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Title: Basic Engineering Mathematics- solving quadratic equations
Description: Basic Engineering Mathematics- solving quadratic equations

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Chapter 14

Solving quadratic equations
14
...
The value of
the unknown is called the root of the equation
...
For example,
x 2 − 3x + 1 = 0 is a quadratic equation
...
These are:
(a)

by factorization (where possible),

(b) by ‘completing the square’,
(c)

by using the ‘quadratic formula’, or

(d) graphically (see Chapter 19)
...
2 Solution of quadratic equations
by factorization
Multiplying out (x + 1)(x − 3) gives x 2 − 3x + x − 3
i
...
x 2 − 2x − 3
...

If the quadratic expression can be factorized this
provides the simplest method of solving a quadratic
equation
...
e
...
e
...

The technique of factorizing is often one of trial and
error
...
1016/B978-1-85617-697-2
...
Solve the equation x 2 + x − 6 = 0 by
factorization
The factors of x 2 are x and x
...

The only combination to give a middle term of +x is
+3 and −2,
i
...


x 2 + x − 6 = (x + 3)(x − 2)

The quadratic equation x 2 + x − 6 = 0 thus becomes
(x + 3)(x − 2) = 0
Since the only way that this can be true is for either the
first or the second or both factors to be zero, then
either

(x + 3) = 0, i
...
x = −3

or

(x − 2) = 0, i
...
x = 2

Hence, the roots of x 2 + x − 6 = 0 are x = −3 and
x = 2
...
Solve the equation x 2 + 2x − 8 = 0
by factorization
The factors of x 2 are x and x
...

The only combination to give a middle term of +2x is
+4 and −2,
i
...


x 2 + 2x − 8 = (x + 4)(x − 2)

(Note that the product of the two inner terms (4x) added
to the product of the two outer terms (−2x) must equal
the middle term, +2x in this case
...
Solve the equation x 2 − 5x + 6 = 0
Factorizing gives

(x − 3)(x − 2) = 0
x − 3 = 0 or x − 2 = 0

either

(x + 4) = 0, i
...
x = −4

Hence, either

or

(x − 2) = 0, i
...
x = 2

i
...


Hence, the roots of x 2 + 2x − 8 = 0 are x = −4 and
x = 2
...
Determine the roots of
x 2 − 6x + 9 = 0 by factorization
x 2 − 6x + 9 = (x − 3)(x − 3),

i
...
(x − 3)2 = 0

The LHS is known as a perfect square
...

Problem 4
...
e
...
Answers
can always be checked by substitution into the original
equation
...
Solve the equation x 2 + 3x − 4 = 0
(x − 1)(x + 4) = 0
x − 1 = 0 or x + 4 = 0
x = 1 or x = −4

Problem 6
...

By factorizing, 4x 2 − 25 = (2x + 5)(2x − 5), i
...

(2x + 5)(2x − 5) = 0
Hence, either

(2x + 5) = 0, i
...


or

(2x − 5) = 0, i
...


Rearranging gives

x 2 + 2x − 15 = 0

Factorizing gives

(x + 5)(x − 3) = 0

Hence, either

x + 5 = 0 or x − 3 = 0

i
...


x = −5 or x = 3

The factors of 3x 2 are 3x and x
...

Remembering that the product of the two inner terms
added to the product of the two outer terms must equal
−11x, the only combination to give this is +1 and −4,

or x − 4 = 0

i
...


Problem 8
...
Solve the equation 3x 2 − 11x − 4 = 0
by factorization

either

Factorizing gives

103

5
x = − = −2
...
5
2

i
...


3x 2 −11x − 4 = (3x + 1)(x − 4)

The quadratic equation 3x 2 − 11x − 4 = 0 thus
becomes
(3x + 1)(x − 4) = 0
1
3
(x − 4) = 0, i
...
x = 4

Hence, either (3x + 1) = 0, i
...
x = −
or

and both solutions may be checked in the original
equation
...
Solve the quadratic equation
4x 2 + 8x + 3 = 0 by factorizing
The factors of 4x 2 are 4x and x or 2x and 2x
...

Remembering that the product of the inner terms added
to the product of the two outer terms must equal +8x,
the only combination that is true (by trial and error) is
(4x 2 + 8x + 3) = (2x + 3)(2x + 1)

104 Basic Engineering Mathematics
Hence, (2x + 3)(2x + 1) = 0, from which either
(2x + 3) = 0 or (2x + 1) = 0
...
5
2
1
or −0
...

Problem 11
...

The factors of −8 are −4 are +2, or 4 and −2, or −8
and +1, or 8 and −1
...

4
2
Hence, x = −
or x =
5
3
which may be checked in the original equation
...
The roots of a quadratic equation
1
are and −2
...

Hence, if α =

or

1
and β = −2,
3


1
x−
(x − (−2)) = 0
3


1
x−
(x + 2) = 0
3
2
1
x 2 − x + 2x − = 0
3
3
5
2
x2 + x − = 0
3
3
3x 2 + 5x − 2 = 0

Problem 13
...
e
...
e
...
Find the equation in x whose roots
are 1
...
4
If 1
...
4 are the roots of a quadratic equation
then
(x − 1
...
4) = 0
2

i
...


x − 1
...
4x − 0
...
e
...
8x − 0
...

1
...
x 2 + 4x − 32 = 0

3
...
4x 2 − 9 = 0

5
...
8x 2 − 32 = 0

7
...
x 2 + 10x + 25 = 0

9
...
x 2 + 5x + 6 = 0

11
...
x 2 − x − 2 = 0

13
...
y 2 − 9y + 14 = 0

15
...
x 2 − 4x + 4 = 0

17
...
x 2 − 9 = 0

19
...
4z 2 −
16
2
23
...
4x 2 + 12x + 9 = 0
22
...
10x 2 + 3x − 4 = 0

26
...
8x 2 + 13x − 6 = 0

28
...
6x 2 − 5x − 4 = 0

30
...
6x 2 − 5x + 1 = 0

In problems 31 to 36, determine the quadratic
equations in x whose roots are
31
...
2 and −5

33
...
2
...
5

35
...
2
...
7

Solving quadratic equations
14
...


If x 2 = 3 then x = ± 3


If (x + 2)2 = 5 then x + 2 = ± 5 and x = −2 ± 5


If (x − 3)2 = 8 then x − 3 = ± 8 and x = 3 ± 8
Hence, if a quadratic equation can be rearranged so that
one side of the equation is a perfect square and the other
side of the equation is a number, then the solution of
the equation is readily obtained by taking the square
roots of each side as in the above examples
...

