# How do you solve (x-3)^2<=1 using a sign chart?

Feb 18, 2017

The answer is $x \in \left[2 , 4\right]$

#### Explanation:

We need

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

Let's rewrite the inequality

${\left(x - 3\right)}^{2} \le 1$

${\left(x - 3\right)}^{2} - 1 \le 0$

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

$\left(x - 4\right) \left(x - 2\right) \le 0$

Let $f \left(x\right) = \left(x - 4\right) \left(x - 2\right)$

Now, 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 a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$4$$\textcolor{w h i t e}{a a 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}$$x - 4$$\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}$$f \left(x\right)$$\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}$$+$

Therefore,

$f \left(x\right) \le 0$ when $x \in \left[2 , 4\right]$