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9780521740494c06.xml CUAU033-EVANS September 9, 2008 8:36

C H A P T E R

6Algebra II

ObjectivesTo understand equality of polynomials

To use equating coefficients to solve problems

To solve quadratic equations by various methods

To use rates to solve problems

To resolve a rational algebraic expression into partial fractions

To find the coordinates of the points of intersection of linear graphs with

� parabolas

� rectangular hyperbolae

� circles

6.1 Polynomial identitiesPolynomials are introduced in Chapter 3 of Essential Mathematical Methods 1 & 2. A

polynomial function has a rule of the type

y = anxn + an−1xn−1 + . . . a1x + a0, n ∈ N

where a0, a1, . . . an are numbers called coefficients.

The degree of a polynomial is given by the value of n, the highest power of x with non-zero

coefficient.

Two polynomials are equal if they give the same value for all x. If two polynomials are

equal then they are of the same degree, and corresponding coefficients are equal.

For example, if ax + b = cx2 + dx + e, then c = 0, d = a and e = b

if x2 − x − 12 = x2 + (a + b)x + ab then ab = −12 and a + b = −1

This process is called equating coefficients.

Example 1

If the expressions (a + 2b)x2 − (a − b)x + 8 and 3x2 − 6x + 8 are equal for all x, find the

values of a and b.

151

SAMPLE

Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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152 Essential Advanced General Mathematics

Solution

If (a + 2b)x2 − (a − b)x + 8 = 3x2 − 6x + 8 for all x,

then a + 2b = 3 and −(a − b) = −6

Solve as simultaneous equations

a + 2b = 3 . . . 1

−a + b = −6 . . . 2

Add 1 and 2

3b = −3

∴ b = −1

Substitute into 2

∴ a = 5

Example 2

Express x2 in the form c(x − 3)2 + a(x − 3) + d.

Solution

If x2 = c(x − 3)2 + a(x − 3) + d,

then x2 = c(x2 − 6x + 9) + a(x − 3) + d

= cx2 + (a − 6c)x + 9c − 3a + d

which implies c = 1 . . . 1

a − 6c = 0 . . . 2

9c − 3a + d = 0 . . . 3

From 2 a = 6

and from 3 9 − 18 + d = 0

i.e. d = 9

∴ x2 = (x − 3)2 + 6(x − 3) + 9

Example 3

Find the values of a, b, c and d such that

x3 = a(x + 2)3 + b(x + 1)2 + cx + d for all x .

Solution

Expand the right hand side.

a(x3 + 6x2 + 12x + 8) + b(x2 + 2x + 1) + cx + d

Collect like terms.

ax3 + (6a + b)x2 + (12a + 2b + c)x + (8a + b + d)

If x3 = ax3 + (6a + b)x2 + (12a + 2b + c)x + (8a + b + d)

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Chapter 6 — Algebra II 153

then a = 1 . . . 1

6a + b = 0 . . . 2

12a + 2b + c = 0 . . . 3

8a + b + d = 0 . . . 4

Substituting a = 1 into 2 gives

b = −6a

= −6

Substituting a = 1 and b = −6 into 3 gives

12 − 12 + c = 0

∴ c = 0

Substituting a = 1 and b = −6 into 4 gives

8 − 6 + d = 0

∴ d = −2

x3 = (x + 2)3 − 6(x + 1)2 − 2

Example 4

Show that 2x3 − 5x2 + 4x + 1 cannot be expressed in the form a(x + b)3 + c.

Solution

Assume that 2x3 − 5x2 + 4x + 1 can be expressed in the form a(x + b)3 + c

i.e., 2x3 − 5x2 + 4x + 1 = a(x + b)3 + c

Expanding the right hand side

2x3 − 5x2 + 4x + 1 = a(x3 + 3bx2 + 3b2x + b3)c

= ax3 + 3abx2 + 3ab2x + ac

Equating coefficients a = 2 . . . 1

3ab = −5 . . . 2

3ab2 = 4 . . . 3

and ab3 + c = 1 . . . 4

From 2 b = −5

6, but from 3 b =

√2

3=

√6

3We have a contradiction and therefore have shown that 2x3 − 5x2 + 4x + 1 cannot be

expressed in the form a(x + b)3 + c.

Exercise 6A

1 If ax2 + bx + c = 10x2 − 7, find the values of a, b and c.Example 1

2 If (2a − b)x2 + (a + 2b)x + 8 = 4x2 − 3x + 8, find the values of a and b.

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3 If (2a − 3b)x2 + (3a + b)x + c = 7x2 + 5x + 7, find the values of a, b and c.

4 If 2x2 + 4x + 5 = a(x + b)2 + c, find the values of a, b and c.

5 Express x2 in the form c(x + 2)2 + a(x + 2) + d.Example 2

6 Express x3 in the form (x + 1)3 + a(x + 1)2 + b(x + 1) + c.

7 Find the values of a, b and c such that x2 = a(x + 1)2 + bx + c.Example 3

8 a Show that 3x3 − 9x2 + 8x + 2 cannot be expressed in the form a(x + b)3 + c.Example 4

b If 3x3 − 9x2 + 9x + 2 can be expressed in the form a(x + b)3 + c, then find the

values of a, b and c.

