# How do you solve 2x+2>x^2 using a sign chart?

May 23, 2017

The solution is $x \in \left(1 - \sqrt{3} , 1 + \sqrt{3}\right)$

#### Explanation:

Let's rewrite the equation

$2 x + 2 > {x}^{2}$

${x}^{2} - 2 x - 2 < 0$

The roots of the equation

${x}^{2} - 2 x - 2 = 0$

are ${x}_{1}$ and ${x}_{2}$

${x}_{2} = \frac{2 + \sqrt{{2}^{2} - \left(4 \cdot 1 \cdot - 2\right)}}{2} = \frac{2 + \sqrt{12}}{2} = \frac{2 + 2 \sqrt{3}}{2} = 1 + \sqrt{3}$

${x}_{1} = \frac{2 - \sqrt{{2}^{2} - \left(4 \cdot 1 \cdot - 2\right)}}{2} = \frac{2 - \sqrt{12}}{2} = \frac{2 - 2 \sqrt{3}}{2} = 1 - \sqrt{3}$

Let $f \left(x\right) = {x}^{2} - 2 x - 2$

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}$${x}_{1}$$\textcolor{w h i t e}{a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$x - {x}_{2}$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{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}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) < 0$, when $x \in \left({x}_{1} , {x}_{2}\right)$