# How do you solve  sqrt[x-5]-sqrt[x-3]=4 and find any extraneous solutions?

Aug 31, 2017

GIven: $\sqrt{x - 5} - \sqrt{x - 3} = 4$

If you multiply both sides by the conjugate, $\left(\sqrt{x - 5} + \sqrt{x - 3}\right)$, the radicals on the left disappear:

$\left(x - 5\right) - \left(x - 3\right) = 4 \left(\sqrt{x - 5} + \sqrt{x - 3}\right)$

Please observe that the left side simplifies to a negative number:

$- 2 = 4 \left(\sqrt{x - 5} + \sqrt{x - 3}\right)$

Divide both sides by 4 and flip the equation:

$\sqrt{x - 5} + \sqrt{x - 3} = - \frac{1}{2}$

This means that no solution exists.

Think of it this way.

You are starting with

$\sqrt{x - 5}$

(which must be a positive number) and you are subtracting another positive number

$\sqrt{x - 3}$

and you obtain a positive number

4

but, if you add the two numbers, you obtain a negative number, -1/2?

No.