9 Show that constants a, b, c and d can be found such that

n3 = a(n + 1)(n + 2)(n + 3) + b(n + 1)(n + 2) + c(n + 1) + d

10 a Show that no constants can be found such that

n2 = a(n + 1)(n + 2) + b(n + 2)(n + 3)

b Express n2 in the form a(n + 1)(n + 2) + b(n + 1) + c

11 a Express a(x + b)2 + c in expanded form.

b Express ax2 + bx + c in completed square form.

12 Prove that, if ax3 + bx2 + cx + d = (x − 1)2(px + q), then

b = d − 2a and c = a − 2d

13 If 3x2 + 10x + 3 = c(x − a)(x − b) for all values of x, find the values of a, b and c.

14 If n is any number, show that n2 can be expressed in the form

a(n − 1)2 + b(n − 2)2 + c(n − 3)2, and find the values of a, b and c.

15 If x3 + 3x2 − 9x + c is of the form (x − a)2(x − b), show that c = 5 or c = −27 and

find a and b for each of these cases.

6.2 Quadratics and ratesQuadraticsThe general expression of a quadratic function is y = ax2 + bx + c, x ∈ R, a �= 0

The number of solutions to the quadratic equation ax2 + bx + c = 0 can be determined by the

discriminant, �, where � = b2 − 4ac

i If b2 − 4ac > 0, the quadratic equation ax2 + bx + c = 0 has two real solutions

ii If b2 − 4ac = 0, the quadratic equation ax2 + bx + c = 0 has one real solution

iii If b2 − 4ac < 0, the quadratic equation ax2 + bx + c = 0 has no real solutions

A quadratic equation may be solved by factorising, completing the square or using the general

quadratic formula x = −b ± √b2 − 4ac

2a. The following example demonstrates each method.

SAMPLE


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Example 5

Solve the following quadratic equations for x.

a x2 + 3x = 4 b 3x2 + 4x = 2 c 9x2 + 6x + 1 = 0

Solution

a Rearranging the quadratic equation x2 + 3x − 4 = 0

Factorising (x + 4)(x − 1) = 0

Applying the null factor law x + 4 = 0 or x − 1 = 1

Therefore x = −4 or 1

Note: � = 32 − 4 × 1 × (−4) = 25, so there are two real solutions.

Using the TI-NspireUse Solve( ) from the Algebra menu (b

1) as shown.

Using the Casio ClassPadEnter and highlight the equation

2x2 + 5x − 12 = 0, tap Interactive,Equation/inequality, solve and ensure

the variable is set to x.

SAMPLE


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b Rearranging the quadratic equation 3x2 + 4x − 2 = 0

3

(x2 + 4

3x

)− 2 = 0

Add and subtract

(b

2

)2

to ‘complete the square’

3

(x2 + 4

3x +

(2

3

)2

−(

2

3

)2)

− 2 = 0

3

((x + 2

3

)2

− 4

9

)− 2 = 0

3

(x + 2

3

)2

− 4

3− 2 = 0

3

(x + 2

3

)2

= 10

3(x + 2

3

)2

= 10

9

x + 2

3= ±

√10

3

x = −2

3±

√10

3

Therefore x = −2 + √10

3or

−2 − √10

3

Note: � = 42 − 4 × 3 × (−2) = 40, so there are two real solutions.

c Consider 9x2 + 6x + 1 = 0

Using the general quadratic formula x = −b ± √b2 − 4ac

2a

x = −6 ± √62 − 4 × 9 × 1

2 × 9

= −6 ± √0

18

Therefore x = −1

3

More simply 9x2 + 6x + 1 = (3x + 1)2

Note: � = 62 − 4 × 9 × 1 = 0, so there is one real solution.

Example 6

Consider the quadratic equation x2 − 4x = t . Make x the subject and give the values of t for

which real solution(s) to the equation can be found.SAM

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Solution

x2 − 4x = t

Completing the square x2 − 4x + 4 = t + 4

(x − 2)2 = t + 4

x − 2 = ±√t + 4

x = 2 ± √t + 4

For real solutions to exist, t + 4 ≥ 0, i.e., t ≥ −4


41) as shown.

Using the Casio ClassPadEnter and highlight the equation x2 − 4x = t ,

tap Interactive, Equation/inequality, solve and

ensure the variable is set to x.

Note: Variable t is found in the menu in the

keyboard screen.

SAMPLE


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Example 7

A rectangle has an area of 288 cm2. If the width is decreased by 1 cm, and the length increased

by 1 cm, the area would be decreased by 3 cm2. Find the original dimensions of the rectangle.

Solution

Let w and l be the width and length, in centimetres, of the original rectangle.

Therefore wl = 288 . . . 1

The dimensions of the new rectangle are w − 1 and l + 1, and the area is 285 cm2

Therefore (w − 1)(l + 1) = 285 . . . 2

Rearranging 1 to make l the subject, and substituting in 2 gives

(w − 1)

(288

w+ 1

)= 285

288 − 288

w+ w − 1 = 285

w − 288

w+ 2 = 0

w2 + 2w − 288 = 0

Using the general quadratic formula x = −b ± √b2 − 4ac

2a

w = −2 ± √22 − 4 × 1 × −288

2 × 1= −18 or 16

But w > 0, so w = 16. The original dimensions of the rectangle are 16 cm by 18 cm.

