# How do you solve 4x^2-8x+3<0?

Aug 13, 2016

color(green)(x in (1/2,3/2)

#### Explanation:

$4 {x}^{2} - 8 x + 3 < 0$
can be factored as
$\left(2 x - 1\right) \left(2 x - 3\right) < 0$

{: (2x-1 < 0,color(white)("XXX"),2x-3 < 0), ("if " x < 1/2,,"if " x < 3/2) :}

$\left(2 x - 1\right) \left(2 x - 3\right)$ will be less than zero if one but not both of the terms are less than zero.

That is $\left(2 x - 1\right) \left(2 x - 3\right) < 0$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXX}}$if $\frac{1}{2} < x < \frac{3}{2}$