# How do you solve (3x-1)^2 = 12 using the square root property?

Aug 17, 2015

$3 x - 1 = \sqrt{12} \mathmr{and} 3 x - 1 = - \sqrt{12}$
because both $\sqrt{12} \mathmr{and} - \sqrt{12}$ give $12$ when squared.
Since $\sqrt{12} = 2 \sqrt{3} \to$
$3 x = 1 \pm 2 \sqrt{3} \to {x}_{1 , 2} = \frac{1 \pm 2 \sqrt{3}}{3} = \frac{1}{3} \pm \frac{2}{3} \sqrt{3}$