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

Jul 11, 2016

Convert into a quadratic equation and then select test points.

$4 {x}^{2} - 4 x + 1 = 0$

$4 {x}^{2} - 2 x - 2 x + 1 = 0$

$2 x \left(2 x - 1\right) - 1 \left(2 x - 1\right) = 0$

$\left(2 x - 1\right) \left(2 x - 1\right) = 0$

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

$x = \frac{1}{2}$

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

Selecting test points, we find that all values of x satisfy the inequality.

Hence, the solution is $\left\{x | x \in \mathbb{R} |\right\}$.

Hopefully this helps!