# What is the solution set for 2 -sqrt(x+3) =6?

Aug 19, 2015

$x = \emptyset$

#### Explanation:

As it is written, this equation has no solution among real numbers, and here's why that is the case.

For real numbers, you can only take the square root of a positive number, and the result will always be another positive number.

$\textcolor{b l u e}{\sqrt{x} \ge 0 \text{, } \left(\forall\right) x \in \left[0 , + \infty\right)}$

Rearrange the equation to isolate the square root on one side

$- \sqrt{x + 3} = 4$

$\sqrt{x + 3} = - 4$

Since the square root must always be a positive number, your equation does not have a valid solution among real numbers.

$\sqrt{x + 3} \textcolor{red}{\ne} - 4$