Quadratic Equations: Ab 0, Then Either A 0 or B 0. Both A and B May Equal Zero
Quadratic Equations: Ab 0, Then Either A 0 or B 0. Both A and B May Equal Zero
Quadratic Equations: Ab 0, Then Either A 0 or B 0. Both A and B May Equal Zero
is zero.
In symbols:
If ab = 0, then either a = 0 or b = 0. Both a and b may equal zero.
Quadratic equations To solve a quadratic equation:
r .PWFBMMUIFUFSNTUPUIFMFGUIBOETJEFPGUIFFRVBUJPOMFBWJOH[FSP
on the right-hand side.
We have been solving linear equations for some time.
r 'BDUPSJTFUIFFYQSFTTJPOBTBQSPEVDUPGUXPGBDUPST
We will now learn how to apply factorisation to solve
r &RVBUFFBDIGBDUPSUP[FSPBOETPMWFFBDIFRVBUJPO
quadratic equations.
The solutions to these linear equations are the solutions to the quadratic
Quadratic equations turn up routinely in mathematics. equation.
Being able to solve them is a fundamental skill.
For example, finding the distance a rocket or cricket ball
17A Solution of simple quadratic equations
will travel involves quadratic equations.
The ancient Babylonians were solving quadratic equations In Example 1 the quadratic equations are given in factorised form.
over 5000 years ago!
Example 1
Solve the equations: If there is a numerical factor common to all of the coefficients in
the equation, then we can divide both sides of the equation by that
a x2 – 2x = 0 b x2 – 2x = 48 common factor.
c x2 = 3x + 10 d x2 – 7x = 18
Example 3
Solution
In each case we move all terms to the left-hand side and factorise Solve:
the quadratic expression so that a product can be obtained. a 6x2 = 18x b 2x2 – 10x + 12 = 0 c –x2 + 9x – 18 = 0
a x2 – 2x = 0 b x2 – 2x = 48 Solution
x(x – 2) = 0 x2 – 2x – 48 = 0
a 6x2 = 18x
x = 0 or x – 2 = 0 (x – 8)(x + 6) = 0 2
6(x – 3x) = 0
x = 0 or x = 2 x = 8 or x = –6
6x(x – 3) = 0
c x2 = 3x + 10 d x2 – 7x = 18 x(x – 3) = 0 (Divide both sides of the equation by 6.)
x – 3x – 10
2
=0 x – 7x – 18 = 0
2
x = 0 or x – 3 = 0
(x – 5)(x + 2) =0 (x – 9)(x + 2) = 0 x = 0 or x = 3
x = 5 or x = –2 x = 9 or x = –2
b 2x2 – 10x + 12 = 0
2(x2 – 5x + 6) = 0
Notice that in these examples we obtain two solutions to the original x2 – 5x + 6 = 0 (Divide both sides of the equation by 2.)
equation. We can check they are solutions by substitution. For example,
(x – 3)(x – 2) = 0
in part c above:
x – 3 = 0 or x – 2 = 0
Using x = 5: LHS = 52 = 25 x = 3 or x = 2
RHS = 3 × 5 + 10
c –x2 + 9x – 18 = 0
= 25
–1(x2 – 9x + 18) = 0
Therefore LHS = RHS
x2 – 9x + 18 = 0 (Divide both sides of the equation by –1.)
Using x = –2: LHS = (–2)2 = 4
(x – 3)(x – 6) = 0
RHS = 3 × –2 + 10
x – 3 = 0 or x – 6 = 0
=4
x = 3 or x = 6
Therefore LHS = RHS
– Move all terms to the left-hand side of the equation leaving zero Example 3 4 Solve the quadratic equations.
on the right-hand side. a 5x2 – 10x = 0 b 2x2 + 5x = 0 c 4x2 = –12x
– If there is a numerical common factor, then divide both sides of d 3x = 2x2 e 5x = 15x2 f 3a2 + 15a + 18 = 0
the equation by the factor.
