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Quadratic Equations
mc-TY-quadeqns-1
This unit is about the solution of quadratic equations
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We
will look at four methods: solution by factorisation, solution by completing the square, solution
using a formula, and solution using graphs
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature
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Introduction
2
2
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Solving quadratic equations by completing the square
5
4
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Solving quadratic equations by using graphs
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Introduction
This unit is about how to solve quadratic equations
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g
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It may also contain terms
involving x, e
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5x or −7x, or 0
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It can also have constant terms - these are just numbers:
6, −7, 21
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It cannot have terms like
1
x
in it
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b and c can be any numbers including zero
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Key Point
A quadratic equation takes the form
ax2 + bx + c = 0
where a, b and c are numbers
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In this unit we will look at how to solve quadratic equations using four methods:
• solution by factorisation
• solution by completing the square
• solution using a formula
• solution using graphs
Factorisation and use of the formula are particularly important
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Solving quadratic equations by factorisation
In this section we will assume that you already know how to factorise a quadratic expression
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Example
Suppose we wish to solve 3x2 = 27
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By observation there is a common factor of 3 in both terms
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The contents of the brackets are
adjusted accordingly:
3x2 − 27 = 3(x2 − 9) = 0
Notice here the difference of two squares which can be factorised as
3(x2 − 9) = 3(x − 3)(x + 3) = 0
If two quantities are multiplied together and the result is zero then either or both of the quantities
must be zero
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x = −3
Example
Suppose we wish to solve 5x2 + 3x = 0
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There is a common factor of x in both
terms
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These are the two solutions
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A common error that students make is to cancel the
common factor of x in the original equation:
3
5x2 + 3
x=0
so that
5x + 3 = 0
giving x = −
5
But if we do this we lose the solution x = 0
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Example
Suppose we wish to solve x2 − 5x + 6 = 0
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Now
−3 × −2 = 6
− 3 + −2 = −5
so the two numbers are −3 and −2
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To factorise this we seek two numbers which multiply to give −4 (the coefficient of x2 multiplied
by the constant term) and which add together to give 3
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We use these two numbers to write 3x as 4x − x and then
factorise as follows:
2x2 + 3x − 2
2x2 + 4x − x − 2
2x(x + 2) − (x + 2)
(x + 2)(2x − 1)
=
=
=
=
0
0
0
0
from which
x+2=0
or
2x − 1 = 0
so that
x = −2
or
x=
1
2
These are the two solutions
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First of all we write this in the standard form:
4x2 − 12x + 9 = 0
We should look to see if there is a common factor - but there is not
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Now, by inspection,
−6 × −6 = 36
− 6 + −6 = −12
so the two numbers are −6 and −6
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So we can have
quadratic equations for which the solution is repeated
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Example
Suppose we wish to solve x2 − 3x − 2 = 0
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Never
mind how hard you try you will not find any such two numbers
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We need another approach
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Exercise 1
Use factorisation to solve the following quadratic equations
a) x2 − 3x + 2 = 0
b) 5x2 = 20
c) x2 − 5 = 4x
e)
x2 + 19x + 60 = 0
f) 2x2 + x − 6 = 0
d)
g) 2x2 − x − 6 = 0 h)
2x2 = 10x
4x2 = 11x − 6
3
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In order to complete the square we look at the first two terms, and try to write them in the form
(
)2
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So the quadratic equation can be written
2 2
3
3
2
x − 3x − 2 = x −
− −
−2=0
2
2
Simplifying
2
3
9
x−
− −2 = 0
2
4
2
17
3
= 0
−
x−
2
4
2
17
3
=
x−
2
4
√
√
17
17
3
x−
=
or
−
2
2 √
2
√
17
3
17
3
+
or x = −
x =
2
2
2
2
We can write these solutions as
√
√
3 + 17
3 − 17
x=
or
2
2
Again we have two answers
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Approximate values can be obtained using
a calculator
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Exercise 2
a) Show that x2 + 2x = (x + 1)2 − 1
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b) Show that x2 − 6x = (x − 3)2 − 9
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c) Use completing the square to solve x2 − 5x + 1 = 0
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4
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√
−b ± b2 − 4ac
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Key Point
Formula for solving ax2 + bx + c = 0:
x=
−b ±
√
b2 − 4ac
2a
We will illustrate the use of this formula in the following example
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Comparing this with the general form ax2 + bx + c = 0 we see that a = 1, b = −3 and c = −2
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√
−b ± b2 − 4ac
x =
2a p
−(−3) ± (−3)2 − 4 × 1 × (−2)
=
2×1
√
3± 9+8
=
√2
3 ± 17
=
2
These solutions are exact
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Example
Suppose we wish to solve 3x2 = 5x − 1
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We see that a = 3, b = −5 and c = 1
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√
−b ± b2 − 4ac
x =
2a p
−(−5) ± (−5)2 − 4 × 3 × 1
=
2×3
√
5 ± 25 − 12
=
√6
5 ± 13
=
6
Again there are two exact solutions
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Exercise 3
Use the quadratic formula to solve the following quadratic equations
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Solving quadratic equations by using graphs
In this section we will see how graphs can be used to solve quadratic equations
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If the coefficient of x2 is negative the graph will take the
form shown in Figure 1(b)
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Graphs of y = ax2 + bx + c have these general shapes
We will now add x and y axes
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(a)
y
(b)
y
(c)
y
x
x
x
Figure 2
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The horizontal line, the x axis, corresponds to points on the graph where y = 0
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In Figure 2, the graph in (a) never cuts or touches the horizontal axis and so this corresponds
to a quadratic equation ax2 + bx + c = 0 having no real roots
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The graph in (c) cuts the horizontal axis twice, corresponding to the case in which the quadratic
equation has two different roots
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This case is shown in Figure 3
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Graphs of y = ax2 + bx + c when a is negative
Referring to Figure 3: in case (a) there are no real roots
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Case (c) corresponds to there being two real roots
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We consider y = x2 − 3x − 2 and produce a table of values so that we can plot a graph
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From the
graph we observe that solutions of the equation x2 − 3x − 2 = 0 lie between −1 and 0, and
between 3 and 4
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Graph of y = x2 − 3x − 2
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For example to solve x2 − 3x − 2 = 6 we
can simply locate points where the graph crosses the line y = 6 as shown in Figure 5
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Using the graph of y = x2 − 3x − 2 to solve x2 − 3x − 2 = 6
Example
We can use the same graph to solve x2 −3x−5 = 0 by rewriting the equation as x2 −3x−2−3 = 0
and then as x2 − 3x − 2 = 3
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Exercise 4
By plotting the graph y = x2 − 5x + 2, solve the equation x2 − 5x + 2 = 0, giving your answers
to 1 decimal place
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Answers
Exercise 1
a) 1, 2 b) 2, −2 c) 5, −1
d) 0, 5 e) −4, −15 f) −2,
Exercise 2
a) 1, −3 b) 3 ±
√
14 c)
Exercise 3
5+
5±
√
33
b) 2, 43 c)
2
√
g) −2 ± 14 h) 45 repeated
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6, 0
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4, −0
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2, −0
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√
3
2
g) 2, − 32
h) 2,
3
4
12
e)
3±
√
4
17
f) No real roots
d) 6
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3
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