Algebra III – Quadratic Relations Practice Questions 4.1 - Gill ...

Destination Maths Chapter 4 Algebra III – Quadratic Relations

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Algebra III – Quadratic Relations

Practice Questions 4.1

1. Graph the following functions in the given domains.

(i) f (x) = x2 − 4x − 5 in the domain − 2 ≤ x ≤ 6.

f (x) = x2 – 4x – 5

f (–2) = (–2)2 – 4(–2) – 5 = 7 (–2, 7)

f (–1) = (–1)2 – 4(–1) – 5 = 0 (–1, 0)

f (0) = (0)2 – 4(0) – 5 = –5 (0, –5)

f (1) = (1)2 – 4(1) – 5 = –8 (1, –8)

f (2) = (2)2 – 4(2) – 5 = –9 (2, –9)

f (3) = (3)2 – 4(3) – 5 = –8 (3, –8)

f (4) = (4)2 – 4(4) – 5 = –5 (4, –5)

f (5) = (5)2 – 4(5) – 5 = 0 (5, 0)

f (6) = (6)2 – 4(6) – 5 = 7 (6, 7)

(ii) f (x) = 2x2 − 12x + 9 in the domain 0 ≤ x ≤ 6.

f (x) = 2x2 – 12x + 9

f (0) = 2(0)2 – 12(0) + 9 = 9 (0, 9)

f (1) = 2(1)2 – 12(1) + 9 = –1 (1, –1)

f (2) = 2(2)2 – 12(2) + 9 = –7 (2, –7)

f (3) = 2(3)2 – 12(3) + 9 = –9 (3, –9)

f (4) = 2(4)2 – 12(4) + 9 = –7 (4, –7)

f (5) = 2(5)2 – 12(5) + 9 = –1 (5, –1)

f (6) = 2(6)2 – 12(6) + 9 = 9 (6, 9)


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(iii) f (x) = 7 − 3x2 in the domain −2 ≤ x ≤ 2.

f (x) = 7 – 3x2

f (–2) = 7 – 3(–2)2 = –5 (–2, –5)

f (–1) = 7 – 3(–1)2 = 4 (–1, 4)

f (0) = 7 – 3(0)2 = 7 (0, 7)

f (1) = 7 – 3(1)2 = 4 (1, 4)

f (2) = 7 – 3(2)2 = –5 (2, –5)

(iv) f (x) = − x2 + 3x + 12 in the domain −3 ≤ x ≤ 4.

f (x) = –x2 + 3x + 12

f (–3) = –(–3)2 + 3(–3) + 12 = –6 (–3, –6)

f (–2) = –(–2)2 + 3(–2) + 12 = 2 (–2, 2)

f (–1) = –(–1)2 + 3(–1) + 12 = 8 (–1, 8)

f (0) = –(0)2 + 3(0) + 12 = 12 (0, 12)

f (1) = –(1)2 + 3(1) + 12 = 14 (1, 14)

f (2) = –(2)2 + 3(2) + 12 = 14 (2, 14)

f (3) = –(3)2 + 3(3) + 12 = 12 (3, 12)

f (4) = –(4)2 + 3(4) + 12 = 8 (4, 8)


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(v) f (x) = −2x2 − 4x + 9 in the domain −4 ≤ x ≤ 1.

f (x) = –2x2 – 4x + 9

f (–4) = –2(–4)2 – 4(–4) + 9 = –7 (–4, –7)

f (–3) = –2(–3)2 – 4(–3) + 9 = 3 (–3, 3)

f (–2) = –2(–2)2 – 4(–2) + 9 = 9 (–2, 9)

f (–1) = –2(–1)2 – 4(–1) + 9 = 11 (–1, 11)

f (0) = –2(0)2 – 4(0) + 9 = 9 (0, 9)

f (1) = –2(1)2 – 4(1) + 9 = 3 (1, 3)

2. (i) Graph the function g(x) = x2 + 6x + 2 in the domain −8 ≤ x ≤ 2.

g(x) = x2 + 6x + 2

g(–8) = (–8)2 + 6(–8) + 2 = 18

g(–7) = (–7)2 + 6(–7) + 2 = 9

g(–6) = (–6)2 + 6(–6) + 2 = 2

g(–5) = (–5)2 + 6(–5) + 2 = –3

g(–4) = (–4)2 + 6(–4) + 2 = –6

g(–3) = (–3)2 + 6(–3) + 2 = –7

g(–2) = (–2)2 + 6(–2) + 2 = –6

g(–1) = (–1)2 + 6(–1) + 2 = –3

g(0) = (0)2 + 6(0) + 2 = 2

g(1) = (1)2 + 6(1) + 2 = 9

g(2) = (2)2 + 6(2) + 2 = 18


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(ii) Use your graph to estimate the roots of g(x).

The roots are the points where the graph crosses the x-axis.

This occurs at x = –5·65 and x = –0·35

Acceptable answers would be: x = –5·65 ± 0·1 and x = –0·35 ± 0·1

3. A function k is defined as k: x → x2 − 4x + 4.

(i) Find k(−1), k(0), k(1), k(2), k(3), k(4)

k : x → x2 – 4x + 4

k(–1) = (–1)2 – 4(–1) + 4 = 9

k(0) = (0)2 – 4(0) + 4 = 4

k(1) = (1)2 – 4(1) + 4 = 1

k(2) = (2)2 – 4(2) + 4 = 0

k(3) = (3)2 – 4(3) + 4 = 1

k(4) = (4)2 – 4(4) + 4 = 4

(ii) Use your answers to part (i) to graph the function k(x).

(iii) Use your graph to estimate the roots of k(x).

The graph has a turning point at the point (2, 0)

This means that it has two roots at this value. So the roots are x = 2, x = 2


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4. (i) Graph the function ℎ(x) = −3x2 + 12x + 5 in the domain −1 ≤ x ≤ 5.

h(x) = –3x2 + 12x + 5

h(–1) = –3(–1)2 + 12(–1) + 5 = –10

h(0) = –3(0)2 + 12(0) + 5 = 5

h(1) = –3(1)2 + 12(1) + 5 = 14

h(2) = –3(2)2 + 12(2) + 5 = 17

h(3) = –3(3)2 + 12(3) + 5 = 14

h(4) = –3(4)2 + 12(4) + 5 = 5

h(5) = –3(5)2 + 12(5) + 5 = –10

(ii) Use your graph to estimate the roots of h(x).


This occurs at x = – 0·38 and x = 4·38

Acceptable answers would be: x = – 0·38 ± 0·1 and x = 4·38 ± 0·1

(iii) List the elements of the domain and range.

Domain → x-values = – 1, 0, 1, 2, 3, 4, 5

Range → y-values = – 10, 5, 14, 17


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5. (i) Graph the function f (x) = x2 − 4x + 5 in the domain − 1 ≤ x ≤ 5.

f(x) = x2 – 4x + 5

f(–1) = (–1)2 – 4(–1) + 5 = 10

f(0) = (0)2 – 4(0) + 5 = 5

f(1) = (1)2 – 4(1) + 5 = 2

f(2) = (2)2 – 4(2) + 5 = 1

f(3) = (3)2 – 4(3) + 5 = 2

f(4) = (4)2 – 4(4) + 5 = 5

f(5) = (5)2 – 4(5) + 5 = 10

(ii) Use your graph to determine the nature of the roots of f (x).


This graph does not cross the x-axis. Therefore, it has no real roots.

6. Sketch the functions which have the following properties:

(i) Positive x2 and roots − 1 and 4

Positive x2 means the graph is U-shaped.

Roots – 1 and 4 means it crosses x-axis at – 1 and 4.

Sketch:


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(ii) Negative x2 and roots 2 and 5∙5

Negative x2 means the graph is ∩-shaped.

Roots at 2 and 5·5 means it crosses the x-axis at 2 and 5·5.

Sketch:

(iii) Negative x2 and roots − 3∙5 and 0

Negative x2 means the graph is ∩-shaped.

Roots at –3·5 and 0 means it crosses the x-axis at –3·5 and 0.

Sketch:

(iv) Positive x2 and roots − 6 and − 0∙5

Positive x2 means the graph is U-shaped.

Roots at – 6 and – 0·5 means it crosses the x-axis at – 6 and – 0·5.

Sketch:


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(v) Positive x2 and roots 1 and 1

Positive x2 means the graph is U-shaped

Roots at 1 and 1 means the graph does not cross the x-axis.

It instead has a turning point at x = 1.

Sketch:


1. The diagram shows the graph of the function

f (x) = x2 − x − 8 in the domain −3 ≤ x ≤ 4.

Use the graph to answer the following questions:


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(i) Find the values of x for which f (x) = 4.

Find x when f(x) = 4:

• Go to 4 on the y-axis.

• Draw a line horizontally across until you touch the graph.

• Draw vertical lines down to the x-axis.

• Read the values off the x-axis

When f(x) = 4, x = –3 and x = 4


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(ii) Estimate the values of x for which f (x) = −4.

Estimate x when f(x) = –4:

• Go to –4 on the y-axis.


• Draw vertical lines up to the x-axis.

• Read the values off the x-axis

When f(x) = –4, x = –1·56 and x = 2·56

Acceptable answers would be: x = –1·56 ± 0·1 and x = 2·56 ± 0·1


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(iii) Estimate the value of f (2·5).

Estimate f(2·5):

• Go to 2·5 on the x-axis.

• Draw a line vertically downwards until you touch the graph.

• Draw a horizontal line across to the y-axis.

• Read the value off the y-axis

f(2·5) = –4·25

Acceptable answers would be: –4·25 ± 0·1


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(iv) Estimate the value of f (−0∙5).

Estimate f(–0·5):

• Go to –0·5 on the x-axis.

• Draw a line vertically downwards until you touch the graph.



f(–0·5) = –7·25

Acceptable answers would be: –7·25 ± 0·1


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(v) Estimate the roots of f (x) (i.e. the values of x for which f (x) = 0).

Estimate the roots of f(x):

• Go to the x-axis.?

• Find the two points where the graph is crossing the x-axis.

• Read these values off the x-axis

Roots: x = – 2·37, 3·37

Acceptable answers would be: x = – 2·37 ± 0·1 and x = 3·37 ± 0·1


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(vi) Estimate the coordinates of the minimum point on the graph.

Estimate the coordinates of the minimum point on the curve:

• Go to the lowest point on the curve.

• Read its x and y coordinates off the axes.

Minimum point = (0·5, – 8·25)

Acceptable answers would be: x = 0·5 ± 0·1 and y = – 8·25 ± 0·1

2. The diagram shows the graph of the function

g(x) = −2x2 + 3x + 6 in the domain −1∙5 ≤ x ≤ 3.

Use the graph to answer the following questions:

(i) Estimate the values of x for which g(x) = 2.

Find x when g(x) = 2:


• Draw a line horizontally across until you

touch the graph.

• Draw vertical lines down to the x-axis.

• Read the values off the x-axis.


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When g(x) = 2, x = – 0·85 and x = 2·35

Acceptable answers would be: x = – 0·85 ± 0·1 and y = 2·35 ± 0·1

(ii) Estimate the value of g(1∙5).

Estimate g(l∙5):

• Go to 1∙5 on the x-axis.

• Draw a line vertically upwards until you touch the graph.



g(1∙5) = 6


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(iii) Estimate the value of g(−0∙5).

Estimate g(–0∙5):

• Go to – 0∙5 on the x-axis.




g(– 0∙5) = 4

(iv) Estimate the roots of g(x).

The roots of g(x) are the points where the graph crosses the x-axis.

Roots: − 1·1 and 2·6


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(v) Estimate the coordinates of the maximum point on the curve:

• Go to the highest point on the curve.

• Read its x and y coordinates off the axes.

Maximum point = (0∙75, 7∙125)

Acceptable answers would be: x = 0·75 ± 0·1 and y = 7·125 ± 0·1

3. Graph the function ℎ(x) = −3x2 + 2x + 7 in the domain −2 ≤ x ≤ 3.

h(x) = –3x2 + 2x + 7

h(–2) = –3(–2)2 + 2(–2) + 7 = –9

h(–1) = –3(–1)2 + 2(–1) + 7 = 2

h(0) = –3(0)2 + 2(0) + 7 = 7

h(1) = –3(1)2 + 2(1) + 7 = 6

h(2) = –3(2)2 + 2(2) + 7 = –1

h(3) = –3(3)2 + 2(3) + 7 = –14


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Use your graph to answer the following questions:

(i) Estimate the values of x for which ℎ(x) = −2.

Find x when h(x) = −2:

• Go to −2 on the y-axis.


• Draw vertical lines up to the x-axis.


When h(x) = −2, x = –1·43 and x = 2·1

Acceptable answers would be: x = –1·43 ± 0·1 and x = 2·1 ± 0·1

(ii) Estimate the value of ℎ(0∙5).

Estimate ℎ(0∙5):

• Go to 0∙5 on the x-axis.





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h(0·5) = 7·25

Acceptable answers would be: 7·25 ± 0·1

(iii) Estimate the value of ℎ(−0∙5).

Estimate ℎ(−0∙5):

• Go to −0∙5 on the x-axis.





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h(−0·5) = 5·25

Acceptable answers would be: 5·25 ± 0·1

(iv) Estimate the roots of ℎ(x).

The roots of h(x) are the points where the graph crosses the x-axis.

Roots: − 1·23 and 1·9

Acceptable answers would be: − 1·23 ± 0·1 and 1·9 ± 0·1


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(v) Estimate the coordinates of the maximum point on the graph.

The maximum point is the highest point on the graph:

( )1 22, 0 33, 7 333 3

⇒ ⋅ ⋅

Acceptable answers would be: x = 0·33 ± 0·1 and y = 7·33 ± 0·1

4. Dhafir invests in an investment fund. The value of the fund, v, in hundreds of euro,

after t months of investment, is given by: v(t) = 50 + 73t − 3t2.