g 4r2 – 4r – 24 = 0 h 5c2 + 40c + 60 = 0 i 3a2 – 48a + 192 = 0
– Factorise the resulting expression as a product of two factors.
j 2n2 – 22n + 60 = 0 k 4s2 = 36s + 280 l –x2 + 24x + 25 = 0
– Equate each factor to zero and solve each equation.
m –x + 42 = x2 n –2x2 + 10x + 48 = 0 o 3x2 – 48x = – 84
t /PUBMMRVBESBUJDFRVBUJPOTIBWFTPMVUJPOT'PSFYBNQMFx2 + 5 = 0 has
no solutions, since x2 + 5 ≥ 5 for all values of x. Hence x2 + 5 will never Example 4 5 Solve each quadratic equation.
equal to zero. a x2 – 1 = 0 b x2 – 25 = 0 c 4x2 – 1 = 0
d 2x2 – 50 = 0 e x2 + 6x + 9 = 0 f x2 + 10x + 25 = 0
g x2 – 12x + 36 = 0 h x2 – 16 = 0 i 16x2 – 25 = 0
Solve x = 3x – 2
In many mathematical problems and applications, equations arise that do x
not initially appear to be quadratic equations. We often need to rearrange Solution
an equation to put it into the standard form ax2 + bx + c = 0.
x = 3x – 2
Example 6 x
x2 = 3x – 2 (Multiply both sides by x.)
Solve: x – 3x + 2
2
=0 (Rearrange.)
(x – 1)(x – 2) =0
a x(x + 5) = 6 b x(x – 3) + x = 1
2 x – 1 = 0 or x – 2 =0
Solution x = 1 or x =2
a x(x + 5) = 6 We can check mentally that these are solutions to the original
equation.
x2 + 5x – 6 = 0 (Rearrange)
(x + 6)(x – 1) = 0
x + 6 = 0 or x – 1 = 0 Example 8
x = –6 or x = 1
We can check that these are the correct solutions by substitution. Solve x – 2 = x5
3
If x = –6, LHS = –6(–6 + 5) = 6 = RHS Solution
If x = 1, LHS = 1(1 + 5) = 6 = RHS
Method 1 Method 2
x(x – 3) x–2 x–2
b +x =1 = x5 5
2 3 3 = x
x(x – 3) + 2x =2 (Multiply both sides by 3) 3x × x – 2 = 3x × x5 x(x – 2) = 3 × 5
3
x2 – 3x + 2x =2 (Rearrange)
(Multiply both sides of (This is called
x2 – x – 2 =0 the equation by 3x.) cross multiplication.)
(x + 1)(x – 2) =0
x + 1 = 0 or x – 2 =0 x(x – 2) = 15
x = –1 or x =2 x2 – 2x = 15
x – 2x – 15 = 0
2
(Rearrange.)
(x + 3)(x – 5) = 0
Some equations involve fractions in which the pronumeral appears in
x + 3 = 0 or x – 5 = 0
the denominator. You will need to take care when solving these. We
always assume that the pronumeral cannot take a value that makes the x = –3 or x = 5
denominator equal to zero. It is important to check that your answers are
in fact correct solutions to the given equation. Note: Cross multiplication is an effective method in the solution of such
equations. It is justified by Method 1.
(x – 4)(x + 3) = 0 m x = 13 – 3 n x = 10
x
–3
3x – 1
x = 4 or x = –3
Example 10 5 Solve:
a 2x + x – 1 = 2 + 22
x
b x + x–4 = x
3 15 3 2 x+4 3
Exercise 17C
c x+1 + 8 =3 d x–1
– x–3
= –2
7 x –2 x+2 x–4 3
Example 6a 1 Solve:
e x(x + 9) + 3 = x(2 – 3x) f 3x(x – 2) = x(x + 1)
a x(x – 6) = –8 b x(x – 7) = –10 c x(x – 5) = –4
d x(x – 8) = –15 e x(x – 3) = 4 f x(x – 6) = 27
g (x – 3)(x – 4) = 12 h (x + 7)(x – 2) + 14 = 0
i (x + 2)(x – 6) – 65 = 0 j (x – 1)(x – 2) = 20
A rectangle has one side 3 cm longer than the other. The rectangle Example 13
has area 28 cm2. How long is the shorter side?
