# How do you solve using the completing the square method 2x^2-8x+3=0?

May 6, 2016

$x = 2 \pm \frac{\sqrt{10}}{2}$

#### Explanation:

Complete the square then use the difference of squares identity:

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

with $a = \left(2 x - 4\right)$ and $b = \sqrt{10}$ as follows.

Multiply by $2$ first to make the leading term a perfect square:

$0 = 2 \left(2 {x}^{2} - 8 x + 3\right)$

$= 4 {x}^{2} - 16 x + 6$

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

$= {\left(2 x - 4\right)}^{2} - 16 + 6$

$= {\left(2 x - 4\right)}^{2} - 10$

$= {\left(2 x - 4\right)}^{2} - {\left(\sqrt{10}\right)}^{2}$

$= \left(\left(2 x - 4\right) - \sqrt{10}\right) \left(\left(2 x - 4\right) + \sqrt{10}\right)$

$= \left(2 x - 4 - \sqrt{10}\right) \left(2 x - 4 + \sqrt{10}\right)$

$= \left(2 \left(x - 2 - \frac{\sqrt{10}}{2}\right)\right) \left(2 \left(x - 2 + \frac{\sqrt{10}}{2}\right)\right)$

$= 4 \left(x - 2 - \frac{\sqrt{10}}{2}\right) \left(x - 2 + \frac{\sqrt{10}}{2}\right)$

Hence:

$x = 2 \pm \frac{\sqrt{10}}{2}$