MEI Core 1 Inequalities and indices
Section 1: Errors and inequalities
Notes and Examples
These notes contain subsections on
 Inequalities
 Linear inequalities
 Quadratic inequalities
Inequalities
Inequalities are similar to equations, but instead of an equals sign, =, they
involve one of these signs:
<
less than
>
greater than

less than or equal to

greater than or equal to
This means that whereas the solution of an equation is a specific value, or
two or more specific values, the solution of an inequality is a range of values.
Inequalities can be solved in a similar way to equations, but you do have to
be very careful, as in some situations you need to reverse the inequality. This
is shown in these examples.
Linear inequalities
A linear inequality involves only terms in x and constant terms.
Example 1
Solve the inequality
3x  1  x  5
You can treat this just
like a linear equation.
Solution
3x  1  x  5
2 x  1  5
2 x  6
Subtract x from each side
Subtract 1 from each side
x  3
Divide both sides by 2
The next example involves a situation where you have to divide by a negative
number. When you are solving an equation, multiplying or dividing by a
negative number is not a problem. However, things are different with
inequalities.
1 of 6
21/12/15 © MEI
MEI C1 Inequalities and indices 1 Notes and Examples
The statement
is clearly true.
-3 < 2
If you subtract something
from each side, it is still true
If you add something to
each side, it is still true
-3 – 4 < 2 – 4
-7 < -2
-3 + 1 < 2 + 1
-2 < 3


If you multiply or divide each side
by a positive number, it is still true
-3  2 < 2  2
-6 < 4

However, if you multiply each
side by a negative number
then things go wrong!
-3  -2 < 2  -2
6 < -4

When you multiply or divide each side by a negative number, you must
reverse the inequality.
The following example demonstrates this. Two solutions are given: in the first
the inequality is reversed when dividing by a negative number, in the second
this situation is avoided by a different approach.
Example 2
Solve the inequality
Solution (1)
1 x  2x  5
1  3x  5
1 x  2x  5
Subtract 2x from each side
Subtract 1 from each side
3x  6
x2
Solution (2)
1 x  2x  5
1  3x  5
6  3x
2 x
x2
Divide both sides by –3,
reversing the inequality.
Add x to each side
Add 5 to each side
Divide both sides by 3
Finish by writing the inequality
the other way round.
2 of 6
21/12/15 © MEI
MEI C1 Inequalities and indices 1 Notes and Examples
You can check that you have the sign the right way round by picking a
number within the range of the solution, and checking that it satisfies the
original inequality. In the above example, you could try x = 1. In the original
inequality you get 0  -3, which is correct.
When solving inequalities, make sure that you use the correct inequality sign.
If the question involves < or >, the solution will also involve < or > (not
necessarily the same one of these as in the question though!). If the question
involves  or , the solution will also involve one of these. You could waste a
mark, for example by using < when it should be .
For more practice in solving linear inequalities, try section 1 of the
Inequalities skill pack.
There is also an Inequalities puzzle, in which you need to cut out all the
pieces and match linear inequalities with their solutions to form a large
hexagon.
Quadratic inequalities
You can solve a quadratic inequality by factorising the quadratic expression,
just as you do to solve a quadratic equation. This tells you the boundaries of
the solutions. There are then several possible approaches to finding a
solution:



sketching a graph
using a number line
using a table
All three of these approaches are shown in Examples 3 and 4 below. The
number line and the table deal with the same idea (looking at the sign of each
factor), but they are set out differently.
In examinations, many candidates have difficulty giving the solution to a
quadratic inequality, even after factorising correctly. Make sure that you
choose an approach that you are comfortable with and get plenty of practice.
Example 3
Solve the inequality x 2  x  6  0
This shows that the graph of
y  x2  x  6 cuts the x-axis at x = -2
and x = 3. Use this information to
Solution 1 – using a graph
x2  x  6  0
( x  2)( x  3)  0
sketch the graph.
3 of 6
21/12/15 © MEI
MEI C1 Inequalities and indices 1 Notes and Examples
8
y
6
4
2
–6
–4
–2
x
2
–2
4
The solution to the inequality is the
negative part of the graph. This is
the part between –2 and 3.
6
–4
–6
–8
The solution is
2  x  3 .
Solution 2 – using a number line
x2  x  6  0
( x  2)( x  3)  0
Open circles
indicate that this
point is not included
The factor x + 2 is positive if x > -2.
The factor x – 3 is positive if x > 3.
-3
-2
-1
0
1
2
3
4
In this case we need the product of the two factors to be negative. This is the case if
one factor only is positive. From the number line, this is when -2 < x < 3.
Solution 3 – using a table
x2  x  6  0
( x  2)( x  3)  0
These are the critical points
– the points at which the sign
may change.
The left-hand side is zero when x = -2 and when x = 3.
x+2
x–3
(x + 2)(x – 3)
x < -2
–
–
+
-2 < x < 3
+
–
–
The table
looks at the
sign of each
factor for
each possible
set of values.
x>3
+
+
+
From the table, the left-hand side of the inequality is negative for -2 < x < 3.
In the next example, the term in x² is negative. It is usually best to start by
multiplying through by -1 and changing the inequality sign, as it is easier to
work with a positive x² term.
Example 4
Solve the inequality 3  5 x  2 x 2  0
4 of 6
21/12/15 © MEI
MEI C1 Inequalities and indices 1 Notes and Examples
This shows that the graph of
Solution
3  5x  2 x2  0
y  2 x2  5x  3 cuts the x-axis at
x = -3 and x = ½. You can now
2 x2  5x  3  0
sketch the graph.
( x  3)(2 x  1)  0
8
y
6
4
2
The solution to the inequality is
the positive part of the graph.
This is in fact two separate parts.
x
–4 –3 –2 –1
–2
1
2
3
4
–4
–6
–8
The solution is x  -3 or x  ½.
Solution 2 – using a number line
3  5x  2 x2  0
2 x2  5x  3  0
( x  3)(2 x  1)  0
The factor x + 3 is positive if x  3 .
The factor 2x – 1 is positive if x  12 .
-3
-2
0
-1
1
2
3
4
In this case we need the product of the two factors to be positive. This is the case if
either both factors are positive or both factors are negative. From the number line, this
is when x  3 or x  12 .
Solution 3 – using a table
3  5x  2 x2  0
2 x2  5x  3  0
( x  3)(2 x  1)  0
The left-hand side is zero when x = -3 and when x = 12 .
x < -3
x+3
2x – 1
(x + 3)(2x – 1)
–
–
+
-3 < x <
+
–
–
1
2
x>
+
+
+
1
2
The table
looks at the
sign of each
factor for
each possible
set of values.
From the table, the left-hand side of the inequality is positive when x  3 or x  12 .
When there are two parts to the solution, as in Example 4, it should be written
as two separate inequalities. You cannot write 3  x  12 or 12  x  3 - these
5 of 6
21/12/15 © MEI
MEI C1 Inequalities and indices 1 Notes and Examples
suggest that the solution is values of x which are less that -3 AND greater
than 12 - there are no such values of x!
Also, as for linear inequalities, make sure that you use the appropriate
inequality sign, according to the question.
For more practice in solving quadratic inequalities, try section 2 of the
Inequalities skill pack.
You can also look at the Solving inequalities video, which uses a range of
approaches.
There is also a Quadratic inequalities puzzle, in which you need to cut out
all the pieces and match quadratic inequalities with their solutions to form a
large triangle.
6 of 6
21/12/15 © MEI