Inequalities » Quadratic inequalities

When solving a quadratic inequality, you solve the corresponding equation first, to calculate the point(s) of intersection. Just like with linear inequalities.
Then you decide on the solution using a sketch or the number line.

You have to know the meaning of the different signs being used.
If necessary, check domain and range for theory about intervals.

Example 1
a2 + 30 < –11a + 2
a2 + 30 = –11a + 2
a2 + 11a + 28 = 0
(a + 4)(a + 7) = 0
a = –4 or a = –7

Continue with sketch Continue with number line
There are two points of intersection and the graph is a upward-opening parabola.

That gives you this sketch:
upward-opening parabola with intersecting falling line

Look to the original inequality:
a2 + 30 < –11a + 2
Parabola < line
That is in between the points of intersection.

Solution: –7 < a < –4
Calculate whether or not the inequality is
correct for numbers outside and in
between the solutions.

Outside the points of intersection:
Take for example a = –8.
(–8)2 + 30 < –11 × –8 + 2
94 < 90
Incorrect!

In between the two solutions:
Take for example a = –6.
(–6)2 + 30 < –11 × –6 + 2
66 < 68
Correct!

You may sketch a number line to help you.
Number line with above -8 'No', above -7 '=', above -6 'Yes' and above -4 '='.

Because 'in between' is correct,
the solution is:
–7 < a < –4


Example 2
–2x2 – 4x – 69 ≤ –28x – 5 
–2x2 – 4x – 69 = –28x – 5 
–2x2 + 24x – 64 = 0 
x2 – 12x + 32 = 0 
(x – 4)(x – 8) = 0
x = 4 of x = 8

Continue with sketch Continue with number line
There are two points of intersection and the graph is a downward-opening parabola.

That gives you this sketch:
downward-opening parabola with rising line

Look to the original inequality:
–2x2 – 4x – 69 ≤ –28x – 5 
parabool ≤ lijn
That is outside the points of intersection!

Solution: x ≤ 4 or x ≥ 8
Calculate whether or not the inequality is
correct for numbers outside and in
between the solutions.

Outside the points of intersection:
Take for example x = 2.
–2 × 22 – 4 × 2 – 69 ≤ –28 × 2 – 5
–85 ≤ –61
Correct!

In between the two solutions:
Take for example x = 5.
–2 × 52 – 4 × 5 – 69  ≤ –28 × 5 – 5
–139 ≤ –145
Incorrect!

You may sketch a number line to help you.
Number line with above 2 'Yes', above 4 '=', above 5 'No' and above 8 '='.

Because 'outside' is correct,
the solution is:
x ≤ 4 or x ≥ 8


Watch out with:


No intersections (see sketch)


Parabola < line or parabola ≤ line
Solution: for every value of x

Parabola > line or parabola ≥ line
Solution: for no value of x



downward-opening parabola with a line above

One point of intersection/tangent* (see sketch)


*A tangent in this context is the point
where the graphs touch.

Parabola < line
Solution: x ≠ 1
(Every value of x except x = 1)

Parabola > line
Solution: for no value of x

Parabola ≤ line
Solution: for every value of x

Parabola ≥ line
Solution: x = 1
downward-opeing parabola with a touching line
tangent at x = 1