(i) Copy and complete the following table:

t(months) 0 4 8 12 16 20 24 28

v(t) (hundreds of

euro)

t 0 4 8 12 16 20 24 28

v(t) 50 294 442 494 450 310 74 – 258


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(ii) Hence, draw the graph of the v(t) function.

(iii) What was the initial sum of money that Dhafir invested?

The initial sum is when 𝑡𝑡 = 0. From the graph or the table this is 𝑣𝑣 = 50.

This represents a sum of 50 × 100 = €5,000


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(iv) Use your graph to find the value of Dhafir’s fund after 10 months.

• Go to 10 on the t-axis.




v(10) = 480 × 100 = €48,000


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(v) After how many months is Dhafir’s fund at a value of €25,000 for the

second time? Give your answer to the nearest month.


• Draw a line horizontally until you touch the graph for a second time.

• Draw a vertical line down to the t-axis.

• Read the value off the t-axis.

t = 21·19 month 21 months, to the nearest month.


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(vi) Estimate the maximum value that Dhafir’s fund reaches.

The maximum value is the highest value the graph reaches.

Maximum value = €494 × 100 = €49,400


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(vii) After how many months is Dhafir’s fund worthless? Give your answer to

the nearest month.

The fund will be worthless when the value is at €0.

This occurs at 25 months.


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5. A company makes a weekly profit of p euro by selling x items, according to the equation:

p(x) = − 0∙5x2 + 40x


x items sold (in 100s)

0 5 10 15 20 25 30 35 40

p(x) (profit in 1,000s

of euro)

x 0 5 10 15 20 25 30 35 40

p(x) 0 187·5 350 487·5 600 687·5 750 787·5 800

(ii) Draw a graph of the profit function, p(x), in the domain 0 ≤ x ≤ 40.


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(iii) Use your graph to estimate the profit made on the sale of 2,200 items.

2,200 ÷ 100 = 22

• Go to 22 on the x-axis.




y = 638

∴ p = €638,000


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(iv) Use your graph to estimate the number of items the company would sell, in

order to make a profit of €450,000.

450,000 ÷ 1000 = 450



• Draw a vertical line down to the x-axis.

• Read the value off the x-axis

x = 13·5

∴ x = 1350 items


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6. During training, a tennis player hits a ball straight up

into the air.

The height of the ball, ℎ, in metres, after t seconds is

modelled by the following quadratic function:

ℎ(t) = −12t2 + 36t + 1


t(sec) 0 0·5 1 1·5 2 2·5 3

h(t) (m)

t(sec) 0 0·5 1 1·5 2 2·5 3

h(t)(m) 1 16 25 28 25 16 1

(ii) Hence, draw the graph of the ℎ(t) function.

Using your graph:

(iii) find the maximum height the ball reached

The highest the graph reaches is 28.

Therefore, the maximum height of the ball is 28 m

(iv) estimate how long it takes until the ball hits the ground

The ball reaches the ground at the point here the height is zero.

h(t) = 0 when t = 3·03 sec.


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(v) estimate the two times when the height of the ball is 10 m.

Give your answers to one decimal place.


• Draw a line horizontally until you touch the graph.

• Draw vertical lines down to the t-axis.

• Read the value off the t-axis

t = 0·28 sec, t = 2·72 sec

Acceptable answers would be: t = 0·28 ± 0·1 and t = 2·72 ± 0·1

7. A diver jumps off a spring board. He starts by going up

into the air and then dives down into the swimming pool

below.

The height of the diver, ℎ, in metres, above the water in

the pool, t seconds after he leaves the spring board, is modelled by the following

quadratic function:

22 12

( ) 105 5

h t t t= − + +


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(i) Graph the function ℎ(t) in the domain 0 ≤ t ≤ 9.

Use your graph to solve the following:

t 0 1 2 3 4 5 6 7 8 9

h(t) 10 12 13·2 13·6 13·2 12 10 7·2 3·6 – 0·8

(ii) Estimate the maximum height that the diver reaches. From the graph, we can see the highest point is when t = 3 seconds.

That is a maximum height = 13·6 m (iii) Find the height of the diver after five seconds.

From the graph, or the table, we can see that when t = 5, the height is 12 m. Therefore, h(5) = 12 m

(iv) Estimate the height of the diver after 7∙5 seconds. Go to 7·5 on the t-axis Draw a vertical line upwards until you touch the graph Draw a horizontal line across from this point to the y-axis. Read the value off the y-axis

h(7·5) = 5·5 m Acceptable answers would be: 5·5 ± 0·1


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(v) How long is the diver in the air for? Give your answer to one decimal place.

To find the time for which the diver is in the air, find the time when the

height is 0 m. Height = 0 m when 𝑡𝑡 = 8·8 sec, therefore the diver is in the

air for 8·8 seconds.

8. The average daytime temperature, T, in Paris can be modelled

by:

T(m) = −0∙8m2 + 10m – 1

where T(m) represents the average temperature (in degrees

celsius) in Paris during month m of the year.

(i) Given that m = 1 represents the start of January, find the average

temperature at the start of each month for a year and hence draw the graph

of this model.

Use your graph to solve the following:

m 1 2 3 4 5 6 7 8 9 10 11 12

T(m) 8·2 15·8 21·8 26·2 29 30·2 29·8 27·8 24·2 19 12·2 3·8

(ii) Write down the value of T(4) and explain its meaning in words.

T(4) = 26·2°C. This means that the average temperature at the start of

April (month 4) is 26·2°C


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(iii) During which two months is the average temperature 18°C?

• Go to 18 on the y-axis

• Draw a horizontal line across until you touch the graph

• Draw vertical lines from these points down to the x-axis


T(m) = 18°C when m = 2·3 and m = 10·2

m = 2 is the start of February, so m = 2·3 is during February

m = 10 is the start of October, so m = 10·2 is during October

(iv) Estimate the highest average temperature that is reached in Paris and state

during which month it occurs.

The highest average temperature is the highest point that the graph reaches.

This is a temperature of 30·25°C, which occurs during month 6 = June.


1. Factorise and hence solve the following quadratic equations:

(i) 8x2 – 16x = 0

8x2 – 16x = 0

8x (x – 2) = 0

∴ 8x = 0 and x – 2 = 0

x = 0 and x = 2


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(ii) 6x2 + 3x = 0

6x2 + 3x = 0

3x(2x + 1) = 0

∴ 3x = 0 and 2x + 1 = 0

x = 0 2x = –1

1 2

x = −

(iii) 16x2 – 9 = 0

16x2 – 9 = 0

(4x)2 – (3)2 = 0

(4x – 3)(4x + 3) = 0

∴ 4x – 3 = 0 and 4x + 3 = 0

4x = 3 and 4x = –3

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x = and 34

x = −

(iv) p2 – 8p + 12 = 0

p2 – 8p + 12 = 0

(p – 2)(p – 6) = 0

∴ p – 2 = 0 and p – 6 = 0

p = 2 and p = 6

(v) 4x2 – 36 = 0

4x2 – 36 = 0

(2x)2 – (6)2 = 0

(2x – 6)(2x + 6) = 0

∴ 2x – 6 = 0 and 2x + 6 = 0

2x = 6 and 2x = –6

x = 3 and x = –3.


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(vi) a2 – a – 12 = 0

a2 – a – 12 = 0

(a – 4)(a + 3) = 0

∴ a – 4 = 0 and a + 3 = 0

a = 4 and a = –3.

(vii) 5k2 – 35k = 0

5k2 – 35k = 0

5k (k – 7) = 0

∴ 5k = 0 and k – 7 = 0

k = 0 and k = 7

(viii) b2 + 10b + 16 = 0

b2 + 10b + 16 = 0

(b + 2)(b + 8) = 0

∴ b + 2 = 0 and b + 8 = 0

b = –2 and b = –8

(ix) x2 + 15x + 36 = 0

x2 + 15x + 36 = 0

(x + 3)(x + 12) = 0

∴ x + 3 = 0 and x + 12 = 0

x = –3 x = –12.

2. Factorise and hence solve the following quadratic equations:

(i) 2c2 – 5c – 12 = 0

2c2 – 5c – 12 = 0

(2c + 3)(c – 4) = 0

∴ 2c + 3 = 0 and c – 4 = 0

2c = –3 c = 4

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c = −


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(ii) 3x2 + 10x + 8 = 0

3x2 + 10x + 8 = 0

(3x + 4)(x + 2) = 0

∴3x + 4 = 0 and x + 2 = 0

3x = – 4 and x = –2

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x = −

(iii) 11t2 – 32t – 3 = 0

11t2 – 32t – 3 = 0

(11t + 1)(t – 3) = 0

∴ 11t + 1 = 0 and t – 3 = 0

11t = – 1 and t = 3

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t = −

(iv) 3a2 + 10a – 8 = 0

3a2 + 10a – 8 = 0

(3a – 2)(a + 4) = 0

∴3a – 2 = 0 and a + 4 = 0

3a = 2 and a = – 4

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a =

(v) 4x2 + 10x + 6 = 0

4x2 + 10x + 6 = 0

(2x + 3)(2x + 2) = 0

∴ 2x + 3 = 0 and 2x + 2 = 0

2x = – 3 and 2x = – 2

3

2

x = − and x = –1


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(vi) 4t2 + 12t + 9 = 0

4t2 + 12t + 9 = 0

(2t + 3)(2t + 3) = 0

∴ 2t + 3 = 0

2t = –3

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t = − one root only.

(vii) 4x2 – 23x + 15 = 0 4x2 – 23x + 15 = 0 (4x – 3)(x – 5) = 0 ∴ 4x – 3 = 0 and x – 5 = 0 4x = 3 and x = 5

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x =

(viii) 9a2 + 12a + 4 = 0

9a2 + 12a + 4 = 0

(3a + 2)(3a + 2) = 0

∴ 3a + 2 = 0

3a = – 2

2

3

a = − one root only

(ix) 12x2 + 32x + 5 = 0

12x2 + 32x + 5 = 0

(6x + 1)(2x + 5) = 0

∴ 6x + 1 = 0 and 2x + 5 = 0

6x = – 1 and 2x = –5

1

6

x = − and 5

2

x = −


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3. Rearrange the following quadratic equations into the form ax2 + bx + c = 0, and

hence factorise and solve.

(i) 4x2 – 46 = 3

4x2 – 46 = 3 subtract 3 from both sides

4x2 – 49 = 0

(2x)2 – (7)2 = 0

(2x – 7)(2x + 7) = 0

∴ 2x – 7 = 0 and 2x + 7 = 0

2x = 7 and 2x = –7

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x = and 72

x = −

(ii) k2 – 7 = 9

k2 – 7 = 9 subtract 9 from both sides

k2 – 16 = 0

(k)2 – (4)2 = 0

(k – 4)(k + 4) = 0

∴ k – 4 = 0 and k + 4 = 0

k = 4 and k = – 4

(iii) 6x2 = 24x

6x2 = 24x subtract 24x from both sides

6x2 – 24x = 0

6x (x – 4) = 0

∴ 6x = 0 and x – 4 = 0

x = 0 and x = 4


40

(iv) 7x2 = 15x – 2

7x2 = 15x – 2 subtract 15x from and add 2 to both sides

7x2 – 15x + 2 = 0

(7x – 1)(x – 2) = 0

∴ 7x – 1 = 0 and x – 2 = 0

7x = 1 and x = 2

17

x =

(v) 7k = 15 – 2k2

7k = 15 – 2k2 add 2k2 to and subtract 15 from both sides

2k2 + 7k – 15 = 0

(2k – 3)(k + 5) = 0

∴ 2k – 3 = 0 and k + 5 = 0

2k = 3 and k = –5

32

k =

(vi) 2x2 = 3x + 2

2x2 = 3x + 2 subtract 3x and 2 from both sides

2x2 – 3x – 2 = 0

(2x + 1)(x – 2) = 0

∴ 2x + 1 = 0 and x – 2 = 0

2x = –1 and x = 2

1

2

x = −


41

(vii) 9a2 = 73a – 8

9a2 = 73a – 8 subtract 73a from and add 8 to both sides

9a2 – 73a + 8 = 0

(9a – 1)(a – 8) = 0

∴ 9a – 1 = 0 and a – 8 = 0

9a = 1 and a = 8

19

a =

(viii) 9h2 + 4 = 15h

9h2 + 4 = 15h subtract 15h from both sides

9h2 – 15h + 4 = 0

(3h – 4)(3h – 1) = 0

∴ 3h – 4 = 0 and 3h – 1 = 0

3h = 4 and 3h = 1

43

h = and 13

h =

(ix) 5x2 – 11x – 3 = 2x + 3

5x2 – 11x – 3 = 2x + 3 subtract 2x and 3 from both sides

5x2 – 13x – 6 = 0

(5x + 2)(x – 3) = 0

∴ 5x + 2 = 0 and x – 3 = 0

5x = –2 and x = 3

2

5

x = −


42

4. Use the quadratic formula to solve the following quadratic equations. Leave your answer in simplest surd form. (i) x2 + 2x – 2 = 0 x2 + 2x – 2 = 0 ⇒ a = 1, b = 2, c = –2

2

2

42

2 2 4(1)( 2)2(1)

2 4 ( 8)2

2 122

2 2 32

1 3

1 3 and 1 3

b b acxa

x x

− ± −=

− ± − −=

− ± − −=

− ±=

− ±=

= − ±

= − + = − −

(ii) a2 + 6a – 1 = 0 a2 + 6a – 1 = 0 ⇒ a = 1, b = 6, c = –1

2

2

42

6 6 4(1)( 1)2(1)

6 36 ( 4)2

6 402

6 2 102

3 10

3 10 and 3 10

b b acaa

a a

− ± −=

− ± − −=

− ± − −=

− ±=

− ±=

= − ±

= − + = − −


43

(iii) x2 – 7x – 4 = 0

x2 – 7x – 4 = 0 ⇒ a = 1, b = –7, c = –4

2

2

42

( 7) ( 7) 4(1)( 4)2(1)

7 49 ( 16)2

7 652

7 65 7 65 and 2 2

b b acxa

x x

− ± −=

− − ± − − −=

± − −=

±=

+ −= =

(iv) p2 + 8p – 4 = 0

p2 + 8p – 4 = 0 ⇒ a = 1, b = 8, c = –4

2

2

42

8 8 4(1)( 4)2(1)

8 64 ( 16)2

8 802

8 4 52

8 4 52

4 2 5

4 2 5 and 4 2 5

b b acpa

p

p p

− ± −=

− ± − −=

− ± − −=

− ±=

− ±=

− ±=

= − ±

= − + = − −


44

(v) 2x2 + 5x + 1 = 0

2x2 + 5x + 1 = 0 ⇒ a = 2, b = 5, c = 1

2

2

42

5 5 4(2)(1)2(2)

5 25 84

5 174

5 17 5 17 and 4 4

b b acxa

x x

− ± −=

− ± −=

− ± −=

− ±=

− + − −= =

(vi) 3k2 – 10k + 5 = 0

3k2 – 10k + 5 = 0 ⇒ a = 3, b = –10, c = 5

2

2

42

( 10) ( 10) 4(3)(5)

2(3)

10 100 606

10 406

10 2 106

5 103

5 10 5 10 and 3 3

b b acka

k k

− ± −=

− − ± − −=

± −=

±=

±=

±=

+ −= =


45

5. Rearrange the following quadratic equations into the form ax2 + bx + c = 0 and

hence solve using the quadratic formula. Give your answer to two decimal places.

