Section B Algebra
5.3 Completing the square
5.4.1 The nature of the roots
The formula for the roots of the quadratic equation is
x = -b ±
√
b2 - 4ac2a
Consider the following three situations:
1 If we have (b2 - 4ac) > 0, then the square root is of a positive number.
There are then two distinct roots.
2 If we have (b2 - 4ac) = 0, then the square root is zero and the formula gives just one value for x. Since a quadratic equation must have two roots, we say that the equation has two coincident roots.
3 If we have (b2 - 4ac) < 0, then the square root is of a negative number.
A new type of number has to be invented to enable such expressions to be solved. The number is referred to as a complex number and the roots are said to be imaginary (the roots in 1 and 2 above are said to be real).
Such numbers are not discussed further in this book.
Example
Have the following quadratic equations real roots?
(a) x2 - 5x - 2 = 0, (b) x2 + 3x + 5 = 0, (c) x2 - 4x + 4 = 0.
(a) Comparing the equation with ax2 + bx + c = 0, then a = 1, b = - 5 and c = - 2 . Thus (b2 - 4ac) is 25 + 8 = +33. It is thus greater than 0 and so there are two distinct real roots.
(b) Comparing the equation with ax2 + bx + c = 0, then a = 1, b = 3 and c = 5. Thus (b2 - 4ac) is 9 - 20 = - 1 1 . It is thus less than 0 and so the roots are imaginary.
(c) Comparing the equation with ax2 + bx + c = 0, then a = 1, b = - 4 and c = 4. Thus (b2 - 4ac) is 16 - 16 = 0. Because it is zero there are two coincident real roots.
Revision
15 Have the following quadratic equations real roots?
(a) 2x2 + 3x + 4 = 0, (b) x2 - 2x - 5 = 0, (c) x2 - 3x + 8 = 0, (d) x2 + 2x + 1 = 0, (e) 2x2 - 5x + 1 = 0.
1 Factorise and hence solve the following quadratic equations:
(a) x2 + x - 2 = 0, (b) x2 + 4x + 3 = 0, (c) x2 - 4x + 3 = 0, (d) x2 + 2x + 1 = 0, (e) x2 - 9 = 0, (f) x2 - 25 = 0, (g) x2 + 3x = 0, Problems
(h) 2x2 + 7x + 5 = 0, (i) 6x2 + 5x + 1 = 0, (j) 6x2 + 19x + 10 = 0, (k) 2x2 - 7x - 15 = 0, (1) 9x2 - 33x + 28 = 0.
2 Determine the quadratic equations giving the following roots:
(a) x = 1, x = - 1 , (b) x = 2, x = 2, (c) x = - 3 , x = - 1 , (d) x = 0, x = 1.
3 Solve the following quadratic equations using (a) the method of completing the square, (b) the formula:
(a) x2 - 2x - 8 = 0, (b) x2 + x - 3 = 0, (c) 2x2 + x - 3 = 0, (d) 3x2 - 4x + 1 = 0, (e) x2 + 2x - 5 = 0, (f) 5x2 + x - 3 = 0, (g) 4x2 + 4x - 15 = 0, (h) 5x2 - 8x + 2 = 0, (i) 4x2 + 5x - 3 = 0.
4 The area of a rectangle is 100 cm2. If it has a width which is 5 cm less than its length, what are the dimensions of the rectangle? Hint: let the length be x and the width is then (x - 5).
5 A rectangle has a perimeter of 26 cm and an area of 30 cm2. What are its dimensions? Hint: let x equal the length of one side, then the length of the other is 13 - x.
6 An object is thrown vertically upwards with a velocity of 50 m/s. The height h of the object after a time t is given by h = 50t - 4.9t2. Hence determine the times when the object is at a height of 100 m.
7 A thermocouple gives an e.m.f. E in μV which depends on the tempera
ture θ in degrees C according to the equation E = 6.88θ - 0.0192θ2. Solve the equation for a thermocouple for which the e.m.f. is 200 μV.
