108 Basic Engineering Mathematics
7. 4x 2 + 6x − 8 = 0
8. 5.6x 2 − 11.2x − 1 = 0
9. 3x(x + 2) + 2x(x − 4) = 8
10. 4x 2 − x(2x + 5) = 14
11.
5
x − 3
+
2
x − 2
= 6
12.
3
x − 7
+ 2x = 7 + 4x
13.
x + 1
x − 1
= x − 3
14.5 Practical problems involving
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 23.6 cm 2
and its width is 3.10 cm shorter than its length.
Determine the dimensions of the rectangle, correct
to 3 significant figures
Let the length of the rectangle be x cm. Then the width
is (x − 3.10) cm.
Area = length × width = x(x − 3.10) = 23.6
i.e.
x
2
− 3.10x − 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 = 6.65 cm or −3.55 cm. The latter solution is
neglected since length cannot be negative.
Thus, length x = 6.65 cm and width = x − 3.10 =
6.65 − 3.10 = 3.55 cm, i.e. the dimensions of the rectangle are 6.65 cm by 3.55 cm.
(Check: Area = 6.65 × 3.55 = 23.6 cm 2 , correct to
3 significant figures.)
Problem 23. Calculate the diameter of a solid
cylinder which has a height of 82.0 cm and a total
surface area of 2.0 m 2
Total surface area of a cylinder
= curved surface area + 2 circular ends
= 2πrh + 2πr
2
(where r = radius and h = height)
Since the total surface area = 2.0 m 2 and the height h =
82 cm or 0.82 m,
2.0 = 2πr(0.82) + 2πr
2
i.e.
2πr
2
+ 2πr(0.82) − 2.0 = 0
Dividing throughout by 2π gives r 2 + 0.82r −
1
π
= 0
Using the quadratic formula,
r =
−0.82 ±
(0.82) 2 − 4(1)
−
1
π
2(1)
=
−0.82 ±
√
1.94564
2
=
−0.82 ± 1.39486
2
= 0.2874 or − 1.1074
Thus, the radius r of the cylinder is 0.2874 m (the
negative solution being neglected).
Hence, the diameter of the cylinder
= 2 × 0.2874
= 0.5748 m or 57.5 cm
correct to 3 significant figures.
Problem 24. The height s metres of a mass
projected vertically upwards at time t seconds is
s = ut −
1
2
gt 2 . Determine how long the mass will
take after being projected to reach a height of 16 m
(a) on the ascent and (b) on the descent, when
u = 30 m/s and g = 9.81 m/s 2
When height s = 16 m, 16 = 30t −
1
2
(9.81)t
2
i.e.
4.905t
2
− 30t + 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
7. 4x 2 + 6x − 8 = 0
8. 5.6x 2 − 11.2x − 1 = 0
9. 3x(x + 2) + 2x(x − 4) = 8
10. 4x 2 − x(2x + 5) = 14
11.
5
x − 3
+
2
x − 2
= 6
12.
3
x − 7
+ 2x = 7 + 4x
13.
x + 1
x − 1
= x − 3
14.5 Practical problems involving
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 23.6 cm 2
and its width is 3.10 cm shorter than its length.
Determine the dimensions of the rectangle, correct
to 3 significant figures
Let the length of the rectangle be x cm. Then the width
is (x − 3.10) cm.
Area = length × width = x(x − 3.10) = 23.6
i.e.
x
2
− 3.10x − 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 = 6.65 cm or −3.55 cm. The latter solution is
neglected since length cannot be negative.
Thus, length x = 6.65 cm and width = x − 3.10 =
6.65 − 3.10 = 3.55 cm, i.e. the dimensions of the rectangle are 6.65 cm by 3.55 cm.
(Check: Area = 6.65 × 3.55 = 23.6 cm 2 , correct to
3 significant figures.)
Problem 23. Calculate the diameter of a solid
cylinder which has a height of 82.0 cm and a total
surface area of 2.0 m 2
Total surface area of a cylinder
= curved surface area + 2 circular ends
= 2πrh + 2πr
2
(where r = radius and h = height)
Since the total surface area = 2.0 m 2 and the height h =
82 cm or 0.82 m,
2.0 = 2πr(0.82) + 2πr
2
i.e.
2πr
2
+ 2πr(0.82) − 2.0 = 0
Dividing throughout by 2π gives r 2 + 0.82r −
1
π
= 0
Using the quadratic formula,
r =
−0.82 ±
(0.82) 2 − 4(1)
−
1
π
2(1)
=
−0.82 ±
√
1.94564
2
=
−0.82 ± 1.39486
2
= 0.2874 or − 1.1074
Thus, the radius r of the cylinder is 0.2874 m (the
negative solution being neglected).
Hence, the diameter of the cylinder
= 2 × 0.2874
= 0.5748 m or 57.5 cm
correct to 3 significant figures.
Problem 24. The height s metres of a mass
projected vertically upwards at time t seconds is
s = ut −
1
2
gt 2 . Determine how long the mass will
take after being projected to reach a height of 16 m
(a) on the ascent and (b) on the descent, when
u = 30 m/s and g = 9.81 m/s 2
When height s = 16 m, 16 = 30t −
1
2
(9.81)t
2
i.e.
4.905t
2
− 30t + 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