Basic Engineering Mathematics, Fifth Edition

108 Basic Engineering Mathematics

7. 4x^2 + 6 x− 8 = 0


8. 
   
   
   
   5. 6 x^2 − 11. 2 x− 1 = 0
   
   
   
   


9. 3x(x+ 2 )+ 2 x(x− 4 )= 8


10. 4x^2 −x( 2 x+ 5 )= 14

11.

5

x− 3

+

2

x− 2

= 6

12.

3

x− 7

+ 2 x= 7 + 4 x

13.

x+ 1

x− 1

=x− 3

14.5 Practical problemsinvolving

quadratic equations

There are many practical problems in which a

quadratic equation has first to be obtained, from given

information, before it is solved.

Problem 22. The area of a rectangle is 23.6cm^2

and its width is 3.10cm shorter than its length.

Determine the dimensions of the rectangle, correct

to 3 significant figures

Let the length of the rectangle bexcm. Then the width

is(x− 3. 10 )cm.

Area=length×width=x(x− 3. 10 )= 23. 6

i.e. x^2 − 3. 10 x− 23. 6 = 0

Using the quadratic formula,

x=

−(− 3. 10 )±

√

(− 3. 10 )^2 − 4 ( 1 )(− 23. 6 )

2 ( 1 )

=

3. 10 ±

√

9. 61 + 94. 4

2

=

3. 10 ± 10. 20

2

=

13. 30

2

or

− 7. 10

2

Hence,x= 6 .65cm or− 3 .55cm. The latter solution is

neglected since length cannot be negative.

Thus, length x= 6 .65cm and width=x− 3. 10 =

6. 65 − 3. 10 = 3 .55cm, i.e.the dimensions of the rect-

angle are 6.65cm by 3.55cm.

(Check: Area= 6. 65 × 3. 55 = 23 .6cm^2 , correct to

3 significant figures.)

Problem 23. Calculate the diameter of a solid

cylinder which has a height of 82.0cm and a total

surface area of 2.0m^2

Total surface area of a cylinder

=curved surface area+2 circular ends

= 2 πrh+ 2 πr^2 (wherer=radius andh=height)

Since the total surface area= 2 .0m^2 and the heighth=

82cm or 0.82m,

2. 0 = 2 πr( 0. 82 )+ 2 πr^2

i.e.^2 πr^2 +^2 πr(^0.^82 )−^2.^0 =^0

Dividing throughout by 2πgivesr^2 + 0. 82 r−

1

π

= 0

Using the quadratic formula,

r=

− 0. 82 ±

√

( 0. 82 )^2 − 4 ( 1 )

(

−π^1

)

2 ( 1 )

=

− 0. 82 ±

√

1. 94564

2

=

− 0. 82 ± 1. 39486

2

= 0 .2874 or− 1. 1074

Thus, the radiusrof the cylinder is 0.2874m (the

negative solution being neglected).

Hence, the diameter of the cylinder

= 2 × 0. 2874

=0.5748mor57.5cm

correct to 3 significant figures.

Problem 24. The heightsmetres of a mass

projected vertically upwards at timetseconds is

s=ut−

1

2

gt^2. Determine how long the mass will

take after being projected to reach a height of 16m

(a) on the ascent and (b) on the descent, when

u=30m/sandg= 9 .81m/s^2

When heights=16m, 16= 30 t−

1

2

( 9. 81 )t^2

i.e. 4. 905 t^2 − 30 t+ 16 = 0

Using the quadratic formula,

t=

−(− 30 )±

√

(− 30 )^2 − 4 ( 4. 905 )( 16 )

2 ( 4. 905 )

=

30 ±

√

586. 1

9. 81

=

30 ± 24. 21

9. 81

= 5 .53 or 0. 59