# How do you solve 9x^2-6x+1<0?

Oct 18, 2016

$9 {x}^{2} - 6 x + 1$ is $\ge 0$
$9 {x}^{2} - 6 x + 1 = \left(3 x - 1\right) \left(3 x - 1\right) = {\left(3 x - 1\right)}^{2}$
$\left({\left(3 x - 1\right)}^{2}\right) \ge 0$