How do you solve #sqrt(8x+1) = 5# and find any extraneous solutions?

1 Answer
Jun 5, 2016

Answer:

Square both sides, solve the linear equation, and plug in the solution to get #x=3#.

Explanation:

The first thing we want to do when solving for equations with square roots is get rid of the square root sign. We do this by squaring both sides of the equation:
#sqrt(8x+1)^2=5^2#
#->8x+1=25#

Now, all we have to do is solve this two-step equation. First, we subtract one from both sides to get:
#8x=24#

We finish off by dividing by #8#, to get the result:
#x=3#

Finally, we check to make sure this solution is not extraneous. That basically means making sure the solution actually works when we plug it back in:
#sqrt(8x+1)=5#
#sqrt(8(3)+1)=5#
#sqrt(24+1)=5#
#sqrt(25)=5#

Because #sqrt(25)=5#, the solution #x=3# is good.