Each term in the sequence 5, 9, 13, 17, … is obtained by adding
Solution 4 to the previous number. The sum S of the first n terms in this
sequence is given by S = 2n2 + 3n. How many terms must we add
Let x cm be the length of the shorter side. The other side has
to make a sum of 90?
length (x + 3) cm.
Area of rectangle is x(x + 3) = 28 (x + 3) cm Solution
x + 3x – 28
2
=0 x cm S = 90, so 2n2 + 3n = 90
(x – 4)(x + 7) =0
2n2 + 3n – 90 = 0 (Find two numbers with product
x – 4 = 0 or x + 7 =0 of 2 × (–90) = –180 and sum of 3.
x = 4 or x = –7 The numbers are 15 and –12.)
Since length is positive, x = –7 makes no sense. Hence the shorter 2n2 + 15n – 12n – 90 = 0
side has length 4 cm and the longer side is 7 cm. n(2n + 15) – 6(2n + 15) = 0
(2n + 15)(n – 6) = 0
2n + 15 = 0 or n – 6 = 0
Example 12
n = – 15 or n = 6
2
The formula for the number D of diagonals of a polygon with Since n is a positive whole number, the solution n = – 15 does not
n sides is D = n (n – 3). How many sides are there in a polygon make sense here. So n = 6.
2
2
with 35 diagonals?
Check: 5 + 9 + 13 + 17 + 21 + 25 = 90.
6 Find the value of x in 14 A, B and C do a piece of work together. A could have done it alone
(3x + 1) cm in six hours longer, B in fifteen hours longer and C in twice the time.
the diagram opposite. (x + 1) cm
How long were all three about it?
(2x + 4) cm
17E Graphs of quadratics
7 A metal sheet is 50 cm wide and 60 cm long. It has squares cut out of
the corners so that it can be folded to form a box with a base area of
In this section, we plot graphs of quadratics. First we complete a table
1200 cm2. Find the length of the side of the squares.
of values for y = x2 for x between –3 and 3.
8 A metal sheet is 25 cm long and 20 cm wide. Squares are cut out of
the corners so that it can be folded to form a box with a base area of x –3 –2 –1 0 1 2 3
336 cm2. Find the dimensions of the box. y=x 2
9 4 1 0 1 4 9
xm
x – 5x + 25 – 25 – 3 = 0
2
(Add and subtract the square of Exercise 17F
4 4
half the coefficient of x.)
x2 – 5x + 25 – 37 = 0
4
4 1 Solve the equations:
25 37
x – 5x + =
2
a x2 – 5 = 0 b x2 – 11 = 0 c x2 – 13 = 0 d x2 – 12 = 0
4 4
5 2 37 e 2x2 – 6 = 0 f 4x2 – 8 = 0 g 50 – 5x2 = 0 h 40 – 8x2 = 0
x– =
2 4
Example
x– 5=
√37
or x – 5 = – 37
√ 16,17 2 Solve the equations by completing the square.
2 2 2 2
5 + √37 5 – √37 a x2 + 2x – 1 = 0 b x2 + 4x + 1 = 0 c x2 – 12x + 23 = 0
x= or x =
2 2
d x2 + 6x + 7 = 0 e x2 – 8x – 1 = 0 f x2 + 10x + 12 = 0
Example 18 3 Solve the equations by completing the square.
a x2 + x – 1 = 0 b x2 – 3x + 1 = 0 c x2 – 5x – 1 = 0
d x2 + 3x – 2 = 0 e x2 + 5x + 1 = 0 f x2 + 9x – 2 = 0
g x2 – 6x + 2 = 0 h x2 – 8x + 11 = 0 i x2 + 10x + 23 = 0
2 Solve the equations: 9 Find two numbers, the sum of whose squares is 74, and whose sum is 12.
a x2 – 3x + 2 = 0 b x2 – 7x + 12 = 0 c x2 – 11x + 10 = 0 10 The perimeter of a rectangular field is 500 m and its area is 14 400 m2.