RatesA rate describes how a certain quantity changes with respect to the change in another quantity

(often time). An example of a rate is ‘speed’. A speed of 60 km/h gives us a measure of how

fast an object is travelling. A further example is ‘flow’, where a rate of 20 L/min is going to fill

an empty swimming pool faster than, say, a rate of 6 L/min.

Many problems are solved using rates, which can be expressed as fractions. For example, a

speed of 60 km/h can be expressed in fraction form as60(km)

1(h).

It is often first necessary to add two or more fractions with different denominators, as shown

in the following examples.SAMPLE


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Example 8

a Express6

x+ 6

x + 8as a single fraction.

b Solve the equation6

x+ 6

x + 8= 2 for x.

Solution

a 6

x+ 6

x + 8= 6(x + 8)

x(x + 8)+ 6x

x(x + 8)

= 6x + 48 + 6x

x(x + 8)

= 12(x + 4)

x(x + 8)

b Since6

x+ 6

x + 8= 12(x + 4)

x(x + 8)

then12(x + 4)

x(x + 8)= 2

12(x + 4) = 2x(x + 8)

6(x + 4) = x(x + 8)

6x + 24 = x2 + 8x

0 = x2 + 2x − 24

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

x = −6 or x = 4

Example 9

A tank is filled by two pipes. The smaller pipe alone will take 24 minutes longer than the larger

pipe alone, and 32 minutes longer than when both pipes are used. How long will each pipe take

to fill the tank alone? How long will it take for both pipes used together to fill the tank?

Solution

Let C cubic units be the capacity of the tank, and x minutes the time it takes for the

larger pipe alone to fill the tank. Therefore the average rate of flow for the larger pipe

isC

xcubic units per minute.

Since the smaller pipe alone takes (x + 24) minutes to fill the tank, the average rate

of flow for the smaller pipe isC

x + 24cubic units per minute.SAMPLE


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The average rate of flow when both pipes are used together is the sum of these two

rates,C

x+ C

x + 24cubic units per minute.

Expressed as a single fraction,C

x+ C

x + 24= C(x + 24) + Cx

x(x + 24)

= 2C(x + 12)

x(x + 24)

The time taken to fill the tank using both pipes is

C ÷ 2C(x + 12)

x(x + 24)= C × x(x + 24)

2C(x + 12)

= x(x + 24)

2(x + 12)

The time taken for the smaller pipe to fill the tank can be also be expressed asx(x + 24)

2(x + 12)+ 32

i.e.x(x + 24)

2(x + 12)+ 32 = x + 24

x(x + 24)

2(x + 12)= x − 8

x(x + 24) = 2(x + 12)(x − 8)

x2 + 24x = 2x2 + 8x − 192

0 = x2 − 16x − 192

0 = (x − 24)(x + 8)

x − 24 = 0 or x + 8 = 0

x = 24 or x = −8 (but x > 0)

It takes 24 minutes for the larger pipe alone to fill the tank, and 48 minutes for the

smaller pipe alone to fill the tank, and 16 minutes to fill the tank using both pipes.

Exercise 6B

1 Solve the following quadratic equations for x.Example 5

a −x2 + 2x = 1 b x2 − 6x + 9 = 0 c 5x2 − 10x = 1

d −2x2 + 4x = 1 e 2x2 + 4x = 7 f 6x2 + 13x + 1 = 0

2 Make x the subject in each of the following and give the values of t for which real

solution(s) to the equation can be found.

Example 6

a 2x2 − 4t = x b 4x2 + 4x − 4 = t − 2

c 5x2 + 4x + 10 = t d t x2 + 4t x + 10 = t

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3 a Solve the quadratic equation x2 + 3x − 9 = 0 for x, giving exact solutions.

b i Solve the quadratic equation x2 + px − 16 = 0 in terms of p.

ii Find the value(s) of p, where 0 ≤ p ≤ 10 and p is a natural number, for which the

quadratic equation in i has a natural number solution.

x m

x m

8 m

6 m

10 m

4 A pole 10 m long leans against a wall. The bottom

of the pole is 6 m from the wall. If the bottom of the

pole is pulled away x m so that the top slides down

by the same amount, find x.

5Example 8 a Express6

x− 6

x + 3as a single fraction.

b Solve the equation6

x− 6

x + 3= 1 for x.

6 The sum of the reciprocals of two consecutive odd numbers is36

323. Form a quadratic

equation and hence determine the two numbers.

Example 9

7 A car travels from town A to town B, a distance of 600 km, in x hours. A plane, travelling

220 km/h faster than the car, takes five and a half hours less to cover the same distance.

a Express, in terms of x, the average speed of the car and the average speed of the plane.

b Find the actual average speed of each of them.

8 A car covers a distance of 200 km at a speed of x km/h. A train covers the same distance at

a speed of (x + 5) km/h. If the time taken by the car is 2 hours more than that taken by the

train, find x.

9 A man travels 108 km, and finds that he could have made the journey in 41

2hours less had

he travelled at an average speed 2 km/h faster. What was the man’s average speed when he

made the trip?