(x + a)2 = x 2 + 2ax + a 2
Thus, in order to make the quadratic expression x 2 + 2ax
into a perfect square, it is necessary to add (half the
 2
2a
2
coefficient of x) , i
...

or a 2
2
For example, x 2 + 3x becomes a perfect square by
 2
3
adding
, i
...

2

 2 
3 2
3
x 2 + 3x +
= x+
2
2
The method of completing the square is demonstrated
in the following worked problems
...
Solve 2x 2 + 5x = 3 by completing
the square
The procedure is as follows
...
Hence,
2x 2 + 5x − 3 = 0
(ii) Make the coefficient of the x 2 term unity
...

Hence,
2x 2

5x

2
5
x2 + x −
2

2
i
...


+

3
=0
2
3
=0
2

105

(iii) Rearrange the equations so that the x 2 and x
terms are on one side of the equals sign and the
constant is on the other side
...
In this case the coefficient of x is
2
 2
5
Half the coefficient squared is therefore
4
Thus,
 2
 2
5
3
5
5
x2 + x +
= +
2
4
2
4
The LHS is now a perfect square, i
...



 2
5 2 3
5
x+
= +
4
2
4
(v) Evaluate the RHS
...
Thus,

 


5 2
49
x+
=
4
16
5
7
x + =±
i
...

4
4
(vii) Solve the simple equation
...
5
i
...

4 4 4 2
12
5 7
and
x = − − = − = −3
4 4
4
Hence, x = 0
...
e
...
5 and −3
...
Solve 2x 2 + 9x + 8 = 0, correct to 3
significant figures, by completing the square
Making the coefficient of x 2 unity gives
9
x2 + x + 4 = 0
2
9
2
Rearranging gives
x + x = −4
2

106 Basic Engineering Mathematics
Adding to both sides (half the coefficient of x)2 gives
 2  2
9
9
9
=
−4
x2 + x +
2
4
4
The LHS is now a perfect square
...
031

9
x = − ± 1
...
e
...
22 or −3
...

Problem 17
...
6y 2 + 3
...
75 = 0, correct
to 3 decimal places
4
...
5y − 1
...
5
1
...
6
4
...
75
3
...
6
4
...
5
3
...
75
3
...
6
9
...
6
9
...
Thus,


3
...
2

2
= 0
...
5 √
y+
= 0
...
7246830
9
...
5
Hence,
y =−
± 0
...
2
i
...


y = 0
...
105

Now try the following Practice Exercise
Practice Exercise 55 Solving quadratic
equations by completing the square
(answers on page 346)
Solve the following equations correct to 3 decimal
places by completing the square
...

3
...


x 2 + 4x + 1 = 0
3x 2 − x − 5 = 0
4x 2 − 11x + 3 = 0

2
...

6
...
4 Solution of quadratic equations
by formula
Let the general form of a quadratic equation be given
by ax 2 + bx + c = 0, where a, b and c are constants
...
e
...
)
In summary,

−b ± b2 − 4ac
2
if ax + bx + c = 0 then x =
2a
This is known as the quadratic formula
...
Solve x 2 + 2x − 8 = 0 by using the
quadratic formula
Comparing x 2 + 2x − 8 = 0 with ax 2 + bx + c = 0
gives a = 1, b = 2 and c = −8
...
36 or −1
...

Problem 21
...
Solve
the quadratic formula

x 2 + x − 2 + 12 = 28x − 28

Hence, x =

−2
1
24
or
, i
...
x = 4 or x = −
6
6
3

Problem 20
...
Hence,

−7 ± 72 − 4(4)(2)
x=
2(4)

−7 ± 17 −7 ± 4
...
123
−7 + 4
...
0208
=
=
2
2

− 4 = 0 by using

Comparing 3x 2 − 11x − 4 = 0 with ax 2 + bx + c = 0
gives a = 3, b = −11 and c = −4
...
e
...

2
2
3x 2 − 11x

107

Hence,

x=

27 + 24
...
5104
2

or

x=

27 − 24
...
4896
2

Hence, x = 25
...
490, correct to 4 significant
figures
...

1
...


5
...
86x − 1
...


2x 2 − 7x + 4 = 0
3
4x + 5 =
x
5
(2x + 1) =
x −3

4
...

6
...


4x 2 + 6x − 8 = 0

8
...
6x 2 − 11
...


3x(x + 2) + 2x(x − 4) = 8

10
...

12
...


5
2
+
=6
x −3 x −2
3
+ 2x = 7 + 4x
x −7
x +1
= x −3
x −1

14
...

Problem 22
...
6 cm2
and its width is 3
...

Determine the dimensions of the rectangle, correct
to 3 significant figures

Problem 23
...
0 cm and a total
surface area of 2
...
0 m2 and the height h =
82 cm or 0
...
0 = 2πr(0
...
e
...
82) − 2
...
82r − = 0
π
Using the quadratic formula,



−0
...
82)2 − 4(1) − π1
r=
2(1)

−0
...
94564 −0
...
39486
=
=
2
2
= 0
...
1074
Thus, the radius r of the cylinder is 0
...

Hence, the diameter of the cylinder

Let the length of the rectangle be x cm
...
10) cm
...
10) = 23
...
e
...
10x − 23
...
10) ± (−3
...
6)
x=
2(1)

3
...
61 + 94
...
10 ± 10
...
10
13
...
65 cm or −3
...
The latter solution is
neglected since length cannot be negative
...
65 cm and width = x − 3
...
65 − 3
...
55 cm, i
...
the dimensions of the rectangle are 6
...
55 cm
...
65 × 3
...
6 cm 2 , correct to
3 significant figures
...
2874
= 0
...
5 cm
correct to 3 significant figures
...
The height s metres of a mass
projected vertically upwards at time t seconds is
1
s = ut − gt 2
...
81 m/s2
1
When height s = 16 m, 16 = 30t − (9
...
e
...
905t − 30t + 16 = 0
Using the quadratic formula,

−(−30) ± (−30)2 − 4(4
...
905)

30 ± 586
...
21
=
=
= 5
...
59
9
...
81

Solving quadratic equations
Hence, the mass will reach a height of 16 m after
0
...
53 s on the descent
...
A shed is 4
...
0 m wide
...
If the area of the path is 9
...
1 shows a plan view of the shed with its
surrounding path of width t metres
...
2 and l = 15
...
2 = πr(15
...
3πr − 482
...
e
...
3r −

or

Using the quadratic formula,

−15
...
0 m

4
...
0 1 2t)

SHED

Figure 14
...
0 × t ) + 2t (4
...
50 = 4
...
0t + 4t 2

−(12
...
0)2 − 4(4)(−9
...
0 ± 296
...
0 ± 17
...
e
...
2
−4
π



Now try the following Practice Exercise
Practice Exercise 57 Practical problems
involving quadratic equations (answers on
page 346)
1
...
Deter2
mine the time taken to complete 4 radians if
ω is 3
...
60 rad/s2
...


