# How do you solve x^2- 1/4x - 3/64=64 by completing the square?

Oct 21, 2017

$x = \frac{1 \pm 10 \sqrt{41}}{8}$.

#### Explanation:

To complete the square, you can apply the formula ${x}^{2} + a x = {\left(x + \frac{1}{2} a\right)}^{2} - {\left(\frac{1}{2} a\right)}^{2}$.

You have to substitute $a = \frac{- \frac{1}{4}}{2} = - \frac{1}{8}$ to solve this problem.

${x}^{2} - \frac{1}{4} x - \frac{3}{64} = 64$
${x}^{2} - \frac{1}{4} x = 64 + \frac{3}{64}$
${\left(x - \frac{1}{8}\right)}^{2} - {\left(- \frac{1}{8}\right)}^{2} = 64 + \frac{3}{64}$
${\left(x - \frac{1}{8}\right)}^{2} = 64 + \frac{3}{64} + {\left(- \frac{1}{8}\right)}^{2}$
${\left(x - \frac{1}{8}\right)}^{2} = 64 + \frac{4}{64} = \frac{1025}{16}$

Notice that $1025 = {5}^{2} \cdot 41$.
The root of the equation is
$x - \frac{1}{8} = \pm \sqrt{\frac{1025}{16}}$
$x - \frac{1}{8} = \pm \frac{5 \sqrt{41}}{4}$
$x = \frac{1 \pm 10 \sqrt{41}}{8}$.