10 A bus is due to reach its destination 75 km away at a certain time. The bus usually travels

with an average speed of x km/h. Its start is delayed by 18 minutes but, by increasing its

average speed by 12.5 km/h, the driver arrives on time.

a Find x. b How long did the journey actually take?

11 Ten minutes after the departure of an express train, a slow train starts, travelling at an

average speed of 20 km/h less. The slow train reaches a station 250 km away 3.5 hours

after the arrival of the express. Find the average speed of each of the trains.

12 When the average speed of a car is increased by 10 km/h the time taken for the car to

make a journey of 105 km is reduced by 15 minutes. Find the original average speed.

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13 A tank can be filled with water by two pipes running together in 111

9minutes. If the larger

pipe alone takes 5 minutes less to fill the tank than the smaller pipe, find the time that each

pipe will take to fill the tank.

14 At first two different pipes running together will fill a tank in20

3minutes. The rate that

water runs through each of the pipes is then adjusted. If one pipe, running alone, takes

1 minute less to fill the tank at its new rate, and the other pipe, running alone, takes

2 minutes more to fill the tank at its new rate, then the two running together will fill the

tank in 7 minutes. Find in what time the tank will be filled by each pipe running alone at

the new rates.

15 The journey between two towns by one route consists of 233 km by rail followed by

126 km by sea. By a second route the journey consists of 405 km by rail followed by

39 km by sea. If the time taken for the first route is 50 minutes longer than for the second

route, and travelling by rail is 25 km/h faster than travelling by sea, find the average speed

by rail and the average speed by sea.

16 A freighter sailing due north at 12 km/h sights a cruiser straight ahead at an unknown

distance and speeding due east at unknown speed. After 15 minutes the vessels are 10 km

apart and then, 15 minutes later, they are 13 km apart. (Assume both travel at constant

speeds.) How far apart are the vessels when the cruiser is due east of the freighter?

17 A cask A, of capacity 20 litres, is filled with wine. A certain quantity of wine from A is

poured into a cask B which also has a capacity of 20 litres. Cask B is then filled with

water. After this cask A is filled with some of the mixture from cask B. A further20

3litres

of the mixture now in A is poured back into B, and the two casks now have the same

amount of wine. How much wine was first taken out of cask A?

6.3 Partial fractionsA rational function is the quotient of two polynomials. If g(x) and h(x) are polynomials,

f (x) = g(x)

h(x)is a rational function,

e.g., f (x) = x2 + 1

x2 − 1

If the degree of g(x) < the degree of h(x), theng(x)

h(x)is a proper fraction.

If the degree of g(x) ≥ the degree of h(x), theng(x)

h(x)is an improper fraction.

It is convention to consider rational functions for their maximal domain. For example,x2 + 1

x2 − 1is only considered for R\{−1, 1}.

SAMPLE


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A rational algebraic function may be expressed as a sum of separate functions by resolving

it into what are called partial fractions. This can help in the sketching of graphs of these

functions or performing other mathematical procedures such as integration.

Proper fractionsFor proper fractions, the technique used for obtaining partial fractions depends on the type of

factors in the denominator of the original algebraic fraction. Only examples where the

denominators have factors that are either 1st degree (linear) or 2nd degree (quadratic) will be

considered.

For every linear factor (ax + b) in the denominator, there will be a partial fraction of the

formA

ax + bFor every repeated linear factor (cx + d)2 in the denominator, there will be partial

fractions of the formB

cx + dand

C

(cx + d)2

For every irreducible quadratic factor (ax2 + bx + c) in the denominator, there will be a

partial fraction of the formDx + E

ax2 + bx + cTo resolve an algebraic fraction into its partial fractions, firstly write a statement of identity

between the original fraction and a sum of the appropriate number of partial fractions. Then

express the sum of the partial fractions as a single fraction and note that the numerators of both

sides are equivalent. By choosing an appropriate value(s) for x and/or equating coefficients, the

values of the introduced constants A, B, C, etc. can be found.

Example 10

Resolve3x + 5

(x − 1)(x + 3)into partial fractions.

Solution

Since the denominator has two linear factors, there will be two partial fractions of the

formA

x − 1and

B

x + 3

Let3x + 5

(x − 1)(x + 3)= A

x − 1+ B

x + 3(x ∈ R\{1, −3})

Express the right hand side as a single fraction.

3x + 5

(x − 1)(x + 3)= A(x + 3) + B(x − 1)

(x − 1)(x + 3)

∴ 3x + 5

(x − 1)(x + 3)= (A + B)x + 3A − B

(x − 1)(x + 3)

∴ 3x + 5 = (A + B)x + 3A − B

SAMPLE


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Equate the coefficients,

i.e., A + B = 3

3A − B = 5

Solving these equations simultaneously,

4A = 8

i.e., A = 2

and B = 1

Therefore3x + 5

(x − 1)(x + 3)= 2

x − 1+ 1

x + 3

Using the TI-NspireUse Expand( ) from the Algebra menu (b

) as shown.

Notice that you can access the fraction

template by typing / .

Using the Casio ClassPadTap Interactive, Transformation, expand and the Partial Fraction button. Enter the

expression and the variable.