8 The bending moment M at a point a distance x along a beam is given by:
M = wx(L - x) 2
Determine the values of x when M = 5 0 N m, w = 5 N/m and L = 20 m.
9 The surface area A of a cylinder is given by A = 2πr2h + 2πr2, where r is the radius and h the height. What height cylinder is required if, with r = 5 cm, the area is to be 125 cm2?
10 The current i, in amperes, through a circuit element is related to the voltage v, in volts, across it by the equation i = 0.001v2 + 0.010v.
Determine the value of v when i is 0.004 A.
11 A square sheet of metal has a square hole cut from it so that the border all round the hole has a width of 4 mm. If the hole has an area of one quarter of the area of the original square, determine the length of the side of the square.
12 A right-angled triangle has a hypotenuse of length 18 mm and a total perimeter of length 40 mm. Determine the lengths of the other two sides.
13 A right-angled triangle is to have a hypotenuse of length 10 cm. If, for the other two sides one of them is to be longer than the other by 3 cm, what are the lengths of the sides?
14 Determine the two numbers which have a difference of 5 and a product of 266.
15 A rectangle has a length which is 3 cm greater than its width. If the area of the rectangle is 28 cm2, what are its length and width?
16 A piece of wire of length 40 cm is to be bent into a rectangular shape to enclose an area of 100 cm2. What will be the lengths of the sides of the rectangle?
17 Determine the number which added to its square gives 90.
18 A rectangular sheet of metal is 14 cm by 17 cm. Strips of the same width are cut off one side and one end. If the area remaining is 108 cm2, what is the width of the pieces removed?
19 What are the two consecutive numbers which have a product of 156?
20 The length of a factory floor is to be 50 m more than its width and the total area 266 m2. What will be the width and length?
21 Have the following quadratic equations real roots?
(a) x2 + 3x + 1 = 0, (b) x2 - 5x - 2 = 0, (c) x2 + 2x + 4 = 0, (d) x2 + 3x + 4 = 0, (e) 2x2 + 4x + 2 = 0.
6.1 Introduction In the earlier chapters in this section, the algebraic techniques developed have been illustrated by showing their use in engineering problems. In this chapter the emphasis is on the engineering with the algebraic techniques used to support the development of an engineering topic. The aim is to put the algebraic techniques in an engineering context. The engineering topics in this chapter are the development and use of the equations for straight-line motion and basic electrical circuit analysis.
6.2 Straight line motion Consider an object moving along a straight line with a uniformly accelerated motion. If u is the initial velocity, i.e. at time t = 0, and v the velocity after some time t, then the change in velocity in the time interval t is (v - u). Hence the acceleration a is:
a = v - u t
Multiplying both sides of the equation by t gives at = v - u. Hence the equation can be manipulated to give
v = u + at [Equation 1]
If the object, in its straight line motion, covers a distance s in a time t, then the average velocity in that time interval is s/t. With an initial velocity of u and a final velocity of v at the end of the time interval, the average velocity is:
average velocity = u + v 2 Hence
s
t = u + v 2
Multiplying both sides of the equation by t gives:
s =
(
u + v 2)
tSubstituting for v by using the equation v = u + at gives:
s =
(
u + (u + at)2
)
t =(
2u + at 2)
tHence:
s = ut + ½at2 [Equation 2]
Consider the equation v = u + at. Squaring both sides of this equation gives:
v2 = (u + at)2 = u2 + 2uat + a2t2
We can rewrite this equation as v2 = u2 + 2a(ut + ½at2) Hence:
v2 = u2 + 2as [Equation 3]
The equations [1], [2] and [3] are referred to as the equations for straight-line motion. The following examples illustrate their use in solving engineering problems.
Example
An object moves in a straight line with a uniform acceleration. If it starts from rest and takes 12 s to cover 100 m, what is the acceleration?
