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Solving Quadratic Inequalities0%

Polynomials & Quadratics · Topic 10 of 10

Solving Quadratic Inequalities

Video lesson · from 9:055 worked examples

One lesson video covers all of Polynomials & Quadratics, so it opens at 9:05 for this topic — not from the beginning.

Theory

Quadratic inequalities can come in one of the following forms:

ax^2 + bx + c \gt 0 \qquad ax^2 + bx + c \geq 0

ax^2 + bx + c \lt 0 \qquad ax^2 + bx + c \leq 0

To solve a quadratic inequality we must determine which part of the graph lies above or below the x-axis. We can determine this via a sketch of the parabola.

y = ax² + bx + c where a > 0

y = ax² + bx + c where a < 0

⚠️ Common Examiner Traps

Sketch the parabola: find the roots, decide whether it opens up or down, then read off where it is above or below the axis. Guessing the direction of the inequality is the main way marks go here.
Two regions need two inequalities: an answer such as " $x \lt 1$ or $x \gt 4$ " cannot be compressed into a single chain.
A negative $x^2$ coefficient flips the shape: the parabola opens downwards, so the regions swap over.
Rearrange to zero first: you cannot read off the regions until one side is zero.
Watch strict versus inclusive: $\geq$ includes the roots themselves, $\gt$ does not.

Worked examples

Example 1

Solve $(x-3)(x+5) \ge 0$ .

First, identify the roots of the corresponding equation $(x-3)(x+5) = 0$ :

$x = 3$ or $x = -5$ .

The parabola $y = (x-3)(x+5) = x^2 + 2x - 15$ has a positive $x^2$ coefficient (it's a "U" shape).

We want where this graph is $\ge 0$ (above or on the x-axis).

From a sketch, the graph is above the x-axis for $x \le -5$ or $x \ge 3$ .

Example 2

Solve $x^2-4x-21 < 0$ .

First, factorise to find the roots:

x^2-4x-21 = (x-7)(x+3) = 0

Roots are $x = 7$ and $x = -3$ .

The parabola is "U" shaped ( $a > 0$ ).

We want $< 0$ (below the x-axis).

From a sketch, the graph is below the x-axis between the roots.

The solution is $-3 < x < 7$ .

Example 3

Solve $4-9x^2 < 0$ .

First, find the roots of $4 - 9x^2 = 0$ :

(2 - 3x)(2 + 3x) = 0

Roots are $x = \frac{2}{3}$ and $x = -\frac{2}{3}$ .

The parabola $y = 4 - 9x^2$ has a negative $x^2$ coefficient (it's an "n" shape or inverted "U").

We want $< 0$ (below the x-axis).

From a sketch, the graph is below the x-axis passing outside the roots.

The solution is $x < -\frac{2}{3}$ or $x > \frac{2}{3}$ .

Alternatively, multiplying by -1 gives $9x^2 - 4 > 0$ . This produces the same result using a positive "U" shaped parabola.

Example 4

Find the values of $q$ for which $x^2+(q-4)x+\frac{1}{2}q=0$ has non-real roots.

For non-real roots, we require $b^2 - 4ac < 0$ .

$a = 1$
$b = q - 4$
$c = \frac{1}{2}q$

(q - 4)^2 - 4(1)(\frac{1}{2}q) < 0

q^2 - 8q + 16 - 2q < 0

q^2 - 10q + 16 < 0

Now we have a quadratic inequality to solve. Find the roots of $q^2 - 10q + 16 = 0$ :

(q - 8)(q - 2) = 0

Roots are $q = 8$ and $q = 2$ .

The parabola $y = q^2 - 10q + 16$ is "U" shaped. We want where it is $< 0$ (below the horizontal axis).

This occurs between the roots, so the solution is $2 < q < 8$ .

Example 5

Where is $y=\frac{1}{3}x^3+2x^2-5x+3$ increasing?

A function is increasing when its derivative is greater than zero ( $\frac{dy}{dx} > 0$ ).

First, find the derivative:

\frac{dy}{dx} = x^2 + 4x - 5

Set up the inequality:

x^2 + 4x - 5 > 0

Find the roots of $x^2 + 4x - 5 = 0$ :

(x + 5)(x - 1) = 0

Roots are $x = -5$ and $x = 1$ .

The parabola is "U" shaped. We want $> 0$ (above the x-axis).

This occurs outside the roots. So the function is increasing when $x < -5$ or $x > 1$ .