The power P developed in an electrical circuit is given by P = 10I − 8I 2 , where I is
the current in amperes
...
5 watts in
the circuit
...


The area of a triangle is 47
...
3 cm more than its
base length
...


4
...
Determine the distance
x
between the supports when the sag is 20 m
...


The acid dissociation constant K a of ethanoic
acid is 1
...
Using the Ostwald dilution law,

Hence,
t=



Hence, radius r = 6
...
21 cm, which is
meaningless and is thus ignored)
...
9106)
= 13
...


4t 2+ 12
...
50 = 0

or

(15
...
2
=0
π

2

−15
...
0461 −15
...
12123
=
=
2
2

t

i
...


109

t = 0
...
65058 m
...
651 m or 65 cm correct to
the nearest centimetre
...
If the total surface area of a solid
cone is 486
...
3 cm,
determine its base diameter
...


110 Basic Engineering Mathematics
x2
Ka =
, determine x, the degree of
v(1 − x)
ionization, given that v = 10 dm3
...


A rectangular building is 15 m long by 11 m
wide
...
If the
area of the path is 60
...


7
...
0 m3
...
80 m
...


The bending moment M at a point in a beam
3x(20 − x)
, where x metres
is given by M =
2
is the distance from the point of support
...


9
...
In the
layout of a number of courts an area of ground
must be allowed for at the ends and at the
sides of each court
...


10
...
When connected in parallel their total resistance is 8
...
If one of the resistors has a resistance
of Rx , ohms,
(a) show that Rx2 − 40Rx + 336 = 0 and

For a simultaneous solution the values of y must be
equal, hence the RHS of each equation is equated
...
e
...
e
...
)
Hence, the simultaneous solutions occur when
3
x = − , y = −16 and when x = 1, y = −1
...

Now try the following Practice Exercise

14
...
An algebraic method
of solution is shown in Problem 27; a graphical solution
is shown in Chapter 19, page 160
...
Determine the values of x and y
which simultaneously satisfy the equations
y = 5x − 4 − 2x 2 and y = 6x − 7

Practice Exercise 58 Solving linear and
quadratic equations simultaneously
(answers on page 346)
Determine the solutions of the following simultaneous equations
...
y = x 2 + x + 1
2
...


2x 2 + y = 4 + 5x
x+y =4

y = 2x − 1

List of formulae
Laws of indices:

Areas of plane figures:

a m × a n = a m+n
a m/n =


n m
a

am
an

= a m−n (a m )n = a mn

a −n =

Quadratic formula:
If ax 2 + bx + c = 0

1
an

Area = l × b

(i) Rectangle

a0 = 1
b


−b ± b2 − 4ac
x=
2a

then

Equation of a straight line:

l

(ii) Parallelogram Area = b × h

y = mx + c

Definition of a logarithm:
If y = a x

then

h

x = loga y

Laws of logarithms:
log(A × B) = log A + log B
 
A
= log A − log B
log
B

b

(iii) Trapezium

log An = n × log A

1
Area = (a + b)h
2
a

Exponential series:
ex = 1 + x +

x2 x3
+
+···
2! 3!

h

(valid for all values of x)
b

Theorem of Pythagoras:
b 2 = a 2 + c2

(iv) Triangle

Area =

1
×b×h
2

A

c

B

b
h

a

C
b

List of formulae
Area = πr 2 Circumference = 2πr

(v) Circle

(iii) Pyramid
If area of base = A and
perpendicular height = h then:

r

s



Volume =

1
× A×h
3

r

2π radians = 360 degrees

Radian measure:

h

For a sector of circle:
θ◦
(2πr) = rθ
360

(θ in rad)

1
θ◦
(πr 2 ) = r 2 θ
360
2

(θ in rad)

s=

arc length,

shaded area =

Equation of a circle, centre at origin, radius r:
x 2 + y2 = r 2
Equation of a circle, centre at (a, b), radius r:

Total surface area = sum of areas of triangles
forming sides + area of base
(iv) Cone
1
Volume = πr2 h
3
Curved Surface area = πrl

(x − a)2 + (y − b)2 = r 2

Total Surface area = πrl + πr2

Volumes and surface areas of regular
solids:
l

(i) Rectangular prism (or cuboid)

h

Volume = l × b × h
Surface area = 2(bh + hl + lb)

r

(v) Sphere

l

h

4
Volume = πr3
3
Surface area = 4πr2

b

(ii) Cylinder
Volume = πr2 h
Total surface area = 2πrh + 2πr2
r

r

h

337

338 Basic Engineering Mathematics
Areas of irregular figures by approximate
methods:
Trapezoidal rule


 
width of 1 first + last
Area ≈
interval
2 ordinate

+ sum of remaining ordinates
Mid-ordinate rule
Area ≈ (width of interval)(sum of mid-ordinates)
Simpson’s rule

 

1 width of
first + last
Area ≈
ordinate
3 interval




sum of even
sum of remaining
+4
+2
ordinates
odd ordinates

For a general sinusoidal function y = A sin (ωt ± α),
then
A = amplitude
ω = angular velocity = 2π f rad/s
ω
= frequency, f hertz


= periodic time T seconds
ω
α = angle of lead or lag (compared with
y = A sin ωt )

Cartesian and polar co-ordinates:
If co-ordinate (x, y) = (r, θ) then

y
r = x 2 + y 2 and θ = tan−1
x
If co-ordinate (r, θ) = (x, y) then
x = r cosθ and y = r sin θ

Mean or average value of a waveform:
area under curve
length of base
sum of mid-ordinates
=
number of mid-ordinates

mean value, y =

Triangle formulae:
Sine rule:
Cosine rule:

b
c
a
=
=
sin A sin B
sin C
a 2 = b2 + c2 − 2bc cos A

B

b

a

If a = first term and d = common difference, then the
arithmetic progression is: a, a + d, a + 2d,
...

