# How to solve (x^2 - 4x +3)(x^2 -3)<0?

Jul 10, 2015

$x \epsilon \left(- \sqrt{3} , - 1\right)$
or
$x \epsilon \left(\sqrt{3} , 3\right)$

#### Explanation:

If $\left({x}^{2} - 4 x + 3\right) \left({x}^{2} - 3\right) < 0$
then one (but not both) of the terms must be less than zero.

Case 1: $\left({x}^{2} - 4 x + 3 < 0\right)$ and $\left({x}^{2} - 3 > 0\right)$
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 4 x + 3 < 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 3\right) \left(x - 1\right) < 0$
$\textcolor{w h i t e}{\text{XXXX}}$again one of the terms must be less than zero (and the other greater)
$\textcolor{w h i t e}{\text{XXXX}}$Since $x - 3 < x - 1$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x - 3 < 0$ and $x - 1 > 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x < 3$ and $x > 1$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x \epsilon \left(1 , 3\right)$

and
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 3 > 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow {x}^{2} > 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow x > \sqrt{3}$

$\Rightarrow \sqrt{3} < x < 3$

Case 2: $\left({x}^{2} - 4 x + 3 > 0\right)$ and $\left({x}^{2} - 3 < 0\right)$
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 4 x + 3 > 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x - 3\right) \left(x - 1\right) > 0$
$\textcolor{w h i t e}{\text{XXXX}}$both terms must be negative or both terms must be positive
$\textcolor{w h i t e}{\text{XXXX}}$and since $x - 3 < x - 1$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$x - 3 > 0$ or $x - 1 < 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow x > 3$ or $x < - 1$

and
$\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} - 3 < 0$
$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow {x}^{2} < 3$
$\textcolor{w h i t e}{\text{XXXX}}$$\Rightarrow - \sqrt{3} < x < \sqrt{3}$

$\Rightarrow - \sqrt{3} < x < - 1$

Combining: Case 1 or Case 2
$- \sqrt{3} < x < - 1$
or
$\sqrt{3} < x < 3$