# How do you solve using the completing the square method 2x^2-9x-17=0?

Mar 27, 2016

$x = \frac{9}{4} \pm \frac{\sqrt{217}}{4}$

#### Explanation:

In addition to completing the square, I will use the difference of squares identity:

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

with $a = \left(4 x - 9\right)$ and $b = \sqrt{217}$.

To cut down on the need for fractions, multiply through by ${2}^{3} = 8$ first to get:

$0 = 16 {x}^{2} - 72 x - 136$

$= {\left(4 x - 9\right)}^{2} - 81 - 136$

$= {\left(4 x - 9\right)}^{2} - 217$

$= {\left(4 x - 9\right)}^{2} - {\left(\sqrt{217}\right)}^{2}$

$= \left(\left(4 x - 9\right) - \sqrt{217}\right) \left(\left(4 x - 9\right) + \sqrt{217}\right)$

$= \left(4 x - 9 - \sqrt{217}\right) \left(4 x - 9 + \sqrt{217}\right)$

$= 16 \left(x - \frac{9 + \sqrt{217}}{4}\right) \left(x - \frac{9 - \sqrt{217}}{4}\right)$

So $x = \frac{9}{4} \pm \frac{\sqrt{217}}{4}$