The n’th term is: arn−1
Sum of n terms, Sn =

A
c

Arithmetic progression:

a (1 − r n )
a (r n − 1)
or
(1 − r )
(r − 1)

If − 1 < r < 1, S∞ =

a
(1 − r )

C

Area of any triangle
1
= × base × perpendicular height
2
1
1
1
= ab sin C or
ac sin B or
bc sin A
2
2
2

a +b+c
= [s (s − a) (s − b) (s − c)] where s =
2

Statistics:
Discrete data:

#
mean, x¯ =

x

n
)
(
 #

(x − x¯ )2

standard deviation, σ =
n

List of formulae
Grouped data:

#
fx
mean, x¯ = #
f

)
 #$
%(

f (x − x)
¯ 2

#
standard deviation, σ =
f

Standard integrals
y
axn
cos ax

Standard derivatives
sin ax

y or f(x)

dy
= or f (x)
dx

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1
x

eax
1
x

&

a

y dx
x n+1
+ c (except when n = −1)
n +1

1
sin ax + c
a
1
− cos ax + c
a
1 ax
e +c
a
ln x + c

339

Answers

Answers to practice exercises
Chapter 1

Chapter 2

Exercise 1 (page 2)
1
...

7
...

13
...

17
...
16 m
3
...
£565
6
...
−36 121
9
...
1487
12
...
−70872
15
...
25 cm
d = 64 mm, A = 136 mm, B = 10 mm

1
...

4
...

7
...


5
...

13
...


Exercise 2 (page 5)
(a) 468 (b) 868
2
...

(a) 259 (b) 56
8
...


2
...

6
...

10
...


(a) 12 (b) 360
(a) 90 (b) 2700
(a) 3 (b) 180
(a) 15 (b) 6300
(a) 14 (b) 53 900

Exercise 4 (page 8)
2
...
68

3
...
5

DOI: 10
...
00040-5

3
...

10
1
14
...
1
21
6
...


16
...


2
6
8
2
...

35
9
11
3
1
7
...

5
13
5
3
2
12
...
3
2
5
12
4
3
1
17
...
13
4
9
(a) £60, P£36, Q£16

7
...

15
...


22
9
8
25
3
16
4
27
17
60

4
...

12
...

20
...
5
9
...
−33
10
...
1
2
12
14
...


71
8
11
15
3
16
51
8
52
17
20

19
...

15
5
...
4
20
...
2880 litres

Exercise 7 (page 14)
1
18
7
6
...
2
1
...
22
11
...

(a) 4 (b) 24
(a) 10 (b) 350
(a) 2 (b) 210
(a) 5 (b) 210
(a) 14 (b) 420 420

2
...

3
...

7
...


1
7
4
11
17
30
43
77
9
1
40

1
...
4

1
9
19
7
...
2
20
2
...
1
8
...
7
3

4
...

15
5
...
0
...
14
...
1
...


13
20

7
40
21
1
141
(b)
(c)
(d)
(e)
25
80
500
11
3
7
...
10
25
200
11
1
7
(b) 4
(c) 14
(d) 15
40
8
20
2
...
4
40
41
10
...
(a)

11
...
625

9
250

3
...


6
125

9
...

4
...

10
...


2
...
11
...
185
...
8307
5
...
1581
6
...
571
5
...
1
...
0
...
068
11
...
5 ×10
12
...
5 ×103
−6
−3
4
...
202
...
18
...
6
...
0
...
11
...
0
...
14
...
1
...
2
...
65
...
0
...
329
...
18
...
43
...
72
...
12
...
−124
...
4
...
0
...


9
10

4
...
732
8
...
0
...
0
...
0
...
−0
...
0
...
0
...
5
...
2
...
0
...
0
...
0
...
998 2
...
544
3
...
02 4
...
42
456
...
434
...
626
...
1591
...
444 10
...
62963
11
...
563 12
...
455
13
...
8
...
(a) 24
...
812
(a) 0
...
0064 17
...
4˙ (b) 62
...
4
...
0
...
3
...
13
...
50
...
53
...
36
...
12
...
0
...
46
...
1
...
2
...
2
...
30
...
0
...
219
...
5
...
5
...
52
...
0
...
25
...
591
...
69
...
17
...

4
...

10
...
11927
6
...
0944
10
...
325

2
...
30
...
84
...
10
...
2
...


Exercise 10 (page 19)

1
...

9
...

16
...


2
...
0
...
137
...
19
...
515
...
15
...
52
...
0
...
80
...
295
...
59 cm2
159 m/s
0
...

5
...

11
...
78 mm
0
...
8 m2
281
...

6
...

12
...
5
5
...
5
2
...

4
...

10
...


£589
...
508
...
V = 2
...
5
5
...
81 A 6
...
79 s
E = 3
...
I = 12
...
s = 17
...
184 cm2 11
...
327
(a) 12
...
p
...

(c) 13
...
15 h

342 Basic Engineering Mathematics
Exercise 26 (page 43)

Chapter 5

1
...
£66 3
...
450 g 5
...
56 kg
6
...
00025 (b) 48 MPa 7
...
76 litre

Exercise 21 (page 34)
1
...
32%
2
...
4% 3
...
7% 4
...
4%
5
...
5%
6
...
20
7
...
0125 8
...
6875
9
...
462% 10
...
2% (b) 79
...
(b), (d), (c), (a) 12
...

14
...
A = , B = 50%, C = 0
...
30,
2
17
3
F = , G = 0
...
85, J =
10
20

Exercise 27 (page 45)
1
...
170 fr
3
...
8 mm
4
...
(a) 159
...
5 gallons
6
...
4 MPa
7
...
2 mm 8
...

3
...

7
...

14
...
8 kg 2
...
72 m
(a) 496
...
657 g
(a) 14%
(b) 15
...
49% 11
...
2%
2
...
5
...
73 s 4
...
36% 6
...
76 g
9
...
17% 13
...
3
...
25%
37
...
7%

1
...
5 weeks
2
...
(a) 9
...
12 (c) 0
...
50 minutes
5
...
375 m2 (c) 24 × 103 Pa

Chapter 7
Exercise 29 (page 48)

Exercise 23 (page 38)
1
...

9
...

14
...

