# How do you solve z^2-6z+7<2?

Jan 18, 2017

The answer is z in ] 1,5 [

#### Explanation:

Let's rearrange the inequation

${z}^{2} - 6 z + 7 < 2$

${z}^{2} - 6 z + 5 < 0$

Let's factorise

$\left(z - 5\right) \left(z - 1\right) < 0$

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

Now. we can do the sign chart

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

$\textcolor{w h i t e}{a a a a}$$z - 1$$\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}$$z - 5$$\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}$$f \left(z\right)$$\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,

$f \left(z\right) < 0$, when z in ] 1,5 [