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

1 Answer
Jul 31, 2015

Answer:

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.