Question

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

Answer 1

With the usual meaning for `sqrt(5x)` as the principal square root, this equation has no solutions.

Explanation:

The usual definition given is that
`y = sqrtx` if and only if (both `y^2 = x` and `y >= 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 `abs(2x) = -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(5x) = -5`

`(sqrt(5x))^2 = (-5)^2`

`5x = 25`

`x = 5`

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

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