# How do you solve x^2-1>=4x using a sign chart?

Dec 14, 2016

The answer is x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[

#### Explanation:

Let's rewrite the equation

${x}^{2} - 4 x - 1 \ge 0$

Let $f \left(x\right) = {x}^{2} - 4 x - 1$

We need the values of $x$, when $f \left(x\right) = 0$

that is, ${x}^{2} - 4 x - 1 = 0$

Let's calculate $\Delta = 16 - 4 \cdot 1 \cdot - 1 = 20$

$\Delta > 0$. there are 2 real solutions

${x}_{1} = \frac{4 + \sqrt{20}}{2} = \frac{4 + 2 \sqrt{5}}{2} = 2 + \sqrt{5}$

${x}_{2} = \frac{4 - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = 2 - \sqrt{5}$

Now, we can do our 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 a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$\left(2 - \sqrt{5}\right)$$\textcolor{w h i t e}{a a a a}$$\left(2 + \sqrt{5}\right)$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - 2 + \sqrt{5}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a a}$$+$$\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}$$x - 2 - \sqrt{5}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\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}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a}$$+$

Therefore,

$f \left(x \ge 0\right)$ when x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[

graph{x^2-4x-1 [-11.25, 11.25, -5.625, 5.625]}