SAMPLE


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Example 11

Resolve2x + 10

(x + 1)(x − 1)2into partial fractions.

Solution

Since there is a repeated linear factor and a single linear factor, there are three partial

fractions such that

2x + 10

(x + 1)(x − 1)2= A

x + 1+ B

x − 1+ C

(x − 1)2

∴ 2x + 10

(x + 1)(x − 1)2= A(x − 1)2 + B(x + 1)(x − 1) + C(x + 1)

(x + 1)(x − 1)2

Therefore 2x + 10 = A(x − 1)2 + B(x + 1)(x − 1) + C(x + 1)

To find A, B and C, a combination of methods will be used.

First let x = 1

∴ 2(1) + 10 = C(1 + 1)

∴ 12 = 2C

∴ C = 6

Let x = −1

∴ 2(−1) + 10 = A(−1 − 1)2

∴ 8 = 4A

∴ A = 2

Substitute these values for A and C.

∴ 2x + 10 = 2(x − 1)2 + B(x + 1)(x − 1) + 6(x + 1) . . . 1

= 2(x2 − 2x + 1) + B(x2 − 1) + 6(x + 1)

= (2 + B)x2 + 2x + 8 − B

Now by equating coefficients,

2 + B = 0

and 8 − B = 10

∴ B = −2

So2x + 10

(x + 1)(x − 1)2= 2

x + 1− 2

x − 1+ 6

(x − 1)2

The value of B may also be found by substituting x = 0 into equation 1 .

In the exercises for this section, the following result is established: that it is impossible

to find A and C such that

2x + 10

(x + 1)(x − 1)2= A

(x + 1)+ C

(x − 1)2

SAMPLE


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Example 12

Resolvex2 + 6x + 5

(x − 2)(x2 + x + 1)into partial fractions.

Solution

The denominator contains a quadratic factor, which cannot be reduced to linear

factors (an irreducible quadratic), as well as a single linear factor.

∴ x2 + 6x + 5

(x − 2)(x2 + x + 1)= A

x − 2+ Bx + C

x2 + x + 1

∴ x2 + 6x + 5

(x − 2)(x2 + x + 1)= A(x2 + x + 1) + (Bx + C)(x − 2)

(x − 2)(x2 + x + 1)∴ x2 + 6x + 5 = A(x2 + x + 1) + (Bx + C)(x − 2)

Let x = 2

22 + 6(2) + 5 = A(4 + 2 + 1)

21 = 7A

A = 3

Also x2 + 6x + 5 = A(x2 + x + 1) + (Bx + C)(x − 2) . . . 1

= A(x2 + x + 1) + Bx2 − 2Bx + Cx − 2C

= (A + B)x2 + (A − 2B + C)x + A − 2C

Since A = 3, then

x2 + 6x + 5 = (3 + B)x2 + (3 − 2B + C)x + 3 − 2C

Equating coefficients

3 + B = 1

∴ B = −2

and 3 − 2C = 5

∴ C = −1

[checking: 3 − 2B + C = 3 − 2(−2) + (−1) = 6]

Thereforex2 + 6x + 5

(x − 2)(x2 + x + 1)= 3

x − 2+ −2x − 1

x2 + x + 1

orx2 + 6x + 5

(x − 2)(x2 + x + 1)= 3

x − 2− 2x + 1

x2 + x + 1

Note: The values of B and C can also be found by substituting x = 0 and x = 1

respectively in equation 1 .

Improper fractionsImproper algebraic fractions can be expressed as a sum of partial fractions by first dividing the

denominator into the numerator to produce a quotient and a proper fraction. The resulting

proper fraction can then be resolved into its partial fractions using the techniques outlined

above.

SAMPLE


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Example 13

Expressx5 + 2

x2 − 1in partial fractions.

Solution

Dividing throughx3 + x

x2 − 1)

x5 + 2

x5 − x3

x3 + 2x3 − x

x + 2

∴ x5 + 2

x2 − 1= x3 + x + x + 2

x2 − 1

Expressingx + 2

x2 − 1= x + 2

(x − 1)(x + 1)as partial fractions,

x5 + 2

x2 − 1= x3 + x − 1

2 (x + 1)+ 3

2 (x − 1)

Using the TI-NspireUse Expand( ) from the Algebra menu (b

) as shown.

Notice that you can access the fraction

template by typing / .

Using the Casio ClassPadTap Interactive, Transformation, expand and the

Partial Fraction button. Enter the expression and

the variable.SAMPLE


P1: FXS/ABE P2: FXS



Partial fractions are summarised through examples, as follows.

Linear factors3x − 4

(2x − 3)(x + 5)= A

2x − 3+ B

x + 5

Repeated linear factors

3x − 4

(2x − 3)(x + 5)2= A

2x − 3+ B

x + 5+ C

(x + 5)2

Irreducible quadratic factors

3x − 4

(2x − 3)(x2 + 5)= A

2x − 3+ Bx + C

x2 + 5

If f (x) = g(x)

h(x)and the degree of g(x) is greater than or equal to the degree of h(x) then

division must be performed first.