Find the lengths of the sides.
d x2 + 29x – 30 = 0 e x2 – 5x – 14 = 0 f x2 – x – 90 = 0
11 The base and height of a triangle are x + 3 and 2x – 5. If the area of
g x2 – 5x – 24 = 0 h x2 – 11x + 18 = 0 i x2 – x – 12 = 0
the triangle is 20, find x.
j x2 – 11x + 28 = 0 k x2 + 9x – 10 = 0 l x2 + x – 110 = 0
12 Two positive numbers differ by 7 and the sum of their squares is 169.
3 Solve the equations: Find the numbers.
a x2 + 18x + 81 = 0 b 9x2 – 16 = 0 c 4x – x2 = 0 13 A rectangular field, 70 m long and 50 m wide has a path of uniform
d 3x – 12x – 36 = 0 e x – 8x + 16 = 0
2 2 2
f 9f – 36f + 11 = 0 width around it. If the area of the path is 1024 m2, find the width of
the path.
g 12y2 + 21 = –32y h 7x2 = 28 i x2 – 64 = 0
4 Solve the equations:
2 2
a 5d 2 – 10 = 0 b 2y – 5 = 0 c 3(x – 10) – 12 = 0
3 5
d y2 – 8y + 3 = 0 e 1 = m2 – m f 3n + 3 = n2
5 Plot each of the following graphs for the x-values specified.
a y = x2 – 2x – 3, –2 ≤ x ≤ 4 b y = x2 + 2x – 3, –4 ≤ x ≤ 2
c y = x2 – 3, –2 ≤ x ≤ 2 d y = x2 + 2x – 8, –5 ≤ x ≤ 3
6 Solve the equations:
a x2 – 8x = 1 b x + x1 – 13 = 0
3 6
c 2
+ x + 3 = 10 d 1
+ 1 =4
x+3 2 3 x–3 x–5 3
7 2 4x x–5 4x + 7
e – =1 f + =
3x – 4 x + 2 9 x+3 19
0 x 0 x 0 x
6 x=4 7 10 cm 8 21 cm × 16 cm × 2 cm 9 x = 3 –3 3 –2 1 3 –1 2 3
2
10 V = 12 11 20 km/h 12 75 km/h, 80 km/h 13 6 km/h 14 3 hours y = –x y = 2 – x2 –6 –3
–9 –7 y = x – x2 y = 2x – x 2
Exercise 17E 4 a y b x = –1 or x = 2
y = x2 – x – 2
1 a b c y 4
y y y=x –4 2 c The solutions to x2 – x – 2 = 0 are where the graph
5
y = x2 + 1 y = x2 – 1 crosses the x-axis, as this is where y = 0.
10
8
–2 2
1 –1
0 x 0 x
–3 0 3 –2 –1 2 3
–3 0 3 x –3 –1 1 3 x
–4 –2
(0.5, –2.25)
vertex (0, 1) vertex (0, –1) vertex (0, –4)
5 a b x = –3 or x = 1
d e f y y = x2 – x – 6
y 2 y y c The solutions to x2 – x – 6 = 0 are where the graph
y=x +2 y = x2 – 3 y = x2 + 3
11 6 12 crosses the x-axis, as this is where y = 0.
–2
0
3 x
2 3
–6
–3 0 3 x
–3 0 3 x –3 0 3 x
–3
vertex (0, 2) vertex (0, –3) vertex (0, 3) (0.5, –6.25)
2 a b c y
y y = x2 – 4x + 3 y y 6 a b x = –3 or x = 1
3 y = x2 + 2x – 3 y = 2x2 – 2 y = x2 + 2 x – 3
5 6
2 c The solutions to x2 + 2x – 3 = 0 are where the graph
1 crosses the x-axis, as this is where y = 0.
