Quadratic inequalities

In a nutshell

Sometimes, an inequality will include a quadratic expression. A sketch of the quadratic can be used to help find the required set of solutions.

To solve a quadratic inequality, follow this procedure:

1.	Find the critical values of the quadratic. These are the solutions to the equation $f(x)=0$
2.	Use the critical values to sketch the graph.
3.	Use the graph to determine the solutions by seeing which part is above or below the $x$ -axis.

Find the set of values of $x$ for which $x^2-8x+12 \gt 0$ . Give your answer in set notation.

Set the quadratic equal to $0$ :

$x^2-8x+12=0$ 

Solve using the quadratic formula to find the critical values:

$x = \dfrac{8 \pm \sqrt{(-8)^2 - 4(1)(12)}}{2(1)}$ 

$x=2,\ x=6$ 

Sketch the function:

The question asks for values which make the quadratic greater than $0$ , therefore take the values above the $x$ -axis:

$x\lt2\ or\ x\gt 6$

In set notation:

$\underline{\{x:x\lt2\}\ \cup\ \{x:x\gt6\}}$ 

Find the set of values of $x$ for which $x^2-6x+5 \gt0$ and $2x-1 \gt 3$ . Give your answer in set notation.

Set the quadratic to equal $0$ :

$x^2-6x+5=0$ 

Solve using the quadratic formula:

$x = \dfrac{6 \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)}$ 

$x=1,\ x=5$ 

Sketch the function:

For the quadratic equation to be greater than $0$ :

$x \lt1 \ or\ x\gt5$ 

Solve the linear inequality:

$\begin{aligned}2x-1&\gt3\\2x&\gt4\\x&\gt2\end{aligned}$ 

The two inequalities overlap when $x \gt 5$ :

$\underline{\{x:x\gt5\}}$ 