Question

How do I find the extraneous solution of `y-5=4sqrt(y)`?

Answer 1

The answer is `y=1`

First, let's square both members:
`y-5=4sqrt(y) => (y-5)^2=16y => y^2-26y+25=0`

Then let's solve the equation:
`y_1,y_2=13+-sqrt(169-25)=13+-12 => y_1=1, y_2=25`

We know there are no other roots because of the fundamental theorem of algebra.
`y_2` is an acceptable answer because `25-5=4*sqrt(25)` does make sense, but `y_1` is not acceptable because `-4=4sqrt(1)` does not make any sense because we chose to consider the positive branch of the square root function.

So the extraneous solution is `y_2`