16
...
5%
2
...
£310
4
...
£20 000 7
...
45 8
...
25
£39
...
£917
...
£185 000 12
...
2%
A 0
...
9 kg, C 0
...
3 t
20 000 kg (or 20 tonnes)
13
...
5 mm 17
...
27
6
...
128
7
...
100 000
8
...
24
10
...
96
9
...
1 6
...
16 4
...
01 10
...
76
7
...
1000 9
...
36 13
...
34 15
...
25
1
1
1
17
...
49 19
...
5
20
...
128

2
...
36 : 1

2
...
5 : 1 or 7 : 2 3
...
96 cm, 240 cm 5
...
£3680, £1840, £920 7
...
£2172

Exercise 31 (page 52)
1
...
9

Exercise 25 (page 42)
1
...
76 ml
3
...
12
...
14
...
25 000 kg

147
148
17
13
...


2
...
±3
10
...
64

19
56

32
25
1
7
...


11
...
4

1
2

4
...
±
13
1
12
...

3
...

7
...

11
...

15
...

19
...

23
...

27
...

4
...

8
...

12
...

16
...

20
...

24
...

28
...

y
20
...
(x − y)(a + b)
1
...

5
...

9
...

13
...


2x(y − 4z)
2x(1 + 2y)
4x(1 + 2x)
x(1 + 3x + 5x 2 )
r(s + p + t )


2 p q 2 2 p2 − 5q
2x y(y + 3x + 4x 2 )
7y(4 + y + 2x)
2r
18
...

t
21
...
(a − 2b)(2x + 3y)

2
...

6
...

10
...

14
...


1
...
2

6
...
2

11
...
2

4
...
6
1
8
...
6
18
...
−10
17
...
0
14
...
12y 2 − 3y
4
...
1

7
...
6a 2 + 5a −
11
...
9x 2 +

1
...
−2

3
...
15

7
...
5

11
...
2

16
...
6

21
...
−3

1
4
1
3

10
...
10a 2 − 3a + 2

15
...
5

1
2

1
3

4
...
12

9
...
13

13
...
−11

15
...
3

19
...
10

23
...
±4

Exercise 44 (page 79)
1
...

5
...


10−7
2
...
3
...
8  (b) 30 
digital camera battery £9, camcorder battery £14
800 
7
...
12 cm, 240 cm2
4
...
30 kg

2
...
004
5
...
12 m, 8 m

3
...
£312, £240
9
...
5 N

Chapter 12
Exercise 46 (page 84)
1
...
3

Exercise 43 (page 76)

Exercise 41 (page 72)
1
...
4b − 15b2
3
5
...
2

c
p
V
5
...
r =

3
...
a =
t
1
6
...
x =
m
2
...
R =
I
5
13
...
T =

10
...
C =

ω ωL −

12
...
f =

13
...
λ =

R2


5

345

+ , 63
...
r =
or 1 −
S
S
2
...
f =

AL
3F − AL
or f = F −
3
3

4
...
t =

R − R0
R0 α

6
...
b =
9
...
R =

t 2g
4π 2


360 A
12
...
080
14
...
L =

11
...
a =
m −n
3(x + y)
3
...
b = √
1 − a2
a( p2 − q 2 )
7
...
t2 = t1 +
11
...
725




v 2 − 2as

M
+ r4
π
mrCR
4
...
r =
x+y



2
...
965
10
...
03L
8
...
5
p = 2, q = −1
x = 3, y = 2
a = 5, b = 2
s = 2, t = 3
m = 2
...
5
x = 2, y = 5

2
...

6
...

10
...

14
...


x = 3, y = 4
x = 4, y = 1
x = 1, y = 2
a = 2, b = 3
x = 1, y = 1
x = 3, y = −2
a = 6, b = −1
c = 2, d = −3

Exercise 50 (page 94)

13
...

3
...

7
...

11
...

15
...

3
...

7
...

4
...

8
...
30, b = 0
...
x = , y =
2
4
1
1
3
...
c = 3, d = 4
3
7
...
a = , b = −
3
2
4
...
r = 3, s =
2
8
...

3
...

6
...


a = 0
...
5, c = 3
α = 0
...
56 
a = 4, b = 10

2
...
47, I2 = 4
...
£15 500, £12 800
7
...
40
9
...
5, F2 = −4
...
x = 2, y = 1, z = 3
3
...
x = 2, y = −2, z = 2
4
...

7
...

10
...


x = 2, y = 4, z = 5 6
...
x = −4, y = 3, z = 2
x = 1
...
5, z = 4
...

4
...

8
...


1
...
0
...
905 A
0
...
38 m
1
...
835 m or 18
...

5
...

9
...
84 cm
0
...
78 cm
7m

Chapter 14
Exercise 58 (page 110)

Exercise 54 (page 104)
1
...
−1
...
5
7
...

13
...


4
−2 or −3
4 or −3
2

19
...
4 or −8
4
5
...
−5
11
...
2 or 7
17
...
−1
...
x = 1, y = 3 and x = −3, y = 7

6
...
x = , y = − and −1 , y = −4
5
5
3
3

9
...

15
...


23
...
5

1
4
25
...

5

1
1
26
...

or −
3
2

1
2 or −1
−4
3 or −3

1
8
1
24
...

8
5
30
...


22
...
x 2 − 4x + 3 = 0
33
...
x 2 − 36 = 0

3
...
3
7
...
−2
9
...

2
3
1
11
...
10 000 13
...
9 15
...
0
...

18
...
4

or −2
3
2

32
...
4x 2 − 8x − 5 = 0
36
...
7x − 1
...
4

3
...
−3

5
...
−3
...
268
3
...
468 or −1
...
2
...
307

1
8

3
...
−3
...
637
4
...
290 or 0
...
−2
...
351

1
...
log 12

2
...
log 500

3
...
log 100

4
...
log 6

9
...
log 1 = 0 11
...
log 243 or log 35 or 5 log3
13
...
log 64 or log26 or 6 log2

Exercise 56 (page 107)
1
...

5
...

9
...

13
...
637 or −3
...
781 or 0
...
608 or −1
...
851 or −2
...
481 or −1
...
167
4
...
438

2
...

6
...

10
...


0
...
792
0
...
693
1
...
232
2
...
086
4
...
676
7
...
641

15
...
5
16
...
5
19
...
x = 2

17
...
5 18
...
a = 6 22
...
1
...
3
...
0
...
6
...
2
...
3
...
2
...
−0
...
316
...

2
...

4
...


(a)
0
...
0988
(a) 4
...
04106
2
...


