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Unit F6 Section 3
Inequalities Involving Quadratic Terms

Inequalities involving x2 rather than x can still be solved. For example, the inequality

x2 < 9

will be satisfied by any number between −3 and 3. So the solution is written as

−3 < x < 3

If the inequality had been x2 > 9, then it would be satisfied if x was greater than 3 or if x was less than −3. So the solution will be

x > 3   or   x < −3

The end points of the intervals are defined as = ±3.

Note

For this type of inequality it is very easy to find the end points but care must be taken when deciding whether it is the region between the points or the region outside the points which is required. Testing a point in a region will confirm whether your answer is correct.

For example, for x2 > 9, test x = 2, which gives 4 > 9. This is not true, so the region between the points is the wrong region; the region outside the points is needed.

Worked Examples

1

Show on a number line the solutions to:

(a)

x2 ≥ 16

The solution to x2 ≥ 16 is

x ≤ −4   or   x ≥ 4

which is shown below.
(b)

x2 < 25

The solution of x2 < 25 is

−5 < x < 5

which is shown below.
2

Find the solutions of the inequalities

(a)

x2 + 6 > 15

By subtracting 6 from both sides, the inequality

x2 + 6 > 15

becomes

x2 > 9

Then the solution is

x < −3   or   x > 3
(b)

3x2 − 7 ≤ 41

Begin with the inequality

3x2 − 7 ≤ 41

Adding 7 to both sides gives

3x2 ≤ 48

Dividing both sides by 3 gives

x2 ≤ 16

Then the solution is

−4 ≤ x ≤ 4
3

Solve the inequality

x2 − 3x − 4 > 0

The left-hand side of the inequality can be factorised to give

(x − 4)(x + 1) > 0

The inequality will be equal to 0 when x = 4 and x = −1. This gives the end points of the region as x = 4 and x = −1, as shown below.

Points in each region can now be tested.

x = 2 gives −2 × 3 > 0 or −6 > 0 This is not true.
x = −2 gives −6 × −1 > 0 or 6 > 0 This is true.
x = 5 gives 1 × 6 > 0 or 6 > 0. This is true.

So the inequality is satisfied for values of x greater than 4, or for values of x less than −1. This gives the solution

x < −1   or   x > 4

Exercises

Find the solutions of the following inequalities:

Type your answers separated in each case, as given.

(a)
x2 + 6 ≥ 22 x ,   x
(b)
3x2 − 4 ≥ 8 x ,   x
(c)
5x2 − 20 < 105 x ,   x
(d)
4x2 < 1 x ,   x
(e)
9x2 ≥ 4 x ,   x
(f)
25x2 − 2 ≥ 2 x ,   x
(g)
36x2 + 7 ≤ 11 x ,   x
(h)
2(x2 − 5) < 8 x ,   x
(i)
≥ 53 x ,   x
(j)
10 − x2 > 6 x ,   x
(k)
15 − 2x2 ≤ −3 x ,   x
(l)
10 ≤ 12 − 8x2 x ,   x

Find the solutions of the following inequalities.

Type your answers separated in each case, as given.

(a)
(x − 2)(x + 3) ≥ 0 x ,   x
(b)
(x − 5)(x − 2) ≤ 0 x ,   x
(c)
x(x − 5) > 0 x ,   x
(d)
x2 − 6x ≤ 0 x ,   x
(e)
x2 − 7x + 10 < 0 x ,   x
(f)
x2 + x − 12 > 0 x ,   x
(g)
2x2 − x − 1 ≥ 0 x ,   x
(h)
2x2 + x − 6 ≤ 0 x ,   x

The area, A, of the square shown satisfies the inequality

9 ≤ A ≤ 16

Find an inequality which x satisfies and solve it, giving an inequality for the possible dimensions of the square.

Type <= to get ≤ .

(a)

Write down an expression, in terms of x, for the area, A, of the rectangle below.

A =
(b)

If the area, A, of the rectangle satisfies the inequality

32 ≤ A ≤ 200

write down an inequality for x and solve it.

(c)

What is the maximum (horizontal) length of the rectangle?

(d)

What is the minimum (vertical) width of the rectangle?

Solve the following inequalities for x.

(a)

1 + 3x < 7

(b)

x2 < 1

Investigation

Find the number of points (x, y) where x and y are positive integers which lie on the line 3x + 4y = 29 .