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

May 8, 2015

The answer is $y = 1$

First, let's square both members:
$y - 5 = 4 \sqrt{y} \implies {\left(y - 5\right)}^{2} = 16 y \implies {y}^{2} - 26 y + 25 = 0$

Then let's solve the equation:
${y}_{1} , {y}_{2} = 13 \pm \sqrt{169 - 25} = 13 \pm 12 \implies {y}_{1} = 1 , {y}_{2} = 25$

We know there are no other roots because of the fundamental theorem of algebra.
${y}_{2}$ is an acceptable answer because $25 - 5 = 4 \cdot \sqrt{25}$ does make sense, but ${y}_{1}$ is not acceptable because $- 4 = 4 \sqrt{1}$ does not make any sense because we chose to consider the positive branch of the square root function.

So the extraneous solution is ${y}_{2}$