# How do you solve and write the following in interval notation: x^2+4x-12>0?

Jun 1, 2017

$x \in \mathbb{R} : \left(- \infty , - 6\right) \cup \left(+ 2 , + \infty\right)$

#### Explanation:

${x}^{2} + 4 x - 12 > 0$

To solve this inequality and write in interval notation we must find the values of $x \in \mathbb{R}$ for which ${x}^{2} + 4 x - 12 > 0$

Let's first consider the critical points for which ${x}^{2} + 4 x - 12 = 0$

${x}^{2} + 4 x - 12 = \left(x + 6\right) \left(x - 2\right)$
which $= 0$ for $x = - 6 \mathmr{and} x = 2$

We know ${x}^{2} + 4 x - 12$ is a parabola with a minimum value (since the coefficient of ${x}^{2} > 0$

Hence, ${x}^{2} + 4 x - 12 > 0$ for all real $x$ from $- \infty$ up to $- 6$ exclusive and from $+ 2$ exclusive up to $+ \infty$

This can be written as: $x \in \mathbb{R} : \left(- \infty , - 6\right) \cup \left(+ 2 , + \infty\right)$

We can see this from the graph of ${x}^{2} + 4 x - 12$ below:

graph{x^2+4x-12 [-26.85, 24.5, -17.23, 8.45]}