(i) x2 + 5x = –1

x2 + 5x = –1

x2 + 5x + 1 = 0 ⇒ a = 1, b = 5, c = 1

2

2

42

5 5 4(1)(1)2(1)

5 25 42

5 212

5 21 5 21 and 2 2

0 21 and 4 79

b b acxa

x x

x x

− ± −=

− ± −=

− ± −=

− ±=

− + − −= =

= − ⋅ = − ⋅


46

(ii) x2 = 3x + 1

x2 = 3x + 1

x2 – 3x – 1 = 0 ⇒ a = 1, b = –3, c = –1

2

2

42

( 3) ( 3) 4(1)( 1)2(1)

3 9 ( 4)2

3 132

3 13 3 13 and 2 2

3 30 and 0 30

b b acxa

x x

x x

− ± −=

− − ± − − −=

± − −=

±=

+ −= =

= ⋅ = − ⋅

(iii) 9a = 4 – a2

9a = 4 – a2

a2 + 9a – 4 = 0 ⇒ a = 1, b = 9, c = – 4

2

2

42

9 9 4(1)( 4)2(1)

9 81 ( 16)2

9 972

9 97 9 97 and 2 2

0 42 and 9 42

b b acaa

x x

x x

− ± −=

− ± − −=

− ± − −=

− ±=

− + − −= =

= ⋅ = − ⋅


47

(iv) k2 – 4 = 7k

k2 – 4 = 7k

k2 – 7k – 4 = 0 ⇒ a = 1, b = –7, c = –4

2

2

42

( 7) ( 7) 4(1)( 4)2(1)

7 49 ( 16)2

7 652

7 65 7 65 and 2 2

7 53 and 0 53

b b acka

k k

k k

− ± −=

− − ± − − −=

± − −=

±=

+ −= =

= ⋅ = − ⋅

(v) 8x = 2 – 5x2

8x = 2 – 5x2

5x2 + 8x – 2 = 0 ⇒ a = 5, b = 8, c = – 2

2

2

42

8 8 4(5)( 2)2(5)

8 64 ( 40)10

8 10410

8 104 8 104 and 10 10

0 22 and 1 82

b b acxa

x x

x x

− ± −=

− ± − −=

− ± − −=

− ±=

− + − −= =

= ⋅ = − ⋅


48

(vi) 4p2 = 3 – 5p

4p2 = 3 – 5p

4p2 + 5p – 3 = 0 ⇒ a = 4, b = 5, c = – 3

2

2

42

5 5 4(4)( 3)2(4)

5 25 ( 48)8

5 738

5 73 5 73 and 8 8

0 44 and 1 69

b b acxa

x x

x x

− ± −=

− ± − −=

− ± − −=

− ±=

− + − −= =

= ⋅ = − ⋅

6. (i) The function f (x) = x2 − x − 6. Solve the equation f (x) = 0.

f(x) = x2 – x – 6 = 0

(x – 3)(x + 2) = 0

x – 3 = 0 and x + 2 = 0

x = 3 and x = –2

(ii) Hence draw a sketch of the graph of f (x).

Roots are –2 and 3, so the graph crosses the x-axis at these values

f (x) = x2 − x – 6 has a positive x2 term, so the graph is U shaped

Let x = 0:

f(0) = 02 – 0 – 6

y = – 6

∴ (0, – 6) is on the graph


49

Sketch:

7. (i) The function g(x) = − x2 + 3x + 4. Solve the equation g(x) = 0.

g(x) = –x2 + 3x + 4 = 0 (multiply all parts by –1)

x2 – 3x – 4 = 0

(x – 4)(x + 1) = 0

x – 4 = 0 and x + 1 = 0

x = 4 and x = –1

(ii) Hence draw a sketch of the graph of g(x).

Roots are –1 and 4, so the graph crosses the x-axis at these values

g(x) = − x2 + 3x + 4 has a negative x2 term, so the graph is ∩ shaped

Let x = 0

g(0) = – (0)2 + 3(0) + 4

g = 4

∴ (0, 4) is on the graph.

Sketch:


50

8. (i) The function ℎ(x) = 2x2 − 3x − 5. Solve the equation ℎ(x) = 0.

h(x) = 2x2 – 3x – 5 = 0

(2x – 5)(x + 1) = 0

2x – 5 = 0 and x + 1 = 0

2x = 5 and x = – 1

x = 2·5 (ii) By replacing ℎ(x) with 2x2 − 3x − 5, express ℎ(x) = 9 in terms of x. ℎ(x) = 2x2 – 3x – 5 = 9

2x2 – 3x – 14 = 0

(iii) Hence, solve the equation ℎ(x) = 9.

2x2 – 3x – 14 = 0

(2x – 7)(x + 2) = 0

2x – 7 = 0 and x + 2 = 0

2x = 7 and x = –2

x = 3·5


51

(iv) Draw a sketch of the graph of ℎ(x).

Roots are –1 and 2·5, so the graph crosses the x-axis at these values

ℎ(x) = 2x2 − 3x – 5 has a positive x2 term, so the graph is U shaped

Let x = 0:

h(0) = 2(0)2 – 3(0) – 5 = –5

y = –5

∴ (0, –5) is on the graph.

9. (i) The function f (x) = −3x2 + 2x + 8. Solve the equation f (x) = 0.

f(x) = –3x2 + 2x + 8 = 0 (multiply all parts by −1)

3x2 – 2x – 8 = 0

(3x + 4)(x – 2) = 0

3x + 4 = 0 and x – 2 = 0

3x = –4 and x = 2

4

3

x = −

(ii) By replacing f(x) with −3x2 + 2x + 8, express f (x) = 3 in terms of x.

–3x2 + 2x + 8 = 3

–3x2 + 2x + 5 = 0


52

(iii) Find the values of x for which f (x) = 3.

–3x2 + 2x + 5 = 0 (multiply all parts by –1)

3x2 – 2x – 5 = 0

(3x – 5)(x + 1) = 0

3x – 5 = 0 and x + 1 = 0

3x = 5 and x = –1

(iv) Draw a sketch of the graph of f (x).

Roots are 4

3

− and 2, so the graph crosses the x-axis at these values

f (x) = −3x2 + 2x + 8 has a negative x2 term, so the graph is ∩ shaped

Let x = 0

f(0) = –3(0)2 + 2(0) + 8 = 0

y = 8

∴ (0, 8) is on the graph

Sketch:

10. Given the function f (x) = x2 + 2x − 3

(i) Solve f (x) = 0.

f(x) = x2 + 2x – 3 = 0

(x + 3)(x – 1) = 0

x + 3 = 0 and x – 1 = 0

x = –3 x = 1

53

x =


53

(ii) Hence, identify which of the following graphs represents the function f (x).

Give a reason for your answer.

(a)

(c)

(b) (d)

The function f (x) = x2 + 2x – 3 has roots x = –3 and x = 1, so the graph

crosses the x-axis at these points.

The function has a positive x2, so it’s graph is a U shape.

Therefore, graph (c) represents the function f (x).


54

11. (i) Verify that 1 7+ is a root of the function f (x) = x2 − 2x − 6.

If 1 7+ is a root of the function, then f (1 7+ ) = 0

f(x) = x2 – 2x – 6

2(1 7) (1 7) 2(1 7) 6

1(1 7) 7(1 7) 2(1 7) 6

(1 7 7 7) (2 2 7) 6

8 2 7 2 2 7 60

f + = + − + −

= + + + − + −

= + + + − + −

= + − − −=

Since 1 7+ satisfies the equation f(x) = 0, then 1 7+ is a root of the

function.

(ii) Find the other root of f (x) = 0.

f(x) = x2 – 2x – 6 a = 1, b = –2, c = –6

2

2

42

( 2) ( 2) 4(1)( 6)2(1)

2 4 ( 24)2

2 282

2 2 72

1 7 and 1 7

b b acxa

x

− ± −=

− − ± − − −=

± − −=

±=

±=

= + −

Therefore, the other root is 1 7−


55

(iii) Hence draw a sketch of the graph of f (x).

Roots: 1 7+ = 3·65 1 7− = –1·65

f(x) = x2 – 2x – 6 has a positive x2 term, so it is U shaped

Let x = 0

f(0) = (0)2 – 2(0) – 6

= 0 – 0 – 6

y = – 6

∴ (0, – 6) is on the graph

Sketch:


56

12. (i) Verify that 1 10− is a root of the function f (x) = − x2 + 2x + 9.

If 1 10− is a root of the function, then f (1 10− ) = 0

Substitute 1 10− in for x:

Since ( )1 10− satisfies the question f(x) = 0, then 1 10− is a root of f(x)

(ii) Find the other root of f (x) = 0.

Solve f(x) = 0: –x2 + 2x + 9 = 0 a = –1 b = 2 c = 9

( ) ( )( )( )

2

2

42

2 2 4 1 92 1

2 4 362

2 402

2 2 102

1 10

b b acxa

− ± −=

− ± − −=

−

− ± +=

−− ±

=−

− ±=

−= ±

∴ second root : 1 10+

( )

( ) ( ) ( )( ) ( ) ( )

( ) ( )( )

2

2

2 9

1 10 2 1 10 9

1 1 10 10 1 10 2 1 10 9

1 10 10 10 2 1 10 9

1 2 10 10 2 2 10 9

1 2 10 10 2 2 10 9

2 10 11 2 10 1

1 10

10

f x x x

f

= − + +

= − − + − +

= − − − − + − +

= − − − + + − +

= − − + + − +

= − + − + − +

= − − +=

−


57

(iii) Hence draw a sketch of the graph of f (x).

Roots : 1 10− = –2·16

1 10+ = 4·16

f(x) = –x2 + 2x + 9 is ∩ shaped, crossing y-axis at 9:


1. Form the quadratic equations with the following roots:

(i) 2 and 4

Roots x = 2 and x = 4

x – 2 = 0 and x – 4 = 0

(x – 2)(x – 4) = 0

x(x – 4) – 2(x – 4) = 0

x2 – 4x – 2x + 8 = 0

x2 – 6x + 8 = 0

(ii) 7 and 3


x – 7 = 0 and x – 3 = 0

(x – 7)(x – 3) = 0

x(x – 3) – 7(x – 3) = 0

x2 – 3x – 7x + 21 = 0

x2 – 10x + 21 = 0


58

(iii) 4 and 6


x – 4 = 0 and x – 6 = 0

(x – 4)(x – 6) = 0

x(x – 6) – 4(x – 6) = 0

x2 – 6x – 4x + 24 = 0

x2 – 10x + 24 = 0

(iv) − 1 and 3

Roots x = – 1 and x = 3

x + 1 = 0 and x – 3 = 0

(x + 1)(x – 3) = 0

x(x – 3) + 1(x – 3) = 0

x2 – 3x + x – 3 = 0

x2 – 2x – 3 = 0

(v) – 2 and 9


x + 2 = 0 and x – 9 = 0

(x + 2)(x – 9) = 0

x(x – 9) + 2(x – 9) = 0

x2 – 9x + 2x – 18 = 0

x2 – 7x – 18 = 0

(vi) 8 and − 2

Roots x = 8 and x = – 2

x – 8 = 0 and x + 2 = 0

(x – 8)(x + 2) = 0

x(x + 2) – 8(x + 2) = 0

x2 + 2x – 8x – 16 = 0

x2 – 6x – 16 = 0


59

(vii) 12 and − 3


x – 12 = 0 and x + 3 = 0

(x – 12)(x + 3) = 0

x(x + 3) – 12(x + 3) = 0

x2 + 3x – 12x – 36 = 0

x2 – 9x – 36 = 0

(viii) − 2 and − 6

Roots x = – 2 and x = – 6

x + 2 = 0 and x + 6 = 0

(x + 2)(x + 6) = 0

x(x + 6) + 2(x + 6) = 0

x2 + 6x + 2x + 12 = 0

x2 + 8x + 12 = 0

(ix) − 7 and − 4


x + 7 = 0 and x + 4 = 0

(x + 7)(x + 4) = 0

x(x + 4) + 7(x + 4) = 0

x2 + 4x + 7x + 28 = 0

x2 + 11x + 28 = 0


60

2. The quadratic equation x2 + px + q = 0 has roots 7 and 2. Find the values of p

and q.

x2 + px + q = 0


x – 7 = 0 and x – 2 = 0

(x – 7)(x – 2) = 0

x(x – 2) – 7(x – 2) = 0

x2 – 2x – 7x + 14 = 0

x2 – 9x + 14 = 0

p = – 9, q = 14

3. The quadratic equation x2 + cx + d = 0 has roots − 6 and − 4. Find the values of c

and d.

