# How do you solve 4x^2-4x+1>=0?

Dec 22, 2016

This inequality holds for all $x \in \mathbb{R}$ since:

$4 {x}^{2} - 4 x + 1 = {\left(2 x - 1\right)}^{2} \ge 0$

#### Explanation:

You might recognise this perfect square trinomial, since:

$441 = {21}^{2}$

So:

$4 {x}^{2} - 4 x + 1 = {\left(2 x - 1\right)}^{2}$