# How do you solve and check for extraneous solutions in sqrt(5x) = -5?

Jul 31, 2015

With the usual meaning for $\sqrt{5 x}$ as the principal square root, this equation has no solutions.

#### Explanation:

The usual definition given is that
$y = \sqrt{x}$ if and only if (both ${y}^{2} = x$ and $y \ge 0$)

This equation asks us, to find a number whose non-negative square root is $- 5$. There is no such number.

(This question is similar to asking us to solve $\left\mid 2 x \right\mid = - 10$. It is built into the definition of the expression on the left that it is a non-negative number.)

But
If the point of the exercise is to investigate extraneous solutions, then:

We might try squaring both sides (to get rid of the square root on the left)

$\sqrt{5 x} = - 5$

${\left(\sqrt{5 x}\right)}^{2} = {\left(- 5\right)}^{2}$

$5 x = 25$

$x = 5$

$5$ is not a solution to the original equation, but is is a solution to the equation: $5 x = 25$.

This is one way that extraneous solutions can be introduced in the process of solving an equation.