110 Basic Engineering Mathematics
K a =
x 2
v(1 − x)
, determine x, the degree of
ionization, given that v = 10 dm 3 .
6. A rectangular building is 15 m long by 11 m
wide. A concrete path of constant width is
laid all the way around the building. If the
area of the path is 60.0 m 2 , calculate its width
correct to the nearest millimetre.
7. The total surface area of a closed cylindrical
container is 20.0 m 3 . Calculate the radius of
the cylinder if its height is 2.80 m.
8. The bending moment M at a point in a beam
is given by M =
3x(20 − x)
2
, where x metres
is the distance from the point of support.
Determine the value of x when the bending
moment is 50 Nm.
9. A tennis court measures 24 m by 11 m. 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. If a border of constant
width is allowed around each court and the
total area of the court and its border is 950 m 2 ,
find the width of the borders.
10. Two resistors, when connected in series, have
a total resistance of 40 ohms. When connected in parallel their total resistance is 8.4
ohms. If one of the resistors has a resistance
of R x , ohms,
(a) show that R 2
x − 40R x + 336 = 0 and
(b) calculate the resistance of each.
14.6 Solution of linear and quadratic
equations simultaneously
Sometimes a linear equation and a quadratic equation
need to be solved simultaneously. An algebraic method
of solution is shown in Problem 27; a graphical solution
is shown in Chapter 19, page 160.
Problem 27. Determine the values of x and y
which simultaneously satisfy the equations
y = 5x − 4 − 2x 2 and y = 6x − 7
For a simultaneous solution the values of y must be
equal, hence the RHS of each equation is equated.
Thus,
5x − 4 − 2x 2 = 6x − 7
Rearranging gives 5x − 4 − 2x
2
− 6x + 7 = 0
i.e.
−x + 3 − 2x
2
= 0
or
2x
2
+ x − 3 = 0
Factorizing gives
(2x + 3)(x − 1) = 0
i.e.
x = −
3
2
or x = 1
In the equation y = 6x − 7,
when x = −
3
2
,
y = 6
−
3
2
− 7 = −16
and when x = 1, y = 6 − 7 = −1
(Checking the result in y = 5x − 4 − 2x 2 :
when x = −
3
2
,
y = 5
−
3
2
− 4 − 2
−
3
2
2
= −
15
2
− 4 −
9
2
= −16, as above,
and when x = 1, y = 5 − 4 − 2 = −1, as above.)
Hence, the simultaneous solutions occur when
x = −
3
2
, y = −16 and when x = 1, y = −1.
Now try the following Practice Exercise
Practice Exercise 58 Solving linear and
quadratic equations simultaneously
(answers on page 346)
Determine the solutions of the following simultaneous equations.
1. y = x 2 + x + 1
2. y = 15x 2 + 21x − 11
y = 4 − x
y = 2x − 1
3. 2x 2 + y = 4 + 5x
x + y = 4
K a =
x 2
v(1 − x)
, determine x, the degree of
ionization, given that v = 10 dm 3 .
6. A rectangular building is 15 m long by 11 m
wide. A concrete path of constant width is
laid all the way around the building. If the
area of the path is 60.0 m 2 , calculate its width
correct to the nearest millimetre.
7. The total surface area of a closed cylindrical
container is 20.0 m 3 . Calculate the radius of
the cylinder if its height is 2.80 m.
8. The bending moment M at a point in a beam
is given by M =
3x(20 − x)
2
, where x metres
is the distance from the point of support.
Determine the value of x when the bending
moment is 50 Nm.
9. A tennis court measures 24 m by 11 m. 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. If a border of constant
width is allowed around each court and the
total area of the court and its border is 950 m 2 ,
find the width of the borders.
10. Two resistors, when connected in series, have
a total resistance of 40 ohms. When connected in parallel their total resistance is 8.4
ohms. If one of the resistors has a resistance
of R x , ohms,
(a) show that R 2
x − 40R x + 336 = 0 and
(b) calculate the resistance of each.
14.6 Solution of linear and quadratic
equations simultaneously
Sometimes a linear equation and a quadratic equation
need to be solved simultaneously. An algebraic method
of solution is shown in Problem 27; a graphical solution
is shown in Chapter 19, page 160.
Problem 27. Determine the values of x and y
which simultaneously satisfy the equations
y = 5x − 4 − 2x 2 and y = 6x − 7
For a simultaneous solution the values of y must be
equal, hence the RHS of each equation is equated.
Thus,
5x − 4 − 2x 2 = 6x − 7
Rearranging gives 5x − 4 − 2x
2
− 6x + 7 = 0
i.e.
−x + 3 − 2x
2
= 0
or
2x
2
+ x − 3 = 0
Factorizing gives
(2x + 3)(x − 1) = 0
i.e.
x = −
3
2
or x = 1
In the equation y = 6x − 7,
when x = −
3
2
,
y = 6
−
3
2
− 7 = −16
and when x = 1, y = 6 − 7 = −1
(Checking the result in y = 5x − 4 − 2x 2 :
when x = −
3
2
,
y = 5
−
3
2
− 4 − 2
−
3
2
2
= −
15
2
− 4 −
9
2
= −16, as above,
and when x = 1, y = 5 − 4 − 2 = −1, as above.)
Hence, the simultaneous solutions occur when
x = −
3
2
, y = −16 and when x = 1, y = −1.
Now try the following Practice Exercise
Practice Exercise 58 Solving linear and
quadratic equations simultaneously
(answers on page 346)
Determine the solutions of the following simultaneous equations.
1. y = x 2 + x + 1
2. y = 15x 2 + 21x − 11
y = 4 − x
y = 2x − 1
3. 2x 2 + y = 4 + 5x
x + y = 4