# How do you solve the inequality x^2+2x-3>=0 and write your answer in interval notation?

Aug 14, 2017

The solution is $x \in \left(- \infty , - 3\right] \cup \left[1 , + \infty\right)$

#### Explanation:

Let's factorise the inequality

${x}^{2} + 2 x - 3 \ge 0$

$\left(x + 3\right) \left(x - 1\right) \ge 0$

Let $f \left(x\right) = \left(x + 3\right) \left(x - 1\right)$

We build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 3$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - 3\right] \cup \left[1 , + \infty\right)$

graph{x^2+2x-3 [-10, 10, -5, 5]}