# How do you solve 7-x^2<=0?

Feb 15, 2017

The answer is x in ]-oo,-sqrt7]uu[sqrt7,+oo[

#### Explanation:

Let's factorise the inequality

$7 - {x}^{2} \le 0$

$\left(\sqrt{7} + x\right) \left(\sqrt{7} - x\right) \le 0$

Let $f \left(x\right) = \left(\sqrt{7} + x\right) \left(\sqrt{7} - x\right)$

Now, we build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \sqrt{7}$$\textcolor{w h i t e}{a a a a}$$\sqrt{7}$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$\sqrt{7} + x$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$\sqrt{7} - x$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a 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 a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$

Therefore,

$f \left(x\right) \le 0$ when x in ]-oo,-sqrt7]uu[sqrt7,+oo[