Exercise 67 (page 134)

0
...
064037
2
...
07482
120
...
446
8
...
08286

Exercise 63 (page 120)
1
...
0601

2
...
389 (b) 0
...

x − 2x 4
3
1
4
...
(a) Horizontal axis: 1 cm = 4 V (or 1 cm = 5 V),
vertical axis: 1 cm = 10 
(b) Horizontal axis: 1 cm = 5 m, vertical axis:
1 cm = 0
...
2 mm
2
...
5 (d) 5
3
...
5
4
...
1 (b) −1
...
The 1010 rev/min reading should be 1070 rev/min;
(a) 1000 rev/min (b) 167 V

1 − 2x 2 −

Exercise 64 (page 122)
1
...
95, 2
...
(a) 28 cm3 (b) 116 min

2
...
65, −1
...
(a) 70◦C (b) 5 minutes

Exercise 65 (page 124)
1
...

3
...

11
...

17
...
55547 (b) 0
...
8941
(a) 2
...
33154 (c) 0
...
4904 4
...
5822 5
...
197
6
...
2
0
...
11
...
1
...
1
...
962 12
...
4
147
...
4
...
3
...
e
...
500 19
...
Missing values: −0
...
25, 0
...
25, 2
...
(a) 4, −2 (b) −1, 0 (c) −3, −4 (d) 0, 4
1
1
1 1
(b) 3, −2 (c) ,
2
2
24 2
4
...
(a) 2,

1
2
2
1
2
5
...
(a) (b) −4 (c) −1
5
6
7
...
(2, 1)

9
...
5, 6)

10
...
(a) 89 cm (b) 11 N (c) 2
...
4 W + 48
12
...
15 W + 3
...
a = −20, b = 412

Exercise 69 (page 144)
1
...
(a) 850 rev/min (b) 77
...
(a) 150◦ C (b) 100
...
(a) 0
...
25L + 12
2
...
21 kPa

3
...
32 volts (b) 71
...
(a) 1
...
293 m 5
...
45 s
6
...
37 N
7
...
04 A (b) 1
...
2
...
£2424

347

9
...
07 A (b) 0
...
5 N (e) 592 N (f) 212 N
4
...
003, 8
...
(a) 22
...
43 s (c) v = 0
...
5
6
...
9L − 0
...
(a) 1
...
89% (c) F = −0
...
21
8
...
00022 (c) 28
...
a = 0
...
3 kPa, 275
...
(a)

Chapter 18

9
...
6, 13
...
6, 0
...
6 or 0
...
x = −1
...
5 (a) −30 (b) 2
...
50
(c) 2
...
8

Exercise 74 (page 161)

Exercise 70 (page 149)
1
...
(a) y (b)



x (c) b (d) a

y
1
(c) f (d) e 4
...
(a) (b) 2 (c) a (d) b
x
x
6
...
5, b = 0
...
78 mm2 7
...
15

1
...
5, y = −5
...
(a) x = −1
...
5 (b) x = −1
...
24
(c) x = −1
...
0

3
...
(a) 950 (b) 317 kN
9
...
4, b = 8
...
4 (ii) 11
...
x = −2
...
5 or 1
...
x = −2, 1 or 3, Minimum at (2
...
1),
Maximum at (−0
...
2)
3
...
x = −2
...
4 or 2
...
x = 0
...
5
6
...
3, 1
...
8
7
...
5

Exercise 71 (page 154)
1
...

3
...

5
...

7
...


(a) lg y (b) x (c) lg a (d) lg b
(a) lg y (b) lg x (c) L (d) lg k
(a) ln y (b) x (c) n (d) ln m
I = 0
...
75 candelas
a = 3
...
5
a = 5
...
6, 38
...
0
R0 = 26
...
42
8
...
08e0
...
4 N, μ = 0
...
0 N, 1
...
5, y = 1
...
3, y = −1
...
4, b = 1
...
82◦ 27
2
...
51◦11 4
...
15 44 17 6
...
72
...
27
...
37◦ 57
10
...
reflex 2
...
acute 4
...
(a) 21◦ (b) 62◦ 23 (c) 48◦56 17

Chapter 19

1
...

5
...


Chapter 20

2
...
x = −1, y = 2
6
...
(a) Minimum (0, 0) (b) Minimum (0, −1)
(c) Maximum (0, 3) (d) Maximum (0, −1)
2
...
4 or 0
...
−3
...
9
4
...
1 or 4
...
−1
...
2
6
...
5 or −2, Minimum at (−1
...
1)
7
...
7 or 1
...
(a) ±1
...
3

6
...
(a) 60◦ (b) 110◦ (c) 75◦ (d) 143◦ (e) 140◦
(f ) 20◦ (g) 129
...
Transversal (a) 1 & 3, 2 & 4, 5 & 7, 6 & 8
(b) 1 & 2, 2 & 3, 3 & 4, 4 & 1, 5 & 6, 6 & 7,
7 & 8, 8 & 5, 3 & 8, 1 & 6, 4 & 7 or 2 & 5
(c) 1 & 5, 2 & 6, 4 & 8, 3 & 7 (d) 3 & 5 or 2 & 8
9
...
a = 69◦ , b = 21◦ , c = 82◦ 11
...
1
...
0
...
40◦55

Exercise 78 (page 173)
1
...
a = 40◦ , b = 82◦, c = 66◦,
d = 75◦, e = 30◦ , f = 75◦
3
...
52◦
5
...
5◦
6
...
40◦, 70◦, 70◦, 125◦, isosceles
8
...
a = 103◦, b = 55◦ , c = 77◦ , d = 125◦,
e = 55◦, f = 22◦, g = 103◦, h = 77◦ ,
i = 103◦, j = 77◦, k = 81◦
10
...
A = 37◦, B = 60◦ , E = 83◦

349

4
3
4
3
2
...
sin A = , tan A =
17
15
112
15
, cos X =
4
...
(a)
(b)
(c)
17
17
15
7
24
6
...
(a) 9
...
625

Exercise 79 (page 176)
1
...
proof

Exercise 84 (page 187)
1
...

5
...

13
...
7550 2
...
846
3
...
52
(a) 0
...
1010 (c) 0
...
33◦
6
...
25◦
7
...
78◦
8
...
41 54 11
...
05
12
...
3586 14
...
803

Exercise 80 (page 178)

Exercise 85 (page 189)

1
...
54 mm, y = 4
...
9 cm, 7
...
(a) 2
...
3 m

1
...
22 (b) 5
...
87 (d) 8
...
595 (f ) 5
...
(a) AC = 5
...
04◦ , ∠C = 30
...
928 cm, ∠D = 30◦, ∠F = 60◦
(c) ∠J = 62◦, HJ = 5
...
59 cm
(d) ∠L = 63◦ , LM = 6
...
37 cm
(e) ∠N = 26◦, ON = 9
...
201 cm
(f ) ∠S = 49◦, RS = 4
...
625 cm

Exercise 81 (page 180)
1–5
...


