# How do you solve (x-9)^2=12?

Mar 30, 2018

$x = - 2 \sqrt{3} + 9 , x = - 2 \sqrt{3} + 9$

#### Explanation:

You could expand the left side, move everything to the left, and apply the quadratic formula; however, taking the root of both sides is far faster:

$\sqrt{{\left(x - 9\right)}^{2}} = \pm \sqrt{12}$

Recall that $\sqrt{{a}^{2}} = a$, and that $\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \sqrt{3} = 2 \sqrt{3}$ :

$x - 9 = \pm 2 \sqrt{3}$

$x = - 2 \sqrt{3} + 9 , x = - 2 \sqrt{3} + 9$

You would have the same result even with applying the quadratic formula -- this method is just faster and cleaner.

Mar 30, 2018

$x = 2 \sqrt{3} + 9 ,$ $- 2 \sqrt{3} + 9$

#### Explanation:

Solve:

${\left(x - 9\right)}^{2} = 12$

Take the square root of both sides.

$x - 9 = \pm \sqrt{12}$

Prime factorize $12$.

$x - 9 = \pm \sqrt{2 \times 2 \times 3} =$

$x - 9 = \pm \sqrt{{2}^{2} \times 3}$

Apply rule: $\sqrt{{a}^{2}} = a$

$x - 9 = \pm 2 \sqrt{3}$

Add $9$ to both sides.

$x = \pm 2 \sqrt{3} + 9$

Solutions for $x$.

$x = 2 \sqrt{3} + 9 ,$ $- 2 \sqrt{3} + 9$