# How do you solve x^4-5x^2<=-4?

Jan 31, 2017

The answer is $x \in \left[- 2 , - 1\right] \cup \left[1 , 2\right]$

#### Explanation:

We need

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

Let rewrite the inequality

${x}^{4} - 5 {x}^{2} + 4 \le 0$

Let's factorise

$\left({x}^{2} - 1\right) \left({x}^{2} - 4\right) \le 0$

$\left(x + 1\right) \left(x - 1\right) \left(x + 2\right) \left(x - 2\right) \le 0$

Let $f \left(x\right) = \left(x + 1\right) \left(x - 1\right) \left(x + 2\right) \left(x - 2\right)$

Let's build a 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}$$- 2$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$+ 1$$\textcolor{w h i t e}{a a a a}$$+ 2$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

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

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

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

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[- 2 , - 1\right] \cup \left[1 , 2\right]$

Jan 31, 2017

By graphical method, $x \in \left[- 1 , - 2\right] \mathmr{and} x \in \left[1. 2\right]$

#### Explanation:

graph{y-x^4+5x^2-4>=0 [-3.03, 3.048, -1.528, 1.51]}

Let y-(x^4-5x^2+4)>=0.

The section of the graph by the x-axis ( y = 0 ) depicts the solution

-$\left({x}^{2} - 5 x + 4\right) \ge 0$, giving

${x}^{2} - 5 x + 4 \le 0$