3
...
54 m

4
...
40 mm

Chapter 21
Exercise 82 (page 182)

Exercise 86 (page 192)

1
...

7
...

11
...


1
...
15 m
2
...
249
...
110
...
53
...
9
...
107
...
9
...
56 m
9
...
24 m
3
...
54 mm
20
...
7
...
11
...
11 mm 8
...
20 cm each (b) 45◦
10
...
81 km
3
...
132
...
94 mm
14
...
sin Z = , cos Z = , tan X = , cos X =
41
41
9
41

1
...
78◦ and 137
...
53◦ and 351
...
(a) 29
...
92◦ (b) 123
...
14◦
3
...
21◦ and 224
...
12◦ and 293
...
t = 122◦7 and 237◦53
5
...
θ = 39◦44 and 219◦44

Exercise 88 (page 202)
1
...
180◦
3
...
120◦



5
...
2, 144
7
...
5, 720◦
7
9
...
6, 360◦ 11
...
5 ms
2
13
...
100 μs or 0
...
625 Hz 16
...
leading

Exercise 91 (page 209)
1
...
2 cm, Q = 47
...
65◦,
area = 77
...
p = 6
...
83◦, R = 44
...
938 m2
3
...
33◦, Y = 52
...
05◦,
area = 27
...
Z = 29
...
50◦ , Z = 96
...
(a) 40 (b) 25 Hz (c) 0
...
29 rad (or 16
...

3
...

7
...
(a) 122
...
80◦ , 40
...
54◦
(a) 11
...
55◦
4
...
4 m
BF = 3
...
0 m 6
...
35 m, 5
...
48 A, 14
...
(a) 75 cm (b) 6
...
157 s
(d) 0
...
94◦ ) lagging 75 sin 40t
3
...
01 s or 10 ms
(d) 0
...
61◦) lagging 300 sin 200πt
4
...
43) volts

π

A or
5
...
524)A
6
...
2 sin(100πt + 0
...
(a) 5A, 50 Hz, 20 ms, 24
...
093A (c) 4
...
375 ms (e) 3
...
C = 83◦ , a = 14
...
9 mm,
area = 189 mm2
2
...
568 cm, a = 7
...
65 cm2
3
...
0 cm,
area = 134 cm2
4
...
08 mm,
area = 185
...
J = 44◦29 , L = 99◦31 , l = 5
...
132 cm2 , or, J = 135◦ 31 , L = 8◦29 ,
l = 0
...
917 cm2
6
...
2 mm,
area = 820
...
19 mm, area = 174
...
80
...
38◦, 40
...
(a) 15
...
07◦
3
...
25 cm, 126
...
19
...
36
...
x = 69
...
130◦ 8
...
66 mm

Chapter 24
Exercise 94 (page 215)
1
...

3
...

5
...

7
...


(5
...
04◦) or (5
...
03 rad)
(6
...
82◦) or (6
...
36 rad)
(4
...
57◦) or (4
...
03 rad)
(6
...
58◦) or (6
...
54 rad)
(7
...
20◦) or (7
...
55 rad)
(4
...
31◦) or (4
...
12 rad)
(5
...
04◦) or (5
...
74 rad)
(15
...
75◦) or (15
...
37 rad)

Exercise 95 (page 217)
(1
...
830)
2
...
917, 3
...
362, 4
...
(−2
...
154)
(−9
...
400)
6
...
615, −3
...
750, −1
...
(4
...
233)
(a) 40∠18◦, 40∠90◦, 40∠162◦, 40∠234◦, 40∠306◦
(b) (38
...
36), (0, 40), (−38
...
36),
(−23
...
36), (23
...
36)
10
...
0 mm
1
...

5
...

9
...
p = 105◦ , q = 35◦
3
...
r = 142◦, s = 95◦

Exercise 97 (page 225)
1
...
91 cm
2
...
7 cm2 3
...
27
...
18 cm
6
...
(a) 29 cm2 (b) 650 mm2
8
...
3
...
6750 mm
11
...
30 cm2
12
...

4
...

9
...

13
...

18
...


113 cm2
2
...
1790 mm2
2
2
802 mm
5
...
1269 m2
2
1548 m
8
...
0 cm (b) 783
...
46 m 10
...
80 cm, 74
...
86 mm (b) 197
...
26
...
67 cm, 54
...
82
...
748
(a) 0
...
2 m2
17
...
47 m2
2
(a) 396 mm (b) 42
...
701
...
74 mm

Exercise 104 (page 237)
1
...
Centre at (3, −2), radius 4
3
...
Circle, centre (0, 0), radius 6

Chapter 27
Exercise 98 (page 226)
1
...

3
...

7
...
27 cm2 (b) 706
...
(a) 20
...
41 mm
(a) 53
...
9 mm2
6
...
89 m

Exercise 99 (page 228)
1
...
1624 mm2 3
...
918 ha (b) 456 m

Exercise 105 (page 243)
1
...

5
...

10
...

14
...

17
...


1
...
5 cm3
3
...
15 cm3 , 135 g
7
...
(a) 35
...
3 cm2
1
...
37
...
63 cm
13
...
709 cm, 153
...
99 cm
16
...
099 m2
8
...
22 m
18
...
5 min
4 cm
20
...
08 m3

Exercise 100 (page 229)
1
...
80 m2

3
...
14 ha

Chapter 26
Exercise 101 (page 231)
1
...
24 cm 2
...
5 mm 3
...
629 cm 4
...
68 cm
5
...
73 cm 6
...
97
...
201
...
0 cm2 2
...
68 cm3 , 25
...
113
...
1 cm2 4
...
131 cm 5
...
2681 mm3 7
...
083 cm3
(b) 20 106 mm2 or 201
...
8
...
(a) 512 × 106 km2 (b) 1
...
664

Exercise 102 (page 232)

Exercise 107 (page 251)



π
(b)
(c)
6
12
4
2
...
838 (b) 1
...
054
3
...
(a) 0◦ 43 (b) 154◦8 (c) 414◦ 53

1
...

4
...


1
...
90 cm2
(a) 56
...
82 cm2
3
...
57 kg
5
...
4 m2
29
...
5 cm3 (b) 84
...
4 cm3 (b) 32
...
0 cm3

352 Basic Engineering Mathematics

7
...

