How do you use the sum of two squares formula to solve 4x^2 -8=0?

May 12, 2015

I am not sure that this is what you want, but...
Considering: $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$
Write:
$4 {x}^{2} - 8 = 0$
As:
$\left(2 x + \sqrt{8}\right) \left(2 x - \sqrt{8}\right) = 0$

So:
$\left(2 x + \sqrt{8}\right) = 0$
$\left(2 x - \sqrt{8}\right) = 0$
${x}_{1} = - \frac{\sqrt{8}}{2}$
${x}_{2} = \frac{\sqrt{8}}{2}$