How do you solve by completing the square for y= (2x^2) - 4x - 7?

May 25, 2015

I will assume that you are trying to find the values of $x$ for which the resulting $y = 0$.

$0 = y = 2 {x}^{2} - 4 x - 7 = 2 \left({x}^{2} - 2 x + 1 - 1\right) - 7 = 2 \left({\left(x - 1\right)}^{2} - 1\right) - 7$

$= 2 {\left(x - 1\right)}^{2} - 2 - 7$

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

Add $9$ to both ends to get:

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

Divide both side by $2$ to get:

${\left(x - 1\right)}^{2} = \frac{9}{2} = \frac{18}{4}$

So $x - 1 = \pm \sqrt{\frac{18}{4}} = \pm \frac{3 \sqrt{2}}{2}$

Add $1$ to both sides to get

$x = 1 \pm \frac{3 \sqrt{2}}{2}$