Exercise 6C

1 Resolve the following rational expressions into partial fractions.Example 10

a5x + 1

(x − 1)(x + 2)b

−1

(x + 1)(2x + 1)c

3x − 2

x2 − 4

d4x + 7

x2 + x − 6e

7 − x

(x − 4)(x + 1)


a2x + 3

(x − 3)2b

9

(1 + 2x)(1 − x)2c

2x − 2

(x + 1)(x − 2)2


a3x + 1

(x + 1)(x2 + x + 1)b

3x2 + 2x + 5

(x2 + 2)(x + 1)c

x2 + 2x − 13

2x3 + 6x2 + 2x + 6

4 Resolve3x2 − 4x − 2

(x − 1)(x − 2)into partial fractions.Example 13

5 Find values of A and C such that

2x + 10

(x + 1)(x − 1)2= A

x + 1+ C

(x − 1)2SAMPLE


P1: FXS/ABE P2: FXS



6 Express each of the following in partial fractions.

a1

(x − 1)(x + 1)b

x

(x − 2)(x + 3)c

3x + 1

(x − 2)(x + 5)

d1

(2x − 1)(x + 2)e

3x + 5

(3x − 2)(2x + 1)f

2

x2 − x

g3x + 1

x3 + xh

3x2 + 8

x(x2 + 4)i

1

x2 − 4x

jx + 3

x2 − 4xk

x3 − x2 − 1

x2 − xl

x3 − x2 − 6

2x − x2

mx2 − x

(x + 1)(x2 + 2)n

x2 + 2

x3 − 3x − 2o

2x2 + x + 8

x(x2 + 4)

p1 − 2x

2x2 + 7x + 6q

3x2 − 6x + 2

(x − 1)2(x + 2)r

4

(x − 1)2(2x + 1)

sx3 − 2x2 − 3x + 9

x2 − 4t

x3 + 3

(x + 1)(x − 1)u

2x − 1

(x + 1)(3x + 2)

6.4 Simultaneous equationsIn this section, methods for finding the coordinates of the points of intersection of a linear

graph with different non-linear graphs are discussed. The non-linear graphs are parabolas,

circles and rectangular hyperbolae. The associated relations have been discussed in Essential

Mathematical Methods 1 and 2.

Example 14

Find the coordinates of the points of intersection of the parabola with equation

y = x2 − 2x − 2 with the straight line with equation y = x + 4.

Solution

Consider x + 4 = x2 − 2x − 2

Then 0 = x2 − 3x − 6

∴ x = 3 ± √9 − 4 × −6 × 1

2

= 3 ± √33

2The points of intersection have

coordinates A

(3 − √

33

2,

11 − √33

2

)and

B

(3 + √

33

2,

11 + √33

2

)

y = x + 4

y = x2 – 2x – 2

xA

4

–4–2

0

y

BSAMPLE


P1: FXS/ABE P2: FXS




1) as shown.

The and can either be typed or found in

the catalog ( 1 ).Use the NavPad to move the cursor up to

the solution and see all the solutions.

Using the Casio ClassPadThe CAS calculator will give exact values for the

points of intersection.

Turn on the screen keyboard, tap 2D and select

the simultaneous equations symbol .

Enter the equations in the spaces provided and

the variables x, y as shown.

Example 15

Find the coordinates of the points of intersection of the circle with equation

(x − 4)2 + y2 = 16 and the line with equation x − y = 0.

x(4, 0)

(4, 4)

0

y

Solution

Rearrange x − y = 0 to make y the subject.

Substitute y = x into the equation of the circle.

i.e., (x − 4)2 + x2 = 16

∴ x2 − 8x + 16 + x2 = 16

i.e., 2x2 − 8x = 0

2x(x − 4) = 0

x = 0 or x = 4

The points of intersection are (0, 0) and (4, 4)

Example 16

Find the point of contact of the line with equation1

9x + y = 2

3and the curve with equation

xy = 1.

SAMPLE


P1: FXS/ABE P2: FXS



xy = – x +

91

y =x

1

32

0

y

Solution

Rewrite the equations as y = −1

9x + 2

3and y = 1

x

Consider −1

9x + 2

3= 1

x

∴ −x + 6 = 9

xand −x2 + 6x = 9

Therefore x2 − 6x + 9 = 0

and (x − 3)2 = 0, i.e. x = 3

The point of intersection is

(3,

1

3

)


1) as shown.

The and can either be typed or found in

the catalog ( 1 ).

The multiplication sign between x and y

is required because the calculator will

consider xy a variable.

Using the Casio ClassPadTurn on the screen keyboard, tap 2D and

select the simultaneous equations symbol

.

Enter the equations in the spaces

provided and the variables x, y as shown.SAMPLE


P1: FXS/ABE P2: FXS



Exercise 6D

1 Find the coordinates of the points of intersection for each of the following.Example 14

a y = x2

y = x

b y − 2x2 = 0

y − x = 0

c y = x2 − x

y = 2x + 1


a x2 + y2 = 178

x + y = 16

b x2 + y2 = 125

x + y = 15

c x2 + y2 = 185

x − y = 3

d x2 + y2 = 97

x + y = 13

e x2 + y2 = 106

x − y = 4


a x + y = 28

xy = 187

b x + y = 51

xy = 518

c x − y = 5

xy = 126

4 Find the coordinates of the points of intersection of the straight line with equation y = 2x

and the circle with equation (x − 5)2 + y2 = 25.

