# How do you solve sqrt(2x-9)=11 and identify any restrictions?

Mar 24, 2017

$x = 65$, restrictions are that $x \ge 4.5$

#### Explanation:

$\sqrt{2 x - 9} = 11$

square both sides

${\left(\sqrt{2 x - 9}\right)}^{2} = {11}^{2}$
$2 x - 9 = 121$
$2 x = 121 + 9 = 130$
$x = \frac{130}{2} = 65$

$\sqrt{2 x - 9}$ should be greater than or equal to zero, since we can't take the square roots of negative values. Restrictions can be found by designating the radicand (the stuff under the square root) as equal to zero.

$2 x - 9 = 0$
$2 x = 9$
$x = 4.5$

An x value of 4.5 will make the radicand 0, which is the minimum number it can be.

Therefore, the restrictions for x are any numbers under 4.5.