# How do you solve 3+sqrt[x]=sqrt[x-3] and find any extraneous solutions?

Feb 26, 2017

No feasible solution.

#### Explanation:

The feasible solutions are for $x \ge 0$ and $x - 3 \ge 0$ or resuming $x \ge 3$

Now

$3 = \sqrt{x - 3} - \sqrt{x} = \sqrt{x} \left(\sqrt{1 - \frac{3}{x}} - 1\right)$ but with $x \ge 3$

$\sqrt{1 - \frac{3}{x}} - 1 < 0$ so no solution for this equation