# How do you solve 16x^2-12x-1=0 by completing the square?

May 23, 2015

${\left(4 x - \frac{3}{2}\right)}^{2} = 16 {x}^{2} - 12 x + \frac{9}{4}$

So

$0 = 16 {x}^{2} - 12 x - 1 = 16 {x}^{2} - 12 x + \frac{9}{4} - \frac{9}{4} - 1$

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

$= {\left(4 x - \frac{3}{2}\right)}^{2} - \frac{13}{4}$

Add $\frac{13}{4}$ to both ends to get:

${\left(4 x - \frac{3}{2}\right)}^{2} = \frac{13}{4} = {\left(\frac{\sqrt{13}}{2}\right)}^{2}$

So $4 x - \frac{3}{2} = \pm \frac{\sqrt{13}}{2}$

Add $\frac{3}{2}$ to both sides to get:

$4 x = \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

Divide both sides by $4$ to get

$x = \frac{3}{8} \pm \frac{\sqrt{13}}{8}$