# How do you solve 16p^2-81p^6<=0 using a sign chart?

Jan 27, 2017

The answer is p in ]-oo, -2/3] uu [2/3, +oo [

#### Explanation:

We need

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

Let's factorise the inequality

$16 {p}^{2} - 81 {p}^{6} = {p}^{2} \left(16 - 81 {p}^{4}\right)$

$= {p}^{2} \left(4 - 9 {p}^{2}\right) \left(4 + 9 {p}^{2}\right)$

$= {p}^{2} \left(2 + 3 p\right) \left(2 - 3 p\right) \left(4 + 9 {p}^{2}\right)$

Let $f \left(p\right) = {p}^{2} \left(2 + 3 p\right) \left(2 - 3 p\right) \left(4 + 9 {p}^{2}\right)$

${p}^{2} \ge 0 , \forall p \in \mathbb{R}$

$4 + 9 {p}^{2} > 0 , \forall p \in \mathbb{R}$

Now we can build the sign chart

$\textcolor{w h i t e}{a a a a}$$p$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{2}{3}$$\textcolor{w h i t e}{a a a a a a}$$\frac{2}{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$2 + 3 p$$\textcolor{w h i t e}{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}$$+$

$\textcolor{w h i t e}{a a a a}$$2 - 3 p$$\textcolor{w h i t e}{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}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(p\right)$$\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,

$f \left(p\right) \le 0$ when p in ]-oo, -2/3] uu [2/3, +oo [