# How do you find the domain of y=sqrt(x^2 - 6 x + 5)?

Jul 30, 2017

The domain is $x \in \left(- \infty , 1\right] \cup \left[5 , + \infty\right)$

#### Explanation:

$y = \sqrt{{x}^{2} - 6 x + 5}$

What is under the square root sign is $\ge 0$

Therefore,

${x}^{2} - 6 x + 5 \ge 0$

We factorise the inequality

$\left(x - 1\right) \left(x - 5\right) \ge 0$

Let $f \left(x\right) = \left(x - 1\right) \left(x - 5\right)$

We 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 a a}$$1$$\textcolor{w h i t e}{a a a a a a a a a}$$5$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\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 a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\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 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}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\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 a a}$$+$

Therefore,

$f \left(x\right) \ge 0$, when $x \in \left(- \infty , 1\right] \cup \left[5 , + \infty\right)$

graph{sqrt(x^2-6x+5) [-10, 10, -5, 5]}