# How do you solve 16x^2-81>=0?

Mar 1, 2017

The solution is x in ]-oo,-9/4]uu[9/4,+oo[

#### Explanation:

We need

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Let's factorise the inequality

$16 {x}^{2} - 81 = \left(4 x + 9\right) \left(4 x - 9\right)$

Let $f \left(x\right) = \left(4 x + 9\right) \left(4 x - 9\right)$

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}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{9}{4}$$\textcolor{w h i t e}{a a a a}$$\frac{9}{4}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$4 x + 9$$\textcolor{w h i t e}{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}$$+$

$\textcolor{w h i t e}{a a a a}$$4 x - 9$$\textcolor{w h i t e}{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}$$+$

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

Therefore,

$f \left(x\right) \ge 0$ when x in ]-oo,-9/4]uu[9/4,+oo[