# How do you solve x^4-10x^2+9>=0 using a sign chart?

Oct 27, 2016

Solution is $- \infty \le x \le - 3$ or $- 1 \le x \le 1$, or $3 \le x \le \infty$ and in interval form it can be written as $\left[- \infty , - 3\right] \cup \left[- 1 , 1\right] \cup \left[3 , \infty\right]$

#### Explanation:

Let us first factorize ${x}^{4} - 10 {x}^{2} + 9$.

${x}^{4} - 10 {x}^{2} + 9 = {x}^{4} - 9 {x}^{2} - {x}^{2} + 9$

= x^2(x^2-9)-1(x^2-9)= (x^2-9)(x^2-1))

= $\left(x + 3\right) \left(x - 3\right) \left(x + 1\right) \left(x - 1\right)$

Hence we have to solve the inequality

$\left(x + 3\right) \left(x - 3\right) \left(x + 1\right) \left(x - 1\right) \ge 0$

From this we know that the product $\left(x + 3\right) \left(x - 3\right) \left(x + 1\right) \left(x - 1\right)$ has to be zero or positive. It is apparent that sign of binomials $\left(x + 3\right)$, $\left(x + 1\right)$, $\left(x - 1\right)$ and $\left(x - 3\right)$ will change around the values $- 3$. $- 1$, $1$ and $3$ respectively. In sign chart we divide the real number line using these values, i.e. below $- 3$, between $- 3$ and $- 1$, between $- 1$ and $1$, between $1$ and $3$ and above $3$ and see how the sign of $\left(x + 3\right) \left(x - 3\right) \left(x + 1\right) \left(x - 1\right)$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X} - 3 \textcolor{w h i t e}{X X X X} - 1 \textcolor{w h i t e}{X X X X} 1 \textcolor{w h i t e}{X X X X} 3$

$\left(x + 3\right) \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X} + i v e$

$\left(x + 1\right) \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X} + i v e$

$\left(x - 1\right) \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X} + i v e \textcolor{w h i t e}{X X} + i v e$

$\left(x - 3\right) \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X} + i v e$

${x}^{4} - 10 {x}^{2} + 9$
$\textcolor{w h i t e}{X X X X X X} + i v e \textcolor{w h i t e}{X x} - i v e \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X} - i v e \textcolor{w h i t e}{X X} + i v e$

It is observed that ${x}^{4} - 10 {x}^{2} + 9 \ge 0$

when either $- \infty \le x \le - 3$ or $- 1 \le x \le 1$, or $3 \le x \le \infty$, which is the solution for the inequality.

In interval notation, this can be written as $\left[- \infty , - 3\right] \cup \left[- 1 , 1\right] \cup \left[3 , \infty\right]$

Oct 27, 2016

The values of x are
$- \infty \le x \le - 3$ and $- 1 \le x \le 1$

#### Explanation:

Let's start by factorising the expression

${x}^{4} - 10 {x}^{2} + 9 = \left({x}^{2} - 1\right) \left({x}^{2} - 9\right) = \left(x + 1\right) \left(x - 1\right) \left(x + 3\right) \left(x - 3\right)$

So we can make the sign chart
$x$$\textcolor{w h i t e}{a a a a a}$-oo$\textcolor{w h i t e}{a a a a}$-3$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$1$$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$x + 3$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$0$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a}$$+$
$x + 1$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a}$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a}$$0$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a}$$+$
You can continue the sign chart
and you wiil find that
${x}^{4} - 10 {x}^{2} + 9 \ge 0$ for $- \infty \le x \le - 3$ and $- 1 \le x \le 1$