# How do you solve sqrtx+9=4?

Aug 11, 2015

$x = \emptyset$

#### Explanation:

As it stands, your equation has no possible solutions among real numbers.

To see why that is the case, isolate the radical term on one side of the equation by adding $- 9$ to both sides

$\sqrt{x} + \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} = 4 - 9$

$\sqrt{x} = - 5$

For real numbers, you have

$\textcolor{b l u e}{\sqrt{x} \ge 0 , \left(\forall\right) x \ge 0}$

Simply put, the square root of a positive real number, which is what must be to begin with, is always positive.

In your case, the square root of $x$ is equal to a negative number, which cannot happen for real numbers.

Therefore, your equation has no solution for any $x \in \mathbb{R}$.