# How do you solve 4x^2 - x = 0 by completing the square?

Mar 27, 2016

$x = 0$ or $x = \frac{1}{4}$

#### Explanation:

As ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

Trying to convert $4 {x}^{2} - x$ in form ${a}^{2} + 2 a b$ we can have

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

This means we should add ${\left(- \frac{1}{4}\right)}^{2}$ on both sides for completing the square.

Hence, $4 {x}^{2} - x = 0$ can be written as

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

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

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

$\left(2 x - \frac{1}{2}\right) \cdot \left(2 x\right) = 0$ or

$2 \left(x - \frac{1}{4}\right) \cdot 2 x = 0$ and dividing by $4$ and rearranging we have

$x \left(x - \frac{1}{4}\right) = 0$ i.e. $x = 0$ or $x = \frac{1}{4}$