x2 + cx + d = 0


x + 6 = 0 and x + 4 = 0

(x + 6)(x + 4) = 0

x(x + 4) + 6(x + 4) = 0

x2 + 4x + 6x + 24 = 0

x2 + 10x + 24 = 0

c = 10, d = 24 4. (i) Form the quadratic equation with roots − 3 and 8. Roots x = − 3 and x = 8

x + 3 = 0 and x – 8 = 0

(x + 3)(x – 8) = 0

x(x – 8) + 3(x – 8) = 0

x2 – 8x + 3x – 24 = 0

x2 – 5x – 24 = 0


61

(ii) Verify your result by factorising and solving this quadratic equation.

x2 – 5x – 24 = 0

(x – 8)(x + 3) = 0

x – 8 = 0 and x + 3 = 0

x = 8 and x = –3

5. For each of the following graphs:

(a)

(i) identify the roots of the function


(ii) form the equation

x + 3 = 0 and x – 1 = 0

(x + 3)(x – 1) = 0

x(x – 1) + 3(x – 1) = 0

x2 – x + 3x – 3 = 0

x2 + 2x – 3 = 0


62

(iii) verify the roots of the equation, algebraically

x = – 3 x = 1

(– 3)2 + 2(– 3) – 3 = 0 12 + 2(1) – 3 = 0

9 – 6 – 3 = 0 1 + 2 – 3 = 0

0 = 0 0 = 0

(iv) given each graph is illustrating a function f (x), write the function in

each case.

f(x) = x2 + 2x – 3

(b)

(i) Roots x = – 4 and x = 2

(ii) x + 4 = 0 and x – 2 = 0

(x + 4)(x – 2) = 0

x(x – 2) + 4(x – 2) = 0

x2 – 2x + 4x – 8 = 0

x2 + 2x – 8 = 0

(iii) x = – 4 x = 2

(– 4)2 + 2(– 4) – 8 = 0 (2)2 + 2(2) – 8 = 0

16 – 8 – 8 = 0 4 + 4 – 8 = 0

0 = 0 0 = 0


63

(iv) The graph is ∩ shape therefore x2 must be negative, so the equation

is multiplied by – 1

x2 + 2x – 8 = 0 (multiply all parts by – 1)

– x2 – 2x + 8 = 0

f(x) = – x2 – 2x + 8

(c)

(i) Roots x = 5 and x = 5

(ii) x – 5 = 0 and x – 5 = 0

(x – 5)(x – 5) = 0

x(x – 5) – 5(x – 5) = 0

x2 – 5x – 5x + 25 = 0

x2 – 10x + 25 = 0

(iii) x = 5

(5)2 – 10(5) + 25 = 0

25 – 50 + 25 = 0

0 = 0

(iv) The graph is ∩ shape, therefore x2 must be negative, so the equation

is multiplied by – 1

x2 – 10x + 25 = 0 (multiply all parts by – 1)

– x2 + 10x – 25 = 0

f(x) = –x2 + 10x – 25


64

(d)

(i) Roots x = 1·5 and x = 6

(ii) x = 1·5 x – 6 = 0

2x = 3

2x – 3 = 0

(2x – 3)(x – 6) = 0

2x(x – 6) – 3(x – 6) = 0

2x2 – 12x – 3x + 18 = 0

2x2 – 15x + 18 = 0

(iii) x = 1·5 x = 6

2(1·5)2 – 15(1·5) + 18 = 0 2(6)2 – 15(6) + 18 = 0

2(2·25) – 22·5 + 18 = 0 2(36) – 90 + 18 = 0

4·5 – 22·5 + 18 = 0 72 – 90 + 18 = 0

0 = 0 0 = 0

(iv) f(x) = 2x2 – 15x + 18


65

6. (i) Form the quadratic equation with roots 5 and − 9.


x – 5 = 0 and x + 9 = 0

(x – 5)(x + 9) = 0

x(x + 9) – 5(x + 9) = 0

x2 + 9x – 5x – 45 = 0

x2 + 4x – 45 = 0


x2 + 4x – 45 = 0

(x + 9)(x – 5) = 0

x + 9 = 0 and x – 5 = 0

x = – 9 x = 5

(iii) The graph of a function, k (x), has roots 5 and − 9 and crosses the y-axis at

the point (0, − 45).

Draw a sketch of the graph of k(x).

(iv) Write down the function, k (x).

k(x) = x2 + 4x – 45


66

7. (i) Form the quadratic equation with roots − 4 and − 11.

Roots: x = – 4 and x = – 11

x + 4 = 0 and x + 11 = 0

(x + 4)(x + 11) = 0

x(x + 11) + 4(x + 11) = 0

x2 + 11x + 4x + 44 = 0

x2 + 15x + 44 = 0


x2 + 15x + 44 = 0

(x + 11)(x + 4) = 0

x + 11 = 0 and x + 4 = 0

x = –11 and x = – 4

(iii) The graph of a function, g (x), has roots − 4 and − 11 and crosses the y-axis

at the point (0, − 44). Draw a sketch of the graph of g (x).

(iv) Write down the function, g(x).

To cross y-axis at y = − 44, graph is ∩ shaped, so must have a negative x2.

Multiply the equation by – 1 to give: g(x) = – x2 – 15x – 44.


67

Practice Question 4.5

1. Solve the following equations:

(i) 3 3 2

2x x+ =

2

2

2

3 3 2 LCM ( )(2 )2

3 3( )(2 ) ( )(2 ) ( )(2 )(2)2

(2 )(3) ( )(3) ( )(2 )(2)6 3 (2 )(2)

9 40 4 90 (4 9)

0 and 4 9 0 4 9

x xx x

x x x x x xx x

x x x xx x x

x xx x

x xx x

x

+ = =

+ =

+ =

+ =

=

= −= −= − =

=9 4

x =


68

(ii) 3 4 5

2 1x x+ =

+

2

2

2

2

3 4 5 LCM (2 )( 1)2 1

3 4(2 )( 1) (2 )( 1) (2 )( 1)(5)2 1

( 1)(3) (2 )(4) (2 )( 1)(5)3 3 8 (2 2 )(5)

3 11 10 100 10 10 11 3

10 3 0(5 3)(2 1) 0

5 3 0 and

x xx x

x x x x x xx x

x x x xx x x x

x x xx x x

x xx x

x

+ = = ++

+ + + = + + + + = +

+ + = +

+ = +

= + − −

− − =− + =

− = 2 1 05 3 2 1

3 1 5 2

xx x

x x

+ == = −

= = −


69

(iii) 5 3 11

4 2 16a a+ =

The lowest common denominator of 4a, 2a and 16 is 16a.

However, in this case, the solution is easier if we let the LCM be (4a)(2a)(16)

2

2

5 3 11 LCM (4 )(2 )(16)4 2 16

5 3 11(4 )(2 )(16) (4 )(2 )(16) (4 )(2 )(16)4 2 16(2 )(16)(5) (4 )(16)(3) (4 )(2 )(11)

(32 )(5) (64 )(3) (8 )(11)160 192 88

352

a aa a

a a a a a aa aa a a a

a a aa a a

a

+ = =

+ =

+ =

+ =

+ =

= 2

2

2

880 88 3520 40 ( 4)

0 and 4 0 4

aa a

a aa a

a aa

= −

= −= −

= − ==


70

(iv) 3 2

1 2 1s s

− =−

( )( )

( )( ) ( )( ) ( )( )( )

( )( ) ( )( ) ( )( )

( )( )

2

2

2

3 2 1 LCM 2 12 1

3 22 1 2 1 2 1 12 1

2 1 3 2 2 1

6 3 2 24 3 2

0 2 5 30 2 3 1

2 3 0 and 1 0 2 3 and 1

3 2

s ss s

s s s s s ss s

s s s s

s s s ss s s

s ss s

s ss s

s

− = = −−

− − − = − − − − = −

− − = −

− = −

= − +

= − −

− = − == =

=

(v) 1 1 2

4 3a a+ =

−

( )( )( )LCM 4 3a a= −

( )( )( ) ( )( )( ) ( )( )( )

( )( )( ) ( )( )( ) ( )( )( )( )( )

( )( )

2

2

2

2

1 1 2 4 3

1 1 24 3 4 3 4 34 3

3 1 4 3 1 4 2

3 3 12 4 2

6 12 2 80 2 14 120 7 6

0 6 16 0 and 1 0

6

a a

a a a a a aa a

a a a a

a a a a

a a aa a

a aa a

a aa

+ =−

− + − = − − + − = −

+ − = −

− = −

= − +

= − +

= − −

− = − == 1a =


71

(vi) 1 3 7

1 2x x+ =

−

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )

( )( )( ) ( )( )( ) ( )( )( )( )( )

( )( )

2

2

2

1 3 7 LCM 1 21 2

1 3 71 2 1 2 1 21 2

1 2 1 2 3 1 7

2 2 6 7

8 2 7 70 7 15 20 7 1 2

7 1 0 and 2 0 7 1 and 2

1 7

x xx x

x x x x x xx x

x x x x

x x x x

x x xx xx x

x xx x

x

+ = = −−

− + − = − − − + = −

− + = −

− = −

= − +

= − −

− = − == =

=


72

2. Solve the following equations:

(i) 3 4

12 1 3 1x x

− =− −

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )( )

( )( ) ( )( ) ( )( )

( )

2

2

2

2

3 4 1 LCM 2 1 3 1 12 1 3 1

3 42 1 3 1 1 2 1 3 1 1 2 1 3 1 1 12 1 3 1

3 1 3 2 1 4 2 1 3 1

9 3 8 4 6 2 3 11 6 5 10 6 600 1

0 and 1 0

x xx x

x x x x x xx x

x x x x

x x x x xx x x

x xx xx x

x x

− = = − −− −

− − − − − = − − − − − − − = − −

− − + = − − +

+ = − +

= −

= −

= −

= − = 1x =


73

(ii) 6 3 7

3 1 5x x− =

− −

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )

( )( )( ) ( )( )( ) ( )( )( )( )( ) ( )( ) ( )( )

( )( )

( )( )

2

2

2

2

6 3 7 LCM 3 1 53 1 5

6 3 73 1 5 3 1 5 3 1 53 1 5

1 5 6 3 5 3 3 1 7

5 5 6 5 15 3 3 3 7

30 30 15 45 4 3 7

15 15 7 28 210 7 43 60 7 1 6

7 1 0 and

x xx x

x x x x x xx x

x x x x

x x x x x

x x x x

x x xx xx x

x

− = = − −− −

− − − − − = − − − − − − − = − −

− − − = − − +

− − + = − +

+ = − +

= − +

= − −

− = 6 0 7 = 1 and 6

1 7

xx x

x

− ==

=


74

(iii) 7 1 42 1p p

+ =+ −

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )( )

( )( ) ( )( ) ( )( )( )( )( )( )( )

( )( )

2

2

2

2

7 1 4 LCM 2 1 12 1

7 12 1 1 2 1 1 2 1 1 42 1

1 7 2 1 2 1 4

7 7 2 2 2 4

8 5 2 4

8 5 4 4 80 4 4 30 2 3 2 1

2 3 0 and 2 1 02 3 and 2 1

p pp p

p p p p p pp p

p p p p

p p p p p

p p p

p p pp pp p

p pp p

p

+ = = + −+ −

+ − + + − = + − + −

− + + = + −

− + + = − + −

− = + −

− = + −

= − −

= − +

− = + == = −

=3 1 and 2 2

p = −


75

(iv) 8 8

32 2z z

+ =− +

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )( )

( )( ) ( )( ) ( )( )( )( )( )

( )( )

2

2

2

8 8 3 LCM 2 2 12 2

8 82 2 1 2 2 1 2 2 1 32 2

2 8 2 8 2 2 3

8 16 8 16 2 3 6

16 3 6 6 1216 3 12

0 3 16 120 3 2 6

3 2 0 and 6 03 2 and 6

2

z zz z

z z z z z zz z

z z z z

z z z z

z z z zz z

z zz z

z zz z

z

+ = = − +− +

− + + − + = − + − + + + − = − +

+ + − = − +

= + − −

= −

= − −

= + −

+ = − == − =

= −3

3. Solve the following equations and leave your answer correct to two decimal

places.

