# How do you solve 2(x - 1)^2 - 8(x - 1)^2= 0?

May 21, 2015

${\left(x - 1\right)}^{2}$ is common to both the terms of the expression on the left hand side

We can write the equation as ${\left(x - 1\right)}^{2} \cdot \left(2 - 8\right) = 0$

$\to - 6 {\left(x - 1\right)}^{2} = 0$

Dividing both sides by $- 6$ we get:

$\to \frac{\cancel{- 6} {\left(x - 1\right)}^{2}}{\cancel{- 6}} = \frac{0}{-} 6$

$\to {\left(x - 1\right)}^{2} = 0$

$\to x - 1 = 0$

-> color(green)(x = 1