–3 1 –1 1
–3
0
1 x
0
1 2 3 4 x –4
0
2 x –2
0
2 x
–1
(2, –1) –3 (0, –2)
–4 –3
(–1, –4)
d e f (–1, –4)
y y y y = x 2 – 2x
8
y = x2 + x – 2
4
y = x2 – 2x – 8 Exercise 17F
7
–2 1
–2 4 1 a x = √5 or x = –√5 b x = √11 or x = –√11 c x = √13 or x = –√13
–3
0
5 x
–3
0
2 x 8
–2 2 4
d x = 2√3 or x = –2√3 e x = √3 or x = –√3 f x = √2 or x = –√2
(–0.5, –2.25) 10 (1, –9) 0 x
(1, –1)
g x = √10 or x = –√10 h x = √5 or x = –√5
g h
y = x 2 + 3x y y 2 a x = –√2 – 1 or x = √2 – 1 b x = –√3 – 2 or x = √3 – 2
4 y = 2x 2 – x 4 c x = 6 – √13 or x = 6 + √13 d x = –√2 – 3 or x = √2 – 3
e x = 4 – √17 or x = 4 + √17 f x = –√13 – 5 or x = √13 – 5
a x = – 5 – 1 or x = 5 – 1 b x=3+ or x = 3 – 5
√ √ √5 √
–3 0 x 3
0 x –1 1 2 2 2 2 2
–4 1 (0.5, –0.5)
c x = 5 + 29 or x = 5 – 29 d x = – 17 – 3 or x = –3
√ √ √ √17
2 2 2 2
(–1.5, –2.25)
e x = – 21 – 5 or x = 21 – 5 f x = – 89 – 9 or x = 89 – 9
√ √ √ √
2 2 2 2
Review exercise
1 a x = –3 or x = 5 b x = 5 or x = 0 c x = –7 or x = 0
d x = 4 or x = 8 e x = –13 or x = – 8 f x = 7 or x = 7
3 2 11
2 a x = 1 or x = 2 b x = 3 or x = 4 c x = 1 or x = 10
d x = –30 or x = 1 e x = –2 or x = 7 f x = –9 or x = 10
g x = –3 or x = 8 h x = 2 or x = 9 i x = –3 or x = 4
j x = 4 or x = 7 k x = –10 or x = 1 l x = –11 or x = 10
3 a x = –9 b x = – 4 or x = 4 c x = 4 or x = 0
3 3
d x = 6 or x = –2 e x=4 f f = 1 or f = 11
3 3
g y = – 3 or y = – 7 h x = 2 or x = –2 i x = –8 or x = 8
2 6
–1 3 –3 1 –2 0 2x
–4 2
–2 0 4 x –4 0 2 x
–5 0 3 x
–3 –3
(0, –3)
(1, –4) (–1, –4)
–8
(–1, –9)
6 a x = – 1 or x = 3 b x = 2 or x = 3 c x = – 7 or x = 3
3 3 2 3
d x = 7 or x = 6 e x = – 10 or x = 3 f x = – 87 or x = 3
2 3 10
7 a x = –√11 – 3 or x = √11 – 3 b x = 2 – 2√2 or x = 2 + 2√2
d x = – 13 – 5 or x =
–5
√ √13
c x = 5 + √5 or x = 5 – √5
2 2
e x = 7 + 29 or x = 7 – 29 f x = – 37 – 3 or x = 37 – 3
√ √ √ √
2 2 2 2
8 a x = 0 or x = 11 b x = – 1 or x = 3 c x = 6 or x = – 22 d x = 4 or x = 4
2 5 3
9 7, 5 10 90 m, 160 m 11 5
12 5, 12 13 4 m