(i) 9 4 1

3 6x x+ =

+

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )

( )( )( ) ( )( )( ) ( )( )( )2

2

2

9 4 1 LCM 3 63 6

9 4 13 6 3 6 3 63 6

6 9 3 6 4 3 1

54 24 72 378 72 3

0 75 721, 75, 72

x xx x

x x x x x xx x

x x x x

x x x xx x x

x xa b c

+ = = ++

+ + + = + + + + = +

+ + = +

+ = +

= − −= = − = −


76

( ) ( ) ( )( )( )( )

2

2

42

75 75 4 1 722 1

75 5625 2882

75 59132

75 9 732

b b acxa

− ± −=

− − ± − −=

± +=

±=

±=

75 9 73 75 9 7375 95 and 0 952 2

x x+ −= = ⋅ = = − ⋅

(ii) 5 4 3

2 5z z− =

+

( )( )( )

( )( )( ) ( )( )( ) ( )( )( )

( )( )( ) ( )( )( ) ( )( )( )2

2

2

5 4 3 LCM 2 52 5

5 4 32 5 2 5 2 52 5

2 5 5 5 4 2 3

25 50 20 3 65 50 3 6

0 3 503, 1, 50

z zz z

z z z z z zz z

z z z z

z z z zz z z

z za b c

− = = ++

+ − + = + + + − = +

+ − = +

+ = +

= + −= = = −


77

( )( )( )( )

2

2

42

1 1 4 3 502 3

1 1 6006

1 6016

b b acxa

− ± −=

− ± − −=

− ± − −=

− ±=

1 601 1 6013 92 and 4 256 6

x x− + − −= = ⋅ = = − ⋅

(iii) 4 7 3

2 1m m+ =

− +

2

2

2

4 7 3 LCM ( 2)( 1)(1)2 1

4 7( 2)( 1)(1) ( 2)( 1)(1) ( 2)( 1)(1)(3)2 1

( 1)(4) ( 2)(7) ( 2)(3 3)4 4 7 14 3 3 6 6

11 10 3 3 60 3 14 4

m mm m

m m m m m mm m

m m m mm m m m m

m m mm m

+ = = − +− +

− + + − + = − + − + + + − = − +

+ + − = + − −

− = − −

= − +

2

2

3, 14, 4

4 2

( 14) ( 14) 4(3)(4)2(3)

14 196 486

14 1486

a b c

b b acxa

= = − =

− ± −=

− − ± − −=

± −=

±=

14 148 14 1484 36 and 0 316 6

x x+ −= = ⋅ = = ⋅


78

(iv) 3 4 2

4 1 2x x− =

+ +

2

2

2

3 4 2 LCM (4 1)( 2)(1)4 1 2

3 4(4 1)( 2)(1) (4 1)( 2)(1) (4 1)( 2)(1)(2)4 1 2

( 2)(3) (4 1)(4) (4 1)(2 4)3 6 16 4 8 16 2 4

2 13 8 18 40 8 31 2

x xx x

x x x x x xx x

x x x xx x x x x

x x xx x

− = = + ++ +

+ + − + + = + + + + + − + = + +

+ − − = + + +

− = + +

= + +

2

2

8, 31, 2

4 2

31 (31) 4(8)(2)2(8)

31 961 6416

31 89716

a b c

b b acxa

= = =

− ± −=

− ± −=

− ± −=

− ±=

31 897 31 8970 07 and 3 8116 16

x x− ± − −= = − ⋅ = = − ⋅


79

4. Solve the following equations and leave your answer in surd form.

(i) 4 5 1

4z z− =

−

2

2

2

4 5 1 LCM ( 4)( )(1)4

4 5( 4)( )(1) ( 4)( )(1) ( 4)( )(1)(1)4

( )(4) ( 4)(5) ( 4)( )4 5 20 4

20 40 3 20

z zz z

z z z z z zz z

z z z zz z z z

z z zz z

− = = −−

− − − = − − − − = −

− + = −

− = −

= − −

2

2

1, 3, 20

4 2

( 3) ( 3) 4(1)( 20)2(1)

3 9 ( 80)2

3 892

a b c

b b acxa

= = − = −

− ± −=

− − ± − − −=

± − −=

±=

3 89 3 89 and

2 2x x

+ −= =


80

(ii) 3 1

11 1n n

+ =+ −

2

2

2

3 1 1 LCM ( 1)( 1)(1)1 1

3 1( 1)( 1)(1) ( 1)( 1)(1) ( 1)( 1)(1)(1)1 1

( 1)(3) ( 1) ( 1)( 1)3 3 1 1

4 2 10 4 1

n nn n

n n n n n nn n

n n n nn n n n n

n nn n

+ = = + −+ −

+ − + + − = + − + − − + + = + −

− + + = + − −

− = −

= − +

2

2

1, 4, 1

4 2

( 4) ( 4) 4(1)(1)2(1)

4 16 42

4 122

2 3

2 3 and 2 3

a b c

b b acxa

x x

= = − =

− ± −=

− − ± − −=

± −=

+=

= ±

= + = −


81

(iii) 6 3 7

1 1 2x x+ =

+ −

2

2

2

6 3 7 LCM ( 1)( 1)(2)1 1 2

6 3 7( 1)( 1)(2) ( 1)( 1)(2) ( 1)( 1)(2)1 1 2( 1)(2)(6) ( 1)(2)(3) ( 1)(7 7)

12 12 6 6 7 7 7 718 6 7 7

0 7 18 1

x xx x

x x x x x xx x

x x x xx x x x x

x xx x

+ = = + −+ −

+ − + + − = + − + − − + + = + −

− + + = − + −

− = −

= − −

2

2

7, 18, 1

4 2

( 18) ( 18) 4(7)( 1)2(7)

18 324 ( 28)14

18 35214

9 2 227

a b c

b b acxa

= = − = −

− ± −=

− − ± − − −=

± − −=

±=

±=

9 2 22 9 2 22 and 7 7

x x+ −= =


82

(iv) 2 1 21 1y y

+ =+ −

2

2

2

2 1 2 LCM ( 1)( 1)(1)1 1

2 1( 1)( 1)(1) ( 1)( 1)(1) ( 1)( 1)(1)(2)1 1

( 1)(2) ( 1) ( 1)(2 2)2 2 1 2 2 2 2

3 1 2 20 2 3 1

y yy y

y y y y y yy y

y y y yy y y y y

y yy y

+ = = + −+ −

+ − + + − = + − + −

− + + = + −

− + + = − + −

− = −

= − −

2

2

2, 3, 1

4 2

( 3) ( 3) 4(2)( 1)2(2)

3 9 84

3 174

a b c

b b acxa

= = − = −

− ± −=

− − ± − − −=

± +=

±=

3 17 3 17 and 4 4

x x+ −= =


83


1. Use a suitable method to solve the following pairs of simultaneous equations:

(i) x = y + 1

x + y2 = 3

Substitute x = y + 1 into the quadratic equation:

x + y2 = 3 ⇒ (y + 1) + y2 = 3

y2 + y – 2 = 0

(y + 2)(y – 1) = 0

y + 2 = 0 and y – 1 = 0

y = – 2 and y = 1

x = y + 1

y = –2 ⇒ x = –2 + 1 = –1 ∴ (– 1, – 2)

y = 1 ⇒ x = 1 + 1 = 2 ∴ (2, 1)

(ii) x = y

x2 + 2xy = 3

Substitute x = y into the quadratic equation:

x2 + 2xy = 3 ⇒ x2 + 2x(x) = 3

x2 + 2x2 = 3

3x2 = 3

x2 = 1

1x = ±

x = – 1 and x = 1.

x = y

x = – 1 ⇒ y = – 1 ∴ ( – 1, – 1)

x = 1 ⇒ y = 1 ∴ (1, 1)


84

(iii) x + y = 5

x2 + y2 = 13

Rearrange the linear equation:

x + y = 5 ⇒ y = 5 – x

Substitute this value in for y, in the quadratic equation:

x2 + y2 = 13 ⇒ x2 + (5 – x)2 = 13

x2 + (5 – x)(5 – x) = 13

x2 + [5(5 – x) – x(5 – x)] = 13

x2 + [25 – 5x – 5x – x2] = 13

2x2 – 10x + 25 = 13

2x2 – 10x + 12 = 0

x2 – 5x + 6 = 0

(x – 2)(x – 3) = 0

x – 2 = 0 and x – 3 = 0

x = 2 and x = 3

x + y = 5 ⇒ y = 5 – x

x = 2 ⇒ y = 5 – 2 = 3 ∴ (2, 3)

x = 3 ⇒ y = 5 – 3 = 2 ∴ (3, 2)


85

2. Use a suitable method to solve the following pairs of simultaneous equations:

(i) a − b = −4

a2 + 4 b2 = 37


a – b = – 4 ⇒ a = b – 4

Substitute a = b – 4 into the quadratic equation:

a2 + 4b2 = 37 ⇒ (b – 4)2 + 4b2 = 37

(b – 4)(b – 4) + 4b2 = 37

[b(b – 4) – 4(b – 4)] + 4b2 = 37

b2 – 4b – 4b + 16 + 4b2 = 37

5b2 – 8b + 16 = 37

5b2 – 8b – 21 = 0

(5b + 7)(b – 3) = 0

5b + 7 = 0 and b – 3 = 0

5b = – 7 and b = 3

75

b = −

( )

4 47 7 27 27 7 4 ,5 5 5 5 5

3 3 4 1 1,3

a b a b

b a

b a

− = − ⇒ = −

= − ⇒ = − − = − ∴ − −

= ⇒ = − = − ∴ −


86

(ii) 2x − y = 5

2 x2 − y2 = 7


2x – y = 5 ⇒ y = 2x – 5

Substitute y = 2x – 5 into the quadratic equation:

2x2 – y2 = 7 ⇒ 2x2 – (2x – 5)2 = 7

2x2 – [(2x – 5)(2x – 5)] = 7

2x2 – [2x(2x – 5) – 5(2x – 5)] = 7

2x2 – [4x2 – 10x – 10x + 25] = 7

2x2 – 4x2 + 10x + 10x – 25 = 7

– 2x2 + 20x – 25 = 7

– 2x2 + 20x – 32 = 0

– x2 + 10x – 16 = 0

x2 – 10x + 16 = 0

(x – 2)(x – 8) = 0

x – 2 = 0 and x – 8 = 0

x = 2 and x = 8

2x – y = 5 ⇒ y = 2x – 5

x = 2 ⇒ y = 2(2) – 5 = 4 – 5 = – 1 ∴ (2, – 1)

x = 8 ⇒ y = 2(8) – 5 = 16 – 5 = 11 ∴ (8, 11)


87

(iii) p − 2q = −4

p2 + q2 = 16


p – 2q = – 4 ⇒ p = 2q – 4

Substitute p = 2q – 4 into the quadratic equation:

p2 + q2 = 16 ⇒ (2q – 4)2 + q2 = 16

(2q – 4)(2q – 4) + q2 = 16

2q(2q – 4) – 4(2q – 4) + q2 = 16

4q2 – 8q – 8q + 16 + q2 = 16

5q2 – 16q + 16 = 16

5q2 – 16q = 0

q(5q – 16) = 0

q = 0 and 5q – 16 = 0

5q = 16

165

q =

2 4 2 40 2(0) 4 0 4 4 ( 4,0)16 16 32 12 12 16 2 4 4 ,5 5 5 5 5 5

p q p qq p

q p

− = − ⇒ = −= ⇒ = − = − = − ∴ −

= ⇒ = − = − = ∴


88

3. Use a suitable method to solve the following pairs of simultaneous equations: (i) x − y = 6

x2 − 2x − y = 10


x – y = 6 ⇒ y = x – 6

Substitute y = x – 6 into the quadratic equation:

x2 – 2x – y = 10 ⇒ x2 – 2x – (x – 6) = 10

x2 – 2x – x + 6 = 10

x2 – 3x – 4 = 0

(x + 1)(x – 4) = 0

x + 1 = 0 and x – 4 = 0

x = – 1 and x = 4

x – y = 6 ⇒ y = x – 6

x = – 1 ⇒ y = – 1 – 6 = –7 ∴ ( – 1, – 7)

x = 4 ⇒ y = 4 – 6 = –2 ∴ (4, – 2)


89

(ii) a − 2b = − 5

a2 + ab − b2 = 5


a – 2b = – 5 ⇒ a = 2b – 5

Substitute a = 2b – 5 into the quadratic equation:

a2 + ab – b2 = 5 ⇒ (2b – 5)2 + (2b – 5)b – b2 = 5

(2b – 5)(2b – 5) + 2b2 – 5b – b2 = 5

2b(2b – 5) – 5(2b – 5) + 2b2 – 5b – b2 = 5

4b2 – 10b – 10b + 25 + 2b2 – 5b – b2 = 5

5b2 – 25b + 25 = 5

5b2 – 25b + 20 = 0

b2 – 5b + 4 = 0

(b – 4)(b – 1) = 0

b – 4 = 0 and b – 1 = 0

b = 4 and b = 1

a – 2b = – 5 ⇒ a = 2b – 5

b = 4 ⇒ a = 2(4) – 5 = 8 – 5 = 3 ∴ (3, 4)

b = 1 ⇒ a = 2(1) – 5 = 2 – 5 = – 3 ∴ (– 3, 1)


90

(iii) x + y = −2

x2 + 2xy + 2y2 = 8


x + y = – 2 ⇒ x = – 2 – y

Substitute x = – 2 – y into the quadratic equation:

x2 + 2xy = 2y2 = 8 ⇒ (– 2 – y)2 + 2(– 2 – y)y + 2y2 = 8

(– 2 – y)(– 2 – y) + (– 4 – 2y)y + 2y2 = 8

– 2(– 2 – y) – y(– 2 – y) – 4y – 2y2 + 2y2 = 8

4 + 2y + 2y + y2 – 4y = 8

y2 + 4 = 8

y2 – 4 = 0

(y + 2)(y – 2) = 0

y = – 2 and y = 2

x + y = – 2 ⇒ x = – 2 – y

y = – 2 ⇒ x = – 2 – (–2) = – 2 + 2 = 0 ∴ (0, – 2)

y = 2 ⇒ x = – 2 – (2) = –2 – 2 = – 4 ∴ (– 4, 2).

4. (i) Solve the following simultaneous equations: 3x − y = −2 and y = x2 + 3x + 2


3x – y = – 2 ⇒ y = 3x + 2

Substitute y = 3x + 2 into the quadratic equation:

y = x2 + 3x + 2 ⇒ 3x + 2 = x2 + 3x + 2

0 = x2

0 x=

0 = x

3x – y = – 2 ⇒ y = 3x + 2

x = 0 ⇒ y = 3(0) + 2 = 0 + 2 = 2 ∴ (0, 2)


91

(ii) Which of the sketches below represents the graphs of the equations in part (i)?


(a)

(b)

(c)

Graph(C)

There is only one solution to the simultaneous equations, which means the

line must be a tangent to the curve (they only have one point of

intersection). And the Point of Intersection, (0, 2) is on the y-axis.


92

5. (i) Solve the following simultaneous equations: x − 3y = − 5 and x2 + y2 = 25


x – 3y = – 5 ⇒ x = 3y – 5

Substitute x = 3y – 5 into the quadratic equation:

x2 + y2 = 25 ⇒ (3y – 5)2 + y2 = 25

(3y – 5)(3y – 5) + y2 = 25

3y(3y – 5) – 5(3y – 5) + y2 = 25

9y2 – 15y – 15y + 25 + y2 = 25

10y2 – 30y + 25 = 25

10y2 – 30y = 0

y2 – 3y = 0

y(y – 3) = 0

y = 0 and y – 3 = 0

y = 3

x – 3y = – 5 ⇒ x = 3y – 5

y = 0 ⇒ x = 3(0) – 5 = 0 – 5 = –5 ∴ (– 5, 0)

y = 3 ⇒ x = 3(3) – 5 = 9 – 5 = 4 ∴ (4, 3)


93

(ii) Which of the sketches below represents the graphs of the equations in

part (i)?


