How do you solve x^2-2x-4>0 using a sign chart?

Jul 14, 2017

The solution is $x \in \left(- \infty , \left(1 - \sqrt{5}\right)\right) \cup \left(\left(1 + \sqrt{5}\right) , + \infty\right)$

Explanation:

We must first find the roots of the quadratic equations

${x}^{2} - 2 x - 4 = 0$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left(- 2\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(- 4\right) = 20$

As $\Delta > 0$, there are $2$ real roots

The roots are

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{2 - \sqrt{20}}{2} = 1 - \sqrt{5}$

${x}_{2} = \frac{2 + \sqrt{20}}{2} = 1 + \sqrt{5}$

Let $f \left(x\right) = {x}^{2} - 2 x - 4$

We can 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}$$\left(x - {x}_{1}\right)$$\textcolor{w h i t e}{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}$$\left(x - {x}_{2}\right)$$\textcolor{w h i t e}{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 $x \in \left(- \infty , \left(1 - \sqrt{5}\right)\right) \cup \left(\left(1 + \sqrt{5}\right) , + \infty\right)$