5 Find the coordinates of the points of intersection of the curves with equation

y = 1

x − 2+ 3 and y = x .

6 Find the coordinates of the points A and B for which the line with equation x − 3y = 0

meets the circle with equation x2 + y2 − 10x − 5y + 25 = 0.

7 Find the coordinates of the points of intersection of the line with equationy

4− x

5= 1 and

the circle with equation x2 + 4x + y2 = 12.

8 Find the coordinates of the points of intersection of the curve with equation

y = 1

x + 2− 3 and the line with equation y = −x .

9 Find the coordinates of the point where the line 4y = 9x + 4 touches the parabola with

equation y2 = 9x .

10 Find the coordinates of the point where the line with equation y = 2x + 3√

5 touches the

circle x2 + y2 = 9.

11 Find the coordinates of the point where the straight line with equation y = 1

4x + 1

touches the curve with equation y = −1

x.

12 Find the coordinates of the points of intersection of the curve with equation y = 2

x − 2and the line y = x − 1.

SAMPLE


P1: FXS/ABE P2: FXS


Review


Chapter summary

The general expression of a quadratic function is y = ax2 + bx + c, x ∈ R

A quadratic equation may be solved by� Factorising

e.g., 2x2 + 5x − 12 = 0

(2x − 3)(x + 4) = 0 ∴ x = 3

2or −4

� Completing the square

e.g., x2 + 2x − 4 = 0

Add and subtract

(b

2

)2

to ‘complete the square’.

x2 + 2x + 1 − 1 − 4 = 0

∴ (x + 1)2 − 5 = 0

∴ (x + 1)2 = 5

∴ x + 1 = ±√

5 ∴ x = −1 +√

5

� Using the general quadratic formula x = −b ± √b2 − 4ac

2a

e.g., −3x2 − 12x − 7 = 0

x = −(−12) ± √(−12)2 − 4(−3)(−7)

2(−3)

= 6 ± √15

−3

The number of solutions to the quadratic equation ax2 + bx + c = 0 can be determined by

the discriminant, �, where � = b2 − 4ac� If b2 − 4ac > 0, the quadratic equation ax2 + bx + c = 0 has two real solutions

� If b2 − 4ac = 0, the quadratic equation ax2 + bx + c = 0 has one real solution

� If b2 − 4ac < 0, the quadratic equation ax2 + bx + c = 0 has no real solutions

A function of the form f (x) = g(x)

h(x), where g(x) and h(x) are polynomials in x, is called a

rational algebraic function, e.g. f (x) = x + 1

x2 − 1Some rational algebraic functions may be expressed as a sum of partial fractions. For every

linear factor (ax + b) in the denominator there will be a partial fraction of the formA

ax + bFor every repeated linear factor (cx + d)2 in the denominator there will be two partial

fractions of the formB

(cx + d)and

C

(cx + d)2

SAMPLE


P1: FXS/ABE P2: FXS


Rev

iew


For every irreducible quadratic factor (ex2 + f x + g) in the denominator there will be a

partial fraction of the formDx + E

(ex2 + f x + g)

e.g.2x + 10

(x + 1) (x − 1)2 may be expressed as partial fractions in the form

A

(x + 1)+ B

(x − 1)+ C

(x − 1)2

where A = 2, B = −2 and C = 6

Multiple-choice questions

1 If x2 is written in the form (x + 1)2 + b(x + 1) + c, then the values of b and c are

A b = 0, c = 0 B b = −2, c = 0 C b = −2, c = 1

D b = 1, c = 2 E b = 1, c = −2

2 If x3 = a(x + 2)3 + b(x + 2)2 + c(x + 2) + d, then the values of a, b, c and d are

A a = 0, b = −8, c = 10, d = −6 B a = 0, b = −6, c = 10, d = −8

C a = 1, b = −8, c = 10, d = −6 D a = 1, b = −6, c = 12, d = −8

E a = 1, b = −8, c = 12, d = −6

3 The quadratic equation 3x2 − 6x + 3 = 0 has

A two real solutions, x = ±1 B one real solution, x = −1

C no real solutions D one real solution, x = 1

E two real solutions, x = 1 and x = 2

4 The quadratic equation whose solutions are 4 and −6 is

A (x + 4)(x − 6) = 0 B x2 − 2x − 24 = 0 C 2x2 + 4x = 48

D −x2 + 2x − 24 = 0 E x2 + 2x + 24 = 0

53

x + 4− 5

x − 2is equal to

A−2

(x + 4)(x − 2)B

2(x + 1)

(x + 4)(x − 2)C

−2(x − 7)

(x + 4)(x − 2)

D2(4x + 13)

(x + 4)(x − 2)E

−2(x + 13)

(x + 4)(x − 2)

64

(x + 3)2+ 2x

x + 1is equal to

A8x

(x + 3)2(x + 1)B

2(3x2 + x + 18)

(x + 3)2(x + 1)C

3x2 + 13x + 18

(x + 3)2(x + 1)

D2(3x2 + 13x + 18)