11
...


(b) 146 cm2 (vi) (a) 86
...
9 cm (b) 38
...
125 cm3
3
2
10
...
5 m
10
...
220
...
1 cm3 , 1027 cm2
2
(a) 1458 litres (b) 9
...
45

147 cm3 , 164 cm2
10480 m3 , 1852 m2
10
...
14 m

14
...
72◦ to the 5 N force
29
...
04◦ to the horizontal
9
...
70◦
9
...
89 m/s at 159
...
62 N at 26
...
07 knots, E 9
...

3
...

7
...

7
...

10
...


2
...
1707 cm2
6
...
(a) 54
...
16◦ (b) 45
...
66◦
2
...
71 m/s at 121
...
55 m/s at 8
...
83
...
6◦ to the vertical
2
...
22
...
78◦ N

Exercise 109 (page 256)
1
...
137
...
4
...
54
...
63
...
4
...
143 m2

Exercise 111 (page 260)
1
...
59 m3

2
...
20
...
(a) 2 A (b) 50 V (c) 2
...
0
...
1 A
5
...
13 cm2 , 368
...

3
...

7
...


i − j − 4k
−i + 7j − k
−3i + 27j − 8k
i + 7
...
6i + 4
...
9k

2
...

6
...

10
...
5j − 10k
2i + 40j − 43k

Chapter 30
Exercise 118 (page 279)

2
...
5 V (b) 3 A
4
...
83 V (b) 0

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Chapter 29
Exercise 119 (page 281)
Exercise 113 (page 266)
1
...

2
...
scalar
4
...
scalar
6
...
vector 8
...
vector

Exercise 114 (page 273)
1
...

3
...

5
...
35 N at 18
...
62◦ to the 12 m/s velocity
16
...
57◦ to the 13 N force
28
...
30◦ to the 18 N force
32
...
80◦ to the 30 m displacement

1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395)

Exercise 120 (page 283)
1
...
5 sin(A + 63
...
(a) 20
...
62) volts
(b) 12
...
33) volts
3
...
395) 4
...
11 sin(ωt + 0
...
8
...
173)

Answers to practice exercises
Exercise 121 (page 284)
11
...
324)
2
...
73 sin(ωt − 0
...
79 sin(ωt − 0
...
695 sin(ωt + 0
...
38 sin(ωt + 1
...
3 sin(314
...
233) V (b) 50 Hz
(a) 10
...
3t + 0
...
(a) 79
...
352)V (b) 150 Hz
(c) 6
...

3
...

5
...

7
...
(a) continuous (b) continuous (c) discrete
(d) continuous
2
...
If one symbol is used to represent 10 vehicles, working correct to the nearest 5 vehicles, gives 3
...
5,
6, 7, 5 and 4 symbols respectively
...
If one symbol represents 200 components, working
correct to the nearest 100 components gives: Mon 8,
Tues 11, Wed 9, Thurs 12 and Fri 6
...

3
...

4
...

5
...

6
...

7
...

P increases by 20% at the expense of Q and R
...
Four rectangles of equal height, subdivided as follows: week 1: 18%, 7%, 35%, 12%, 28%; week 2:
20%, 8%, 32%, 13%, 27%; week 3: 22%, 10%, 29%,
14%, 25%; week 4: 20%, 9%, 27%, 19%, 25%
...

9
...
5◦ , 22
...
5◦, 167
...


353

10
...

11
...
(a) £16 450 (b) 138

Exercise 124 (page 297)
1
...
3–39
...
5–39
...
7–39
...
9–40
...
1–40
...
3–40
...
5–40
...
7–40
...

2
...
35, 39
...
75, 39
...
and
heights of 1, 5, 9, 17,
...
There is no unique solution, but one solution is:
20
...
9 3; 21
...
4 10; 21
...
9 11;
22
...
4 13; 22
...
9 9; 23
...
4 2
...
There is no unique solution, but one solution is:
1–10 3; 11–19 7; 20–22 12; 23–25 11;
26–28 10; 29–38 5; 39–48 2
...
20
...
45 13; 21
...
45 37; 22
...
45 48
6
...
5, 15, 21, 24, 27, 33
...
5
...
3, 0
...
67, 2
...
5 and 0
...

7
...
95 2), (11
...
95 19), (12
...
95
42), (13
...
A graph of cumulative frequency against upper class
boundary having co-ordinates given in the answer
to problem 7
...
(a) There is no unique solution, but one solution is:
2
...
09 3; 2
...
14 10; 2
...
19 11;
2
...
24 13; 2
...
29 9; 2
...
34 2
...
07, 2
...
and heights of 3, 10,
...
095 3; 2
...
195 24; 2
...
295 46; 2
...


(d) A graph of cumulative frequency against upper
class boundary having the co-ordinates given
in part (c)
...
Mean 7
...
Mean 27
...
−2542 A/s
2
...
16 cd/V (b) 312
...
(a) −1000 V/s (b) −367
...
−1
...
(a)
15
2
3
4
...
(a)

5
...


7
...

9
...
(a) 8 x + 8 x3 +
x +c
5

7x 2
+c
2
3
(b) t 8 + c
8
5 4
x +c
(b)
24
3
(b) 2t − t 4 + c
4
(b)

(b) 4θ + 2θ 2 +

5
(a) θ 2 − 2θ + θ 3 + c
2
3
2
3
(b) x 4 − x 3 + x 2 − 2x + c
4
3
2
4
1
(a) − + c
(b) − 3 + c
3x
4x
4√ 5
1√
4 9
(a)
x +c
(b)
x +c
5
9
15 √
10
5
(b)
x +c
(a) √ + c
7
t

7
(b) − cos 3θ + c
3
1
(b) 18 sin x + c
3
−2
(b)
+c
15e5x

13
...
(a) 4x + c

3
sin 2x + c
2
1
11
...
(a) e + c
8
10
...
(a) 1
...
5

2
...
5

3
...
333

4
...
75 (b) 0
...
(a) 10
...
1667

6
...
(a) 1 (b) 4
...
(a) 0
...
638

9
...
09 (b) 2
...
(a) 0
...
099

Exercise 141 (page 334)
1
...
37
...
1
...
proof
6
...
5
9
...
67

3
...
1
10
...
29
Title: Basic Engineering Mathematics- solving quadratic equations
Description: Basic Engineering Mathematics- solving quadratic equations