(a)

(b)

(c)

Graph (b)

The two solutions from the simultaneous equations are the points where the

line and the circle intersect.

Only one of the diagrams appears to have a (– 5, 0) and (4, 3) as the points

of intersection of the line and circle.


1. An advertising billboard is 5 m wider than it is high.

The billboard is 204 m2 of advertising space.


94

(i) Letting x be the height of the billboard, write down the width of the

billboard, in terms of x.

Width = height + 5

Width = x + 5

(ii) Form an equation for the area of the billboard, in terms of x.

Area = height × width

204 = x × (x + 5)

204 = x2 + 5x

(iii) Hence, find the dimensions of the billboard.

Area = 204 = x2 + 5x

0 = x2 + 5x – 204

0 = (x + 17)(x – 12)

x + 17 = 0 and x – 12 = 0

x = – 17 and x = 12

Lengths cannot be negative, so x = – 17 is rejected

x = 12 m = height

x + 5 = 12 + 5 = 17 m = width.


95

2. A triangle has a height that is equal to its base. If the area of the triangle is 32 cm2, find the base and height of the triangle.

2

1Area base height base height21212

b b

b

= × × =

= × ×

=

2

2

1Area 32 ( 2)2

64

648

b

b

bb

= = ×

=

± =± =

Lengths cannot be negative, so b = – 8 is rejected.

The base and height of the triangle are b = h = 8 cm.

3. The length of a rectangle is 7 cm, and the width is 4 cm.

(i) Find the area of the rectangle.

Area = Length × Width

= 7 × 4

= 28 cm2

If both the length and the width are increased by k cm, the area of the

rectangle is increased by 102 cm2.

(ii) Form an equation, in terms of k.

Area = length × width

28 + 102 = (7 + k) × (4 + k)

130 = 7(4 + k) + k(4 + k)

130 = 28 + 7k + 4k + k2

130 = k2 + 11k + 28

0 = k2 + 11k – 102


96

(iii) Solve this equation and hence find the length and width of the larger

rectangle.

0 = k2 + 11k – 102

0 = (k + 17)(k – 6)

k + 17 = 0 and k – 6 = 0

k = –17 and k = 6

Lengths cannot be negative, so k = – 17 is rejected.

Length = 7 + k = 7 + 6 = 13 cm

Width = 4 + k = 4 + 6 = 10 cm.

4. A garden has outside dimensions of 35 m in length

and 27 m wide. A concrete path, of width w m, is

installed all around the inside wall of the garden,

leaving a rectangular lawn in the middle of the

garden, as shown in the diagram.

(i) Find the length and width of the lawn, in terms of w.

Length = 35 – 2w

Width = 27 – 2w

(ii) Write an expression for the area of the lawn, in terms of w.

Area = Length × Width

= (35 – 2w) × (27 – 2w)

= 35(27 – 2w) – 2w(27 – 2w)

= 945 – 70w – 54w + 4w2

Area = 4w2 – 124w + 945


97

(iii) If the area of the lawn is 660 m2, find the value of w.

660 = 4w2 – 124w + 945

0 = 4w2 – 124w + 285

0 = (2w – 5)(2w – 57)

2w – 5 = 0 and 2w – 57 = 0

2w = 5 and 2w = 57

5 57

and 2 2

w w= =

w = 2·5 m and w = 28·5 m

w = 28·5 m is rejected because the width of the lawn is only 27 m.

The path is 2·5 m wide.

5. (i) If x is an even integer, write down the next consecutive even integer, in

terms of x.

x + 2

(ii) If the product of half the smaller number and three times the larger number

is 72, find x.

2

2

2

2

1 3( 2) 7221 (3 6) 722

3 3 7223 6 144

2 482 48 0

( 8)( 6) 0 8 0 and 6 0

8 and 6

x x

x x

x x

x xx x

x xx x

x xx x

× + =

× + =

+ =

+ =

+ =

+ − =+ − =

+ = − == − =


98

6. A rectangular piece of cardboard is 10 cm longer

than it is wide. Squares, of side length 2 cm, are

cut from each corner and then the sides are

folded up to make an open box.

(i) Write down the height of the box.

Height = 2 cm

(ii) Write the length and width of the box, in terms of x.

Length = x + 10 – 4 = x + 6

Width = x – 4

(iii) Find the volume of the box, in terms of x.

Volume = Length × Width × Height

= (x + 6) × (x – 4) × 2

= [x(x – 4) + 6(x – 4)] × 2

= [x2 – 4x + 6x – 24] × 2

= (x2 + 2x – 24) × 2

Volume = 2x2 + 4x – 48


99

(iv) If the volume of the box is 112 cm3, find the value of x.

112 = 2x2 + 4x – 48

0 = 2x2 + 4x – 160 (÷2)

0 = x2 + 2x – 80

0 = (x + 10)(x – 8)

x + 10 = 0 and x – 8 = 0

x = – 10 and x = 8

Dimensions cannot be negative, so x = − 10 is rejected.

Therefore, x = 8

(v) Hence, write down the length and width of the original piece of cardboard.

Length = x + 10 = 8 + 10 = 18 cm

Width = x = 8 cm

7. Three hundred metres of fencing is available to

enclose a rectangular field alongside a river. The

river forms one side of the field, as shown in the

diagram.

(i) Given the width of the field is w, find the length of the field in terms of w.

Perimeter = width + length + width = 300 m

w + l + w = 300

2w + l = 300

Length of the field: l = 300 – 2w

(ii) Find the area of the field, in terms of w.

Area = Length × width

= (300 – 2w) × w

Area = 300w – 2w2


100

(iii) What dimensions will produce an area of 10,000 m2?

10,000 = 300w – 2w2

2w2 – 300w + 10,000 = 0

w2 – 150w + 5000 = 0

(w – 100)(w – 50) = 0

w – 100 = 0 and w – 50 = 0

w = 100 and w = 50

Length: l = 300 – 2w

w = 100 ⇒ l = 300 – 2(100) = 300 – 200 = 100 m

w = 50 ⇒ l = 300 – 2(50) = 300 – 100 = 200 m

The dimensions of the field are either:

Length = 100 m and Width = 100 m

or Length = 200 m and Width = 50 m

8. The cost of producing backpacks with a school’s logo can be

modelled by:

22( ) 12 500

5C b b b= − +

where C(b) represents the cost, in euro, to produce b backpacks.

(i) Find the cost of producing 50 backpacks.

2

2

2( ) 12 50052(50) (50) 12(50) 50052 (2500) 600 50051000 100

(5 00) €90

C b b b

C

C

= − +

= − +

= − +

= −=


101

(ii) Find C(150) and explain in words what this value means.

22(150) (150) 12(150) 50052 (22,500) 1800 50059000 1300

(1 €50) 7700

C

C

= − +

= − +

= −=

The cost of producing 150 backpacks is €7,700

(iii) How many backpacks can a school get with a budget of €3,000?

2

2

2

2

23000 12 500520 12 2500 ( 5)5

0 2 60 12,500 ( 2)0 30 6250

1, 30, 6250

b b

b b

b bb b

a b c

= − +

= − − ×

= − − ÷

= − −= = − = −

2

2

42

( 30) ( 30) 4(1)( 6250)2(1)

30 900 ( 25,000)2

30 25,9002

30 10 259 15 5 2592

b b acxa

− ± −=

− − ± − − −=

± − −=

±=

±= = ±

x = 95·47 ⇒ 95

x = – 65·47 ⇒ reject.

The number of backpacks cannot be negative, so b = – 65 is rejected.

The school can produce 95 backpacks with a budget of €3,000.


102

9. A cyclist travels 45 km at a speed of x km/hr.

A jogger runs 24 km at a speed which is 6 km/hr slower than the cyclist.

(i) Write an expression for the time it takes the cyclist to complete his journey, in

terms of x.

Distance DistanceSpeed = Time =

Time Speed⇒

45

Time = x

(ii) Write an expression for the time it takes the jogger to complete his journey, in

terms of x.

Speed of the jogger = speed of the cyclist – 6

24

Time = 6x −

(iii) The cyclist completes his journey 20 minutes ahead of the jogger. Use this

information to form an equation, in terms of x.

Time for cyclist 20 mins = Time for jogger20Time for cyclist (hr) = Time for jogger6045 1 24 =

3 645 24 1

6 3

x x

x x

−

−

−−

− =−


103

(iv) Solve this equation to find the speed of the cyclist.

2

2

2

45 24 1 LCM ( )( 6)(3)6 3

45 24 1( )( 6)(3) ( )( 6)(3) ( )( 6)(3)6 3

( 6)(3)(45) ( )(3)(24) ( )( 6)(1)135 810 72 6

63 810 60 69 8100 ( 54)( 15)

x xx x

x x x x x xx xx x x x

x x x xx x x

x xx x

x

− = = −−

− − − = − − − − = −

− − = −

− = −

= − += − −

− 54 0 and 15 054 and 15

xx x

= − == =

Reject x = 54 as too high a value since the speed of the jogger is at x – 6 ⇒

54 – 6 = 48 km/hr, which is not possible.

Therefore, the speed of the cyclist is 15 km/hr

10. The revenue from selling digital cameras can be modelled

by

R(x) = −3x2 + 90x

Where R(x) represents the revenue, in thousands of euro,

from selling x thousand digital cameras.

(i) Find the revenue from selling five thousand digital cameras.

R(x) = – 3x2 + 90x

R(5) = – 3(5)2 + 90(5)

= – 3(25) + 450

= – 75 + 450

= 375

R(5) = €375,000


104

(ii) Find R(4) and explain in words what this value means.

R(4) = – 3(4)2 + 90(4)

= – 3(16) + 360

= – 48 + 360

= 312

The revenue from selling four thousand digital cameras would be €312,000.

(iii) How many cameras must the company sell to have a revenue of €600,000?

Revenue = €600,000 ⇒ R(x) = 600.

600 = – 3x2 + 90x

3x2 – 90x + 600 = 0

x2 – 30x + 200 = 0

(x – 10)(x – 20) = 0

x – 10 = 0 and x – 20 = 0

x = 10 and x = 20

The company must sell either 10,000 or 20,000 to make a profit of

€600,000.

11. A rock is dropped from a 180-metre-high cliff, so

that it falls into the sea below.

The height of the rock, ℎ, in metres, above the

surface of the water, after t seconds, is given by

the following function:

ℎ(t) = − 5t2 + 180

(i) Find the height of the rock after 2 seconds.

h(t) = – 5t2 + 180

h(2) = – 5(2)2 + 180

= – 5(4) + 180

= – 20 + 180

h(2) = 160 m


105

(ii) Find the height of the rock after 3∙5 seconds.

h(3·5) = – 5(3·5)2 + 180

= – 5(12·25) + 180

= – 61·25 + 180

h(3·5) = 118·75 m

(iii) After how many seconds will the rock be at a height of 65 m? Give your

answer to one decimal place.

65 = – 5t2 + 180

5t2 – 115 = 0

t2 – 23 = 0

Time cannot be negative so the 23− is rejected.

Time = 23 4 7958= ⋅ s

t = 4·8 s

(iv) After how many seconds will the rock hit the surface of the water?

The rock will be at the surface of the water, when it has a height of 0 m.

h(t) = 0 m

0 = – 5t2 + 180

0 = – t2 + 36

0 = t2 – 36

0 = (t)2 – (6)2

0 = (t + 6)(t – 6)

t + 6 = 0 and t – 6 = 0

t = – 6 and t = 6

Time cannot be negative, t = – 6 is rejected.

Time = 6s to hit the water.

2 23

23

t

t

=

= ±


106

12. (a) The annual net income for a clothing retailer can be modelled by the

function:

I(t) = − 1∙5t2 + 32t − 140

where I(t) represents the annual net income, in millions of euro, t years

since the company was founded in 2000.

(i) Use this model to find the net income the company took in 2005.

Explain your answer in words.

I(t) = – 1·5t2 + 32t – 140

2005 means t = 5:

I(5) = – 1·5(5)2 + 32(5) – 140

= – 1·5(25) + 160 – 140

= – 37·5 + 20

= – 17·5

During 2005, the company had a net income of – €17,500,000

(ii) Use this model to find the net income the company took in 2012.

Explain your answer in words.

2012 means r = 12:

I(12) = – 1·5(12)2 + 32(12) – 140

= – 1·5(144) + 384 – 140

= – 216 + 244

= 28

During 2012, the company had a net income of €28,000,000.


107

(iii) During which years did the company show a net income of €20

million?

I (t) = 20

20 = – 1·5t2 + 32t – 140

1·5t2 – 32t + 160 = 0

3t2 – 64t + 320 = 0

(3t – 40)(t – 8) = 0

3t – 40 = 0 and t – 8 = 0

3t = 40 and t = 8 ⇒ 2008

403

t =

(13 year 4 month) ⇒ 2013

Therefore, the company showed a net income of €20 million, during 2008

and 2013

(b) A second clothing retailer has an annual net income, which can be modelled

by the function

f (t) = 2t + 4

where f (t) represents the annual net income, in millions of euro, t years

since the company was founded in 2000.

(i) Use this model to find the net income this company took in 2005. Explain

your answer in words.

f(t) = 2t + 4

2005 means t = 5:

f(5) = 2(5) + 4

= 10 + 4

= 14

During 2005, the company had a net income of €14,000,000


108

(ii) In which year(s) did the two companies have the same net income? Justify

your answers.

I (t) = f(t)

– 1·5t2 + 32t – 140 = 2t + 4

0 = 1·5t2 – 30t + 144

0 = 3t2 – 60t + 288

0 = t2 – 20t + 96

0 = (t – 12)(t – 8)

t – 12 = 0 and t – 8 = 0

t = 12 and t = 8

∴ The companies had the same income in 2008 and 2012.

