and therefore it lies
entirely above the 𝑥-axis.
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