(x + 3)2(x + 1)E

2(x3 + 6x2 + 11x + 2)

(x + 3)2(x + 1)

7 If7x2 + 13

(x − 1)(x2 + x + 2)is expressed in the form

a

x − 1+ bx + c

x2 + x + 2, then the values of a,

b and c are

A a = 5, b = 0, c = −13 B a = 5, b = 0, c = −10 C a = 5, b = 2, c = −3

D a = 7, b = 2, c = 3 E a = 7, b = 3, c = 13

SAMPLE


P1: FXS/ABE P2: FXS


Review


84x − 3

(x − 3)2is equal to

A3

x − 3+ 1

x − 3B

4x

x − 3− 3

x − 3C

9

x − 3+ 4

(x − 3)2

D4

x − 3+ 9

(x − 3)2 E4

x − 3− 15

(x − 3)2

98x + 7

2x2 + 5x + 2is equal to

A2

2x + 1− 3

x + 2B

2

2x + 1+ 3

x + 2C

−4

2x + 2− 1

x + 1

D−4

2x + 2+ 1

x + 1E

4

2x + 2− 1

x + 1

10−3x2 + 2x − 1

(x2 + 1)(x + 1)is equal to

A2

x2 + 1+ 3

x + 1B

2

x2 + 1− 3

x + 1C

5

x2 + 1+ 2

x + 1

D3

x2 + 1− 2

x + 1E

3

x2 + 1+ 2

x + 1

Short-answer questions (technology-free)

1 If (3a + b)x2 + (a − 2b)x + b + 2c = 11x2 − x + 4, find the values of a, b and c.

2 Express x3 in the form (x − 1)3 + a(x − 1)2 + b(x − 1) + c.

3 Prove that, if ax3 + bx2 + cx + d = (x + 1)2(px + q), then b = 2a + d and c = a + 2d.

4 Prove that, if ax3 + bx2 + cx + d = (x − 2)2(px + q), then b = −4a + 1

4d and

c = 4a − d.

5 Solve the following quadratic equations for x.

a x2 + x = 12 b x2 − 2 = x c −x2 + 3x + 11 = 1

d 2x2 − 4x + 1 = 0 e 3x2 − 2x + 5 = t f t x2 + 4 = t x

6 Solve the equation2

x − 1− 3

x + 2= 1

2for x.

7 Express the following as partial fractions.

a−3x + 4

(x − 3)(x + 2)b

7x + 2

x2 − 4c

7 − x

x2 + 2x − 15d

3x − 9

x2 − 4x − 5

e3x − 4

(x + 3)(x + 2)2f

6x2 − 5x − 16

(x − 1)2(x + 4)g

x2 − 6x − 4

(x2 + 2)(x + 1)

h−x + 4

(x − 1)(x2 + x + 1)i

−4x + 5

(x + 4)(x − 3)j

−2x + 8

(x + 4)(x − 3)

8 Express each of the following in partial fractions.

a14(x − 2)

(x − 3)(x2 + x + 2)b

1

(x + 1)(x2 − x + 2)c

3x3

x2 − 5x + 4

SAMPLE


P1: FXS/ABE P2: FXS


Rev

iew


9 Find the coordinates of the points of intersection for each of the following.

a y = x2

y = −x

b x2 + y2 = 16

x + y = 4

c x + y = 5

xy = 4

10 Find the coordinates of the points of intersection of the line with equation 3y − x = 1 and

the circle with equation x2 + 2x + y2 = 9.

Extended-response questions

1 A train completes a journey of 240 km at a constant speed.

a If it had travelled 4 km/h slower, it would have taken two hours more for the journey.

Find the actual speed of the train.

b If it had travelled a km/h slower, and still taken two hours more for the journey of

240 km, what would have been the actual speed? (Answer in terms of a.) Discuss the

practical possible values of a and also the possible values for the speed of the train.

c If the train had travelled a km/h slower, and taken a hours more for the journey of

240 km, and if a is an integer and the speed is an integer, find the possible values for a

and the speed of the train.

2 An upholsterer purchased some fabric for $a. If he had bought the fabric from another

supplier who charged $b per metre more he would have received b metres less for the same

amount of money.

a How many metres did he purchase, in terms of a and b?

b If a and b, and the number of metres purchased, are natural numbers, find the possible

values of a given a < 100.

3 Two trains are travelling at uniform speeds. The slower train takes a hours longer to cover

b km. It travels 1 km less than the faster one in c hours.

a What is the speed of the faster train, in terms of a, b and c?

b If a, b and c, and the speeds of the trains, are rational numbers, find five sets of values for

a, b and c. Choose and discuss two sensible sets of values.

4 A tank can be filled using two pipes. The smaller pipe alone will take a minutes longer than

the larger pipe alone to fill the tank. Also the smaller pipe will take b minutes longer to fill

the tank than when both pipes are used.

a Find, in terms of a and b, how long it will take each of the pipes to fill the tank.

b If a = 24 and b = 32, find how long it takes for each of the pipes to fill the tank.

c If a and b are consecutive positive integers, find five pairs of values of a and b such that

b2 − ab is a perfect square. Interpret these results in the context of this tank problem.SAM

PLE


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