Solving quadratic equations 109
Hence, the mass will reach a height of 16 m after
0.59 s on the ascent and after 5.53 s on the descent.
Problem 25. A shed is 4.0 m long and 2.0 m wide.
A concrete path of constant width is laid all the way
around the shed. If the area of the path is 9.50 m 2 ,
calculate its width to the nearest centimetre
Figure 14.1 shows a plan view of the shed with its
surrounding path of width t metres.
t
2.0 m
4.0 m
(4.0 1 2t)
SHED
t
Figure 14.1
Area of path = 2(2.0 × t ) + 2t (4.0 + 2t )
i.e.
9.50 = 4.0t + 8.0t + 4t
2
or
4t
2
+ 12.0t − 9.50 = 0
Hence,
t =
−(12.0) ±
(12.0) 2 − 4(4)(−9.50)
2(4)
=
−12.0 ±
√
296.0
8
=
−12.0 ± 17.20465
8
i.e. t = 0.6506 m or − 3.65058 m.
Neglecting the negative result, which is meaningless,
the width of the path, t = 0.651 m or 65 cm correct to
the nearest centimetre.
Problem 26. If the total surface area of a solid
cone is 486.2 cm 2 and its slant height is 15.3 cm,
determine its base diameter.
From Chapter 27, page 245, the total surface area A of
a solid cone is given by A = πrl + πr 2 , where l is the
slant height and r the base radius.
If A = 482.2 and l = 15.3, then
482.2 = πr(15.3) + πr 2
i.e.
πr
2
+ 15.3πr − 482.2 = 0
or
r
2
+ 15.3r −
482.2
π
= 0
Using the quadratic formula,
r =
−15.3 ±
(15.3) 2 − 4
−482.2
π
2
=
−15.3 ±
√
848.0461
2
=
−15.3 ± 29.12123
2
Hence, radius r = 6.9106 cm (or −22.21 cm, which is
meaningless and is thus ignored).
Thus, the diameter of the base = 2r = 2(6.9106)
= 13.82 cm.
Now try the following Practice Exercise
Practice Exercise 57 Practical problems
involving quadratic equations (answers on
page 346)
1. The angle a rotating shaft turns through in t
seconds is given by θ = ωt +
1
2
αt 2 . Determine the time taken to complete 4 radians if
ω is 3.0 rad/s and α is 0.60 rad/s 2 .
2. The power P developed in an electrical circuit is given by P = 10I − 8I 2 , where I is
the current in amperes. Determine the current
necessary to produce a power of 2.5 watts in
the circuit.
3. The area of a triangle is 47.6 cm 2 and its
perpendicular height is 4.3 cm more than its
base length. Determine the length of the base
correct to 3 significant figures.
4. The sag, l, in metres in a cable stretched
between two supports, distance x m apart, is
given by l =
12
x
+ x. Determine the distance
between the supports when the sag is 20 m.
5. The acid dissociation constant K a of ethanoic
acid is 1.8 × 10 −5 mol dm −3 for a particular solution. Using the Ostwald dilution law,
Hence, the mass will reach a height of 16 m after
0.59 s on the ascent and after 5.53 s on the descent.
Problem 25. A shed is 4.0 m long and 2.0 m wide.
A concrete path of constant width is laid all the way
around the shed. If the area of the path is 9.50 m 2 ,
calculate its width to the nearest centimetre
Figure 14.1 shows a plan view of the shed with its
surrounding path of width t metres.
t
2.0 m
4.0 m
(4.0 1 2t)
SHED
t
Figure 14.1
Area of path = 2(2.0 × t ) + 2t (4.0 + 2t )
i.e.
9.50 = 4.0t + 8.0t + 4t
2
or
4t
2
+ 12.0t − 9.50 = 0
Hence,
t =
−(12.0) ±
(12.0) 2 − 4(4)(−9.50)
2(4)
=
−12.0 ±
√
296.0
8
=
−12.0 ± 17.20465
8
i.e. t = 0.6506 m or − 3.65058 m.
Neglecting the negative result, which is meaningless,
the width of the path, t = 0.651 m or 65 cm correct to
the nearest centimetre.
Problem 26. If the total surface area of a solid
cone is 486.2 cm 2 and its slant height is 15.3 cm,
determine its base diameter.
From Chapter 27, page 245, the total surface area A of
a solid cone is given by A = πrl + πr 2 , where l is the
slant height and r the base radius.
If A = 482.2 and l = 15.3, then
482.2 = πr(15.3) + πr 2
i.e.
πr
2
+ 15.3πr − 482.2 = 0
or
r
2
+ 15.3r −
482.2
π
= 0
Using the quadratic formula,
r =
−15.3 ±
(15.3) 2 − 4
−482.2
π
2
=
−15.3 ±
√
848.0461
2
=
−15.3 ± 29.12123
2
Hence, radius r = 6.9106 cm (or −22.21 cm, which is
meaningless and is thus ignored).
Thus, the diameter of the base = 2r = 2(6.9106)
= 13.82 cm.
Now try the following Practice Exercise
Practice Exercise 57 Practical problems
involving quadratic equations (answers on
page 346)
1. The angle a rotating shaft turns through in t
seconds is given by θ = ωt +
1
2
αt 2 . Determine the time taken to complete 4 radians if
ω is 3.0 rad/s and α is 0.60 rad/s 2 .
2. The power P developed in an electrical circuit is given by P = 10I − 8I 2 , where I is
the current in amperes. Determine the current
necessary to produce a power of 2.5 watts in
the circuit.
3. The area of a triangle is 47.6 cm 2 and its
perpendicular height is 4.3 cm more than its
base length. Determine the length of the base
correct to 3 significant figures.
4. The sag, l, in metres in a cable stretched
between two supports, distance x m apart, is
given by l =
12
x
+ x. Determine the distance
between the supports when the sag is 20 m.
5. The acid dissociation constant K a of ethanoic
acid is 1.8 × 10 −5 mol dm −3 for a particular solution. Using the Ostwald dilution law,