13. The density of a substance is defined as its ‘mass per unit

volume’. The density is calculated by dividing the mass of a

sample of the substance by its volume.

(i) Based on the information given above, write a formula for finding the

density of a substance.

MassDensity =

Volume

(ii) Rearrange this formula to make volume the subject of the formula.

MassVolume = Density


109

(iii) Liquid A has a density of D kg/litre. Write an expression, in terms of D, for

the volume of 3 kg of liquid A.

3Vol = A D

(iv) Liquid B has a density 0∙2 kg/litre greater than liquid A. Write an

expression, in terms of D, for the volume of 5∙25 kg of liquid B.

Density of liquid B = D + 0·2

5 25Vol =

0 2B D⋅+ ⋅

(v) When 3 kg of liquid A is combined with 5∙25 kg of liquid B, the total

volume of liquid present is 6∙25 litres. Use this information to write an

equation, in terms of D.

VolA + VolB = 6·25 litres

3 5 25

6 250 2D D

⋅+ = ⋅

+ ⋅

(vi) Hence, find the value of D.

2

2

2

3 5 25 6.25 LCM = ( )( 0 2)(1)0 2

3 5 25( )( 0 2)(1) ( )( 0 2)(1) ( )( 0 2)(1)(6 25)0 2

( 0 2)(3) ( )(5 25) ( )( 0 2)(6 25)3 0 6 5 25 6 25 1 25

8 25 0 6 6 25 1 250 6 25

D DD D

D D D D D DD D

D D D DD D D D

D D DD

⋅+ = + ⋅

+ ⋅⋅ + ⋅ + + ⋅ = + ⋅ ⋅ + ⋅

+ ⋅ + ⋅ = + ⋅ ⋅

+ ⋅ + ⋅ = ⋅ + ⋅

⋅ + ⋅ = ⋅ + ⋅

= ⋅ 7 0 66 25, 7, 0 6

Da b c

− − ⋅= ⋅ = − = − ⋅


110

2

2

42

( 7) ( 7) 4(6 25)( 0 6)2(6 25)

7 49 ( 15)12 5

7 6412 5

7 812 5

b b acxa

− ± −=

− − ± − − ⋅ − ⋅=

⋅

± − −=

⋅±

=⋅

±=

⋅

7 8 15 1 212 5 12 5

and7 8 1 0 0812 5 12 5

D

D

+= = = ⋅

⋅ ⋅

− −= = = − ⋅

⋅ ⋅

[Reject the negative value.]

∴ D = 1·2 kg/litre


111

14. A circle has equation x2 + y2 = 100 and a line has equation x + y = − 2.

Solve these equations simultaneously to find the points where the line intersects

the circle.

x + y = – 2 ⇒ x = – 2 – y

Substitute x = – 2 – y into the quadratic equation:

x2 + y2 = 100 ⇒ (–2 – y)2 + y2 = 100

(–2 – y)(–2 – y) + y2 = 100

[–2(–2 – y) –y(–2 – y)] + y2 = 100

[4 + 2y + 2y + y2] + y2 = 100

2y2 + 4y + 4 = 100

2y2 + 4y – 96 = 0

y2 + y – 48 = 0

(y + 8)(y – 6) = 0

y + 8 = 0 and y – 6 = 0

y = –8 and y = 6

x + y = – 2 ⇒ x = – 2 – y

y = – 8 ⇒ x = – 2 – (– 8) = – 2 + 8 = 6 ∴ (6, – 8)

y = 6 ⇒ x = – 2 – 6 = – 8 ∴ (– 8, 6)

15. (a) During testing, a missile is launched vertically

upwards.

The height of the missile ℎ, after t seconds, can be

modelled by the following function:

ℎ(t) = −5t2 + 50t + 7


112

(i) Copy and complete the following table, showing the height of the

missile at the indicated times:

t(sec) 0 1 2 3 4 5 6 7 8 9 10

h(t)

(m)

t(sec) 0 1 2 3 4 5 6 7 8 9 10

h(t)

(m) 7 52 87 112 127 132 127 112 87 52 7

(ii) Draw a graph of the height of the missile for the first 10 seconds of

its flight.

(iii) The point (3, 112) is on this graph. Explain in words what the

coordinates of this point mean.

After three seconds, the vertical height of the rocket is 112 m


113

(iv) Use your graph to estimate the greatest height that the missile

reaches.

Greatest height = 132 m

(v) Use your graph to estimate two times that the height of the missile is

at 100 m.


• Draw a horizontal line across until you touch the graph.

• Draw vertical lines from here, down from the graph to

the t-axis

• Read off the values on the t-axis

The two times when the rocket is at 100 m are: t = 2·5 s and t = 7·5 s


114

(vi) Use algebra to verify your answers to part (v).

h(t) = –5t2 + 50t + 7 = 100

0 = 5t2 – 50t + 93

a = 5, b = –50, c = 93

2

2

42

( 50) ( 50) 4(5)(93)2(5)

50 2500 186010

50 640 50 8 10 7.53 and = 2.4710 10

b b acxa

s s

− ± −=

− − ± − −=

± −=

± ±= = =

(b) Another missile is launched and it follows the path of the function

f (t) = 10t + 7.

(i) What shape is formed by the graph of f (t)?

It is a linear function, therefore it is a straight line.


115

(ii) Using the same axes and scale as above, graph the function f (t) in

the domain 0 ≤ t ≤ 10.

f (t) = 10t + 7

When t = 0: f (0) = 10(0) + 7 = 7 (0, 7)

When t = 10: f (10) = 10(10) + 7 = 107 (10, 107)

(iii) Find the points where the paths of the two missiles meet.

The points where the two missiles meet are the points where the two

graphs intersect. This occurs at: (0, 7) and (8, 87)


116

(iv) Use algebra to verify your answer to part (iii).

h(t) = f(t)

– 5t2 + 50t + 7 = 10x + 7

0 = 5t2 – 40t

0 = t2 – 8t

0 = t (t – 8)

t = 0 and t – 8 = 0

t = 8

t = 0: h(0) = –5(0)2 + 50(0) + 7 f(0) = 10(0) + 7

= 7 m = 7 m

(0, 7)

t = 8: h(8) = – 5(8)2 + 50(8) + 7 f(8) = 10(8) + 7

= – 5(64) + 400 + 7 = 80 + 7

= – 320 + 400 + 7 = 87 m

= 87 m

(8, 87)

16. The total amount of fresh vegetables consumed per

person per year can be modelled by

v(t) = 4(t − 2∙5)2 + 40

where v(t) is the total amount of fresh vegetables

consumed per person in kilograms per year, for t years

since 2005 (i.e. t = 0 is 2005, t = 1 is 2006, etc.)


117

(i) Express v(t) in the form at2 + bt + c, where a, b, c ∈ ℤ.

v(t) = 4 (t – 2·5)2 + 40

= 4 (t – 2·5)(t – 2·5) + 40

= 4 [t(t – 2·5) – 2·5(t – 2·5)] + 40

= 4 [t2 – 2·5t – 2·5t + 6·25] + 40

= 4 [t2 – 5t + 6·25] + 40

= 4t2 – 20t + 25 + 40

v(t) = 4t2 – 20t + 65

(ii) Draw a graph of this model for the years 2005 to 2015, inclusive.

(iii) The point (6, 89) is on the curve. Explain in words what the coordinates of

this point represent.

When t = 6, which is in 2011, the consumption of fresh vegetables is

89 kg per person per year.


118

(iv) Use the graph to estimate when the vegetable consumption was at 110 kg

per person per year.

Go to 110 on the vertical axis.

Draw a horizontal line across to the graph

Go down to the horizontal graph from here and read off the value.

The consumption was 110 kg after 6·7 years.

(v) Use your graph to estimate the vegetable consumption halfway between

2011 and 2012.

Halfway between 2011 and 2012 will be at the point 6·5 on the horizontal

axis.

• Go to 6·5 on the 𝑥𝑥-axis

• Go up to the graph from this point.

• Go across horizontally from here.

Halfway between 2011 and 2012 the consumption was 104 kg.


119

17. A cargo boat travels along a river, moving logs of wood from one location to

another, a distance 75 km away. The boat then returns to its original site, to be

loaded up for the next day.

The boat is able to travel at a speed of V km/hr in still water and the river has a

steady current of 5 km/hr.

(i) Write an expression for the time it takes the boat to travel the 75 km

downstream.

Downstream means that the boat is moving with the current.

Speed = Speed of the boat in still water + speed of the current

Distance DistanceSpeed = Time =

Time Speed⇒

75Time to go downstream =

5v +

(ii) Write an expression for the time it takes the boat to travel the 75 km

upstream.

Upstream means that the boat is moving against the current.

Speed = Speed of the boat in still water − speed of the current

75Time to go upstream =

5v −


120

(iii) If the boat is travelling for a total of eight hours, find the value of V, the

speed of the boat in still water.

Time to go downstream + Time to go upstream = 8

2

2

2

2

75 75 8 LCM ( 5)( 5)(1)5 5

75 75( 5)( 5)(1) ( 5)( 5)(1) ( 5)( 5)(1)(8)5 5

( 5)(75) ( 5)(75) ( 5)(8 40)75 375 75 375 8 40 40 200

150 8 2000 8 150 2000 4 75 1000

v vv v

v v v v v vv v

v v v vv v v v v

v vv vv v

+ = = + −+ −

+ − + + − = + − + − − + + = + −

− + + = − + −

= −

= − −

= − −= (4 5)( 20)

4 5 0 and 20 04 5 and 20

54

v vv v

v v

v

+ −+ = − =

= − =

= −

Speed cannot be negative, so 54

v = − is rejected.

The speed of the boat in still water is 20 km/hr.


121

(iv) On a given day, the current of the river was 2 km/hr. What difference will

this make to the travelling time of the boat? Give your answer to the nearest

minute.

v = 20

75 752 2

75 7520 2 20 2

75 7522 187 57 hr

7 hr 34 mins 32s .

tv v

t

t

tt

+ =+ −

+ =+ −

+ =

⋅ ==

It would complete the return journey 25 minutes faster.


122

18. A circle has equation x2 + y2 = 169 and a line has equation x − 5y = − 13.

Solve these equations simultaneously to find the points where the line intersects

the circle.


x – 5y = –13 ⇒ x = 5y – 13

Substitute x = 5y – 13 into the quadratic equation

x2 + y2 = 169 ⇒ (5y – 13)2 + y2 = 169

(5y – 13)(5y – 13) + y2 = 169

5y (5y – 13) –13(5y – 13) + y2 = 169

25y2 – 65y – 65y + 169 + y2 = 169

26y2 – 130y + 169 = 169

26y2 – 130y = 0

y2 – 5y = 0

y(y – 5) = 0

y = 0 and y – 5 = 0

y = 5

x – 5y = –13 ⇒ x = 5y – 13

y = 0 ⇒ x = 5(0) – 13 = 0 – 13 = –13 ∴ (–13, 0)

y = 5 ⇒ x = 5(5) – 13 = 25 – 13 = 12 ∴ (12, 5).

Revision and Exam Style Questions – Section A 1. Form the quadratic equation, whose roots are − 3 and 4. Roots x = – 3 and x = 4

x + 3 = 0 and x – 4 = 0

(x + 3)(x – 4) = 0

x(x – 4) + 3(x – 4) = 0

x2 – 4x + 3x – 12 = 0

x2 – x – 12 = 0.


123

2. Solve the following quadratic equations:

(i) 2x2 – 7x – 15 = 0

Method 1:

2x2 – 7x – 15 = 0

(2x + 3)(x – 5) = 0

2x + 3 = 0 x – 5 = 0

2x = – 3 x = 5

x = – 1·5

Method 2:

2x2 – 7x – 15 = 0

2x2 – 7x – 15 = 0 a = 2, b = – 7, c = – 15

2

2

42

( 7) ( 7) 4(2)( 15)2(2)

7 49 ( 120)4

7 1694

7 13 5 and 1 54

b b acxa

− ± −=

− − ± − − −=

± − −=

±=

±= = = − ⋅

(ii) 6x2 = 1 – x 6x2 = 1 – x 6x2 + x – 1 = 0 (3x – 1)(2x + 1) = 0 3x – 1 = 0 and 2x + 1 = 0 3x = 1 2x = –1

13

x = 12

x = −


124

3. Solve k(k + 6) = 16 and hence find the two possible values of 3k2 + 2k − 1. k(k + 6) = 16 k2 + 6k = 16 k2 + 6k – 16 = 0 (k + 8)(k – 2) = 0 k + 8 = 0 and k – 2 = 0 k = – 8 and k = 2

3k2 + 2k – 1 k = –8 3(–8)2 + 2(–8) – 1 3(64) + (–16) – 1 192 – 16 – 1 175

k = 2 3(2)2 + 2(2) – 1 3(4) + 4 – 1 12 + 4 – 1 15

4. Solve 6 1 31 1y y

+ =+ −

2

2

2

6 1 3 LCM ( 1)( 1)(1)1 1

6 1( 1)( 1)(1) ( 1)( 1)(1) ( 1)( 1)(1)(3)1 1

( 1)(6) ( 1) ( 1)(3 3)6 6 1 3 3 3 3

7 5 3 30 3 7 20 (3 1)( 2)

3 1 0 and 2 03

y yy y

y y y y y yy y

y y y yy y y y y

y yy yy y

y y

+ = = + −+ −

+ − + + − = + − + −

− + + = + −

− + + = − + −

− = −

= − += − −

− = − =1 213

y y

y

= =

=


125

5. Solve the following simultaneous equations x + 3 = 2y and xy − 7y + 8 = 0.


x + 3 = 2y ⇒ x = 2y – 3

Substitute x = 2y – 3 into the quadratic equation:

xy – 7y + 8 = 0 ⇒ (2y – 3) y – 7y + 8 = 0

2y2 – 3y – 7y + 8 = 0

2y2 – 10y + 8 = 0

y2 – 5y + 4 = 0

(y – 1) (y – 4) = 0

y – 1 = 0 and y – 4 = 0

y = 1 and y = 4

x = 2y – 3

y = 1 ⇒ x = 2(1) – 3 = 2 – 3 = – 1 → (–1, 1)

y = 4 ⇒ x = 2(4) – 3 = 8 – 3 = 5 → (5, 4)

6. When a number x is subtracted from its square, the result is 42.

Write down an equation in x to represent this information and solve it to calculate

two possible values for x.

x2 – x = 42

x2 – x – 42 = 0

(x – 7)(x + 6) = 0

x – 7 = 0 and x + 6 = 0

x = 7 and x = – 6.


