# How do I solve 4x^2 - 6x + 1 = 0 by completing the square?

Sep 13, 2014

$4 {x}^{2} - 6 x = - 1$

Factor out a $4$

$4 \left({x}^{2} - \frac{6}{4} x\right) = - 1$

Simplify fraction

$4 \left({x}^{2} - \frac{3}{2} x\right) = - 1$

Take half of x term and square the result

${\left(\frac{- \frac{3}{2}}{2}\right)}^{2} = {\left(\left(- \frac{3}{2}\right) \cdot \left(\frac{1}{2}\right)\right)}^{2} = {\left(- \frac{3}{4}\right)}^{2} = \frac{9}{16}$

Add $4 \cdot \frac{9}{16}$ or $\frac{9}{4}$ to the right side because we factored out a $4$.
Add $\frac{9}{16}$ to the left side.

$4 \left({x}^{2} - \frac{3}{2} x + \frac{9}{16}\right) = - 1 + \frac{9}{4}$

$4 {\left(x - \frac{3}{4}\right)}^{2} = - 1 + \frac{9}{4}$

$4 {\left(x - \frac{3}{4}\right)}^{2} = - \frac{4}{4} + \frac{9}{4}$, common denominators $- 1 = - \frac{4}{4}$

$4 {\left(x - \frac{3}{4}\right)}^{2} = \frac{5}{4}$

${\left(x - \frac{3}{4}\right)}^{2} = \frac{\frac{5}{4}}{4}$

${\left(x - \frac{3}{4}\right)}^{2} = \frac{5}{16}$

$\sqrt{{\left(x - \frac{3}{4}\right)}^{2}} = \pm \sqrt{\frac{5}{16}}$

$x - \frac{3}{4} = \pm \sqrt{\frac{5}{16}}$

$x - \frac{3}{4} = \pm \frac{\sqrt{5}}{4}$

$x = \pm \frac{\sqrt{5}}{4} + \frac{3}{4}$

$x = \frac{\pm \sqrt{5} + 3}{4}$