We have u = 0, s = 100 m, t = 12 s and are required to obtain a. Using s
= ut + ½at2:
100 = 0 + ½a x 122 = 0 + 72a
Rearranging this equation to give 72a = 100, then dividing both sides by 72, gives a = 100/72 = 1.4 m/s2.
Example
A package slides down a slope with a constant acceleration. Its velocity increases from 100 mm/s to 320 mm/s in a distance of 600 mm. What is the acceleration?
The question gives u = 100 mm/s, v = 320 mm/s, s = 600 mm and we are required to obtain a. Using v2 = u2 + 2as:
3202 = 1002 + 2a x 600
Transposing terms, we can rewrite this equation as:
1200a = 3202 - 1002 = 102 400 - 10 000 = 92 400
Hence, dividing both sides of the equation by 1200 gives the answer as a = 92 400/1200 = 777 mm/s2.
Example
A car travelling at 25 m/s brakes and slows down with a uniform retar
dation of 1.2 m/s2. How long will it take to come to rest?
A retardation is a negative acceleration in that the final velocity is less than the starting velocity. Thus we have u = 25 m/s, a final velocity of v = 0, retardation a = -1.2 m/s2 and are required to obtain t. Using v = u + at, then:
0 = 25 + (-1.2)t = 25 - 1.2t
Transposing the 1.2t to the other side of the equation gives 1.2t = 25.
Dividing both sides of the equation by 1.2 gives t = 25/1.2 = 20.8 s.
Example
A car is moving with a velocity of 10 m/s. It then accelerates at 0.2 m/s2
for 100 m. What will be the time taken for the 100 m to be covered?
We have u = 10 m/s, a = 0.2 m/s2 and s = 100 m and are required to obtain t. Using s = ut + ½at2:
100 = 10t + ½ x 0.2t2 We can write this equation as:
0.1t2 + 10t - 100 = 0
This is a quadratic equation. Using the formula
t = -b ±
√
b2 - 4ac2a then:
t =
-10 ±
√
100 - 4 x 0.1(-100)2 x 0.1 = -10 ±
√
1400.2 = -10 ± 11.83 0.2
Hence t = -50 - 59.2 = -109.2 s or t = -50 + 59.2 = 9.2 s. Since the negative time has no significance, the answer is 9.2 s. The answer can be checked by substituting it back in the original equation.
Revision
1 The following all refer to an object moving in a straight line with uniform acceleration.
(a) Initially at rest, acceleration 2 m/s2 for 8 s. Determine the distance travelled in that time.
(b) Initial velocity 3 m/s, acceleration 2 m/s2 for 6 s. Determine the velocity at the end of that time.
(c) Initial velocity 3 m/s, final velocity 5 m/s after 2 s. Determine the distance travelled in that time.
(d) Initial velocity 3 m/s, distance travelled 20 m, final velocity 7 m/s.
Determine the time taken.
(e) Initial velocity 10 m/s, acceleration - 4 m/s2, final velocity 2 m/s.
Determine the distance travelled.
(f) Initial velocity 10 m/s, acceleration - 4 m/s2, final velocity 2 m/s.
Determine the time taken.
(g) Acceleration 0.5 m/s2 for 10 s, final velocity 10 m/s. Determine the initial velocity.
2 A train is moving with a velocity of 10 m/s. It then accelerates at a uniform rate of 2.5 m/s2 for 8 s. What is the velocity after the 8 s?
3 An object starts from rest and moves with a uniform acceleration of 2 m/s2 for 20 s. What is the distance moved by the object?
4 A cyclist is moving with a velocity of 1 m/s. He/she then accelerates at 0.4 m/s2 for 100 m. What will be the time taken for the 100 m?
5 Rearrange the equation v2 = u2 + 2as to give (a) s in terms of the other quantities, (b) a in terms of the other quantities, (c) u in terms of the other quantities.
6 A car accelerates from 7.5 m/s to 22.5 m/s at 2 m/s2. What is (a) the time taken, (b) the distance travelled during this acceleration?