126

7. Solve the equation x2 − 6x − 18 = 0, giving your answer in the form a a a± ,

where a ∈ .

2

2

2

6 18 0 1, 6, 18

42

( 6) ( 6) 4(1)( 18)2(1)

6 36 ( 72)2

6 36 722

6 1082

6 6 3 3 3 32

x x a b c

b b acxa

− − = = = − = −

− ± −=

− − ± − − −=

± − −=

± +=

±=

±= = ±

8. A ball rolls down a slope and travels a distance 2

62t

d t= + metres in t seconds.

(i) Find the distance, d, when the time is 3 seconds.

2

2

62

3(3) 6(3)2

9182

18 4 522 5 m

td t

d

= +

= +

= +

= + ⋅= ⋅


127

(ii) Find the time taken for the ball to travel 17 m.

Give your answer correct to two decimal places.

2

2

2

2

2

6 172

12 3412 34 0

1, 12, 34

42

12 12 4(1)( 34)2(1)

12 144 ( 136)2

12 2802

12 2 702

6 702 37 and 14 37

tt

t tt t

a b c

b b acxa

+ =

+ =

+ − == = = −

− ± −=

− ± − −=

− ± − −=

− ±=

− ±=

= − ±= ⋅ − ⋅

Time cannot be negative, so t = –14·37 is rejected.

Time = 2·37 s to travel 17 m.

9. A man jogs at an average speed of x km/hr for (x – 5) hours.

If the man jogged a distance of 24 km,

(i) use the information to form an equation.

Distance

Speed Distance Speed TimeTime

= ⇒ = ×

24 = x (x – 5)

24 = x2 – 5x

0 = x2 – 5x – 24


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(ii) Hence, solve the equation to find x.

0 = (x – 8) (x + 3)

x – 8 = 0 and x + 3 = 0

x = 8 and x = –3

Speed cannot be negative

Therefore, x = 8

10. A quadratic function has a positive coefficient on the x2 term and roots of − 5

and 0. Draw a sketch of this function.

Roots of x = –5 and x = 0, means that the graph crosses the x-axis at these points.

A positive coefficient on the x2 term means that the graph is U shaped.


129

11. Find the roots of the function f (x) = 3x2 + 4x – 6. Give your answers to two

decimal places.

( )2

2

2 4 6 0

42

4 4 4(3)( 6)2(3)

4 16 ( 72)6

4 886

4 2 226

4 2 22 4 2 22 and 6 6

0 89681

3 3, 4

–

,

6x

b

f x

b acx

x

a

x x

x x

a b c+ − =

− ± −=

− ± − −=

− ± − −=

− ±=

− ±=

− + − −= =

= ⋅

= = = =

2 2301 0 90 2 23x x

= − ⋅= ⋅ = − ⋅


130

Revision and Exam Style Questions – Section B More challenging problems

1. (i) Find the roots of the function f (x) = x2 − 2x − 2

( ) 2

2

2

2

42

( 2) ( 2) 4(1)(

2)2(1)

2 4 ( 8)2

2 122

2 2 32

1 3 2

– 2 0 1, –2, –

73 and 0 73

2

b b acxa

x x

f x x x a b c−

− ± −=

− − ± − − −=

± − −=

=

±=

±=

= ±= ⋅ =

=

−

=

⋅

= =

(ii) Hence draw a sketch of the graph of f (x).

Roots of x = –0·73 and x = 2·73, means that the graph crosses the x-axis at

these points.

A positive coefficient on the x2 term means that the graph is U shaped.


131

2. A closed rectangular box has a square base of side x cm.

The height of the box is 7 cm.

The total surface area of the box is 240 cm2.

Write down an equation in x to represent this information

and use it to calculate x.

Surface area = 2(l w) + 2(l h) + 2(w h)

= 2(x x) + 2(x 7) + 2(x 7)

= 2(x2) + 2(7x) + 2(7x)

= 2x2 + 14x + 14x

= 2x2 + 28x

240 = 2x2 + 28x

0 = 2x2 + 28x – 240

0 = x2 + 14x – 120

0 = (x + 20)(x – 6)

x + 20 = 0 and x – 6 = 0

x = –20 and x = 6

Lengths cannot be negative so x = –20 is rejected

∴ x = 6.


132

3. (i) Solve the following simultaneous equations

p + q = 1 and p2 + q2 = 13


p + q = 1 ⇒ p = 1 – q

Substitute p = 1 – q into the quadratic equation:

p2 + q2 = 13 ⇒ (1 – q)2 + q2 = 13

(1 – q)(1 – q) + q2 = 13

1(1 – q) – q(1 – q) + q2 = 13

1 – q – q + q2 + q2 = 13

1 – 2q + 2q2 = 13

2q2 – 2q – 12 = 0

q2 – q – 6 = 0

(q – 3)(q + 2) = 0

q – 3 = 0 and q + 2 = 0

q = 3 and q = –2

p + q = 1 ⇒ p = 1 – q

q = 3 ⇒ p = 1 – 3 = –2 = p

q = – 2 ⇒ p = 1 – (–2) = 1 + 2 = 3 = p

(ii) Hence find the two possible values of 2p − 3q.

p = –2, q = 3: 2p – 3q 2(–2) – 3(3) – 4 – 9 –13

p = 3, q = –2: 2p – 3q 2(3) – 3(–2) 6 – (–6) 12


133

4. Five hundred metres of fencing is available to

enclose a rectangular site along the side of a

motorway, as shown in the diagram. The

motorway forms one side of the site, so fencing

is only needed for three sides of the site.

(i) Given the width of the site is w, find the length of the field in terms of w.

Perimeter = w + w + l = 500

2w + l = 500

l = 500 – 2w

(ii) Find the area of the site, in terms of w.

Area = l × w

= (500 – 2w) × w

Area = 500w – 2w2

(iii) What dimensions will produce an area of 20,000 m2?

20,000 = 500w – 2w2

2w2 – 500w + 20,000 = 0

w2 – 250w + 10,000 = 0

(w – 200)(w – 50) = 0

w – 200 = 0 and w – 50 = 0

w = 200 and w = 50

Therefore, for an area of 20,000 m2:

w = 200 ⇒ l = 500 – 2(200) = 500 – 400 = 100 m

or

w = 50 ⇒ l = 500 – 2(50) = 500 – 100 = 400 m


134

5. A packet of raisins costs €4·00.

(i) If the packet contains x grams of raisins, write an expression in x to

represent the cost, in cent, of one gram of raisins.

€4 00 400c g 400c

400cg

4001g

xxx x

x

⋅ =

=

=

=

During a promotion, the manufacturer adds an extra 50 grams of raisins

into the packets. The cost of the packet of raisins remains at €4·00.

(ii) Write an expression in x to represent the cost, in cent, of 1 gram of the

raisins during the promotion.

( 50)g 400c( 50)g 400c( 50) ( 50)

4001 g( 50)

xxx x

x

+ =+

=+ +

=+

During the promotion, each gram of raisins costs 0·4 c less than before the

promotion.

(iii) Write an equation in x to represent the above information.

Cost per gram before promotion – cost per gram during promotion = 0·4

400 4000 4

50400 400 2

50 5

x x

x x

− = ⋅+

− =+


135

(iv) Solve this equation to find how many grams of raisins are in the box during

the promotion.

LCM = (x)(x + 50)(5)

400 400 2( )( 50)(5) ( )( 50)(5) ( )( 50)(5)5 5

x x x x x xx x

+ − + = + +

(x + 50)(5)(400) – (x)(5)(400) = (x)(x + 50)(2)

2000x + 100000 – 2000x = 2x2 + 100x

0 = 2x2 + 100x – 100000

0 = x2 + 50x – 50000

0 = (x – 200)(x + 250)

x – 200 = 0 and x + 250 = 0

x = 200 and x = – 250

The number of raisins cannot be negative so x = –250 is rejected.

x = 200

x + 50 ⇒ 200 + 50 = 250 g in a packet during the promotion.

6. A square sheet of cardboard measures 8 cm by 8 cm.

A square of side x cm is removed from each corner.

The remaining piece of cardboard is folded to form an open box, as shown.


136

(a) Show that the area, in cm2, of each side of the box is 8x − 2x2.

Length = 8 – 2x

Height = x

Area = Length × Height

= (8 – 2x) × x

= 8x – 2x2

(b) Let f be the function f (x) = 8x − 2x2 .

Evaluate f (x) when x = 0, 1, 2, 3, 4.

Hence, draw the graph of f for 0 ≤ x ≤ 4, x ∈ .

x 8x – 2x2 f(x)

0 8(0) – 2(0)2 0

1 8(1) – 2(1)2 6

2 8(2) – 2(2)2 8

3 8(3) – 2(3)2 6

4 8(4) – 2(4)2 0


137

(c) Use your graph from part (b) to estimate:

(i) the area of a side when x = 0·5

x = 0·5 ⇒ f(x) = 3·5 = Area = 3·5 cm2

(ii) the maximum possible area of a side

From the graph, maximum Area = 8 cm2

(iii) the value of x that gives sides of maximum area

The maximum point on the graph is where f(x) = 8 ⇒ x = 2

(iv) the length and height of a side of maximum area.

The maximum area is when x = 2

Length = 8 – 2x = 8 –2(2) = 8 – 4 = 4 cm

Height = 2 cm


138

7. (i) Solve the following simultaneous equations: x + y = 4 and y = − x2 + 5x − 1


x + y = 4 ⇒ y = 4 – x

Substitute y = 4 – x into the quadratic equation:

y = – x2 + 5x – 1 ⇒ 4 – x = – x2 + 5x – 1

x2 – 6x + 5 = 0

(x – 1)(x – 5) = 0

x – 1 = 0 and x – 5 = 0

x = 1 and x = 5

x + y = 4 ⇒ y = 4 – x

x = 1 ⇒ y = 4 – 1 = 3 ∴ (1, 3)

x = 5 ⇒ y = 4 – 5 = – 1 ∴ (5, – 1)

(ii) Which of the sketches below represents the graphs of the equations in

part (i)?


(a)


139

(b)

(c)

Graph (a)

The points of intersection are (1, 3) and (5, – 1) and Graph (a) is

the only one that has an intersection point with a negative y value.

8. A teacher is organising an outing for her class. The total cost

for the trip comes to €210.

This covers the cost of a bus and entry to a museum.

(i) Taking x to be the total number of students in the class, write an expression

in x to represent the cost of the trip per student.

Total costCost per student = Number of students210Cost per student =

x

If three members of the class decide not to go on the trip, then the total cost

drops to €200.


140

(ii) Write an expression in x to represent the cost of the trip per student in

this case.

Total costCost per student = Number of students200Cost per student =

3x −

The cost of the trip per student, in this case, would be increased by 50 c.

(iii) Write an equation in x to represent the above information.

Cost per student with reduced numbers – original cost per student = 0·5

200 210 1

3 2x x− =

−

(iv) Solve this equation to find the number of students in the class.

200 210 1

LCM ( )( 3)(2)3 2

x xx x

− = = −−

200 210 1( )( 3)(2) ( )( 3)(2) ( )( 3)(2)3 2

x x x x x xx x

− − − = − −

(x)(2)(200) – (x – 3)(2)(210) = (x)(x – 3)

400x – 420x + 1260 = x2 – 3x

1260 – 20x = x2 – 3x

0 = x2 + 17x – 1260

0 = (x – 28)(x + 45)

x – 28 = 0 and x + 45 = 0

x = 28 and x = – 45

x = – 45 is rejected as the number of students cannot be negative

x = 28 = number of students


141

9. A picture frame manufacturer finds that the cost €C of making x picture frames

per day is given by

C(x) = x2 − 22x + 200

(i) Complete the following table, showing the cost of making each number of

picture frames per day.

x 0 2 4 6 8 10 12 14 16 18 20

C(x) (€)

x 0 2 4 6 8 10 12 14 16 18 20

C(x) (€) 200 160 128 104 88 80 80 88 104 128 160

(ii) Hence, sketch the graph of the C(x) function.

(iii) What is the cost if no picture frames are made on a particular day?

If no picture frames were made, then x = 0.

When x = 0, C(x) = €200


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(iv) The graph shows that the production of seven picture frames will cost €95. What other number of picture frames will also cost €95 to produce?

Draw a horizontal line across from €95. At the points where is touches the graph, draw vertical lines down. Read the values off the x-axis

C(x) = €95 when x = 7 and x = 15. Therefore, producing 15 frames will have a cost of €95.

(v) If the manufacturer wants to keep his costs to a minimum, how many picture frames should he produce each day?

From the graph, we can see that the minimum cost occurs when x = 11


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(vi) On a given day, the manufacturer produces 17 picture frames. He sells them

for €12 each. Find the profit he has made that day.

Selling price = 17 × €12

= €204

Production cost: C(17) = €115

Profit = Selling price – Production cost

= 204 – 115

= €89

(vii) Given that the manufacturer sells all of his picture frames for €12, write an

expression for the income he receives, per day, by selling x frames.

Income = 12x

(viii) Hence, write an expression for the profit he makes if he produces and sells

x picture frames on a given day.

Profit = Income – Cost

= 12x – (x2 – 22x + 200)

= 12x – x2 + 22x – 200

Profit = – x2 + 34x – 200

Algebra III – Quadratic Relations Practice Questions 4.1 - Gill ...

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