SOLUTION: √(6x+1)= x-1 Solve. I got 6x+1=x^2-2x+1 Subtracted 6x and 1 and got 0=x^2-8x+0 x(x-8) x=0 x=8 What did I do wrong? Thanks

Algebra ->  Radicals -> SOLUTION: √(6x+1)= x-1 Solve. I got 6x+1=x^2-2x+1 Subtracted 6x and 1 and got 0=x^2-8x+0 x(x-8) x=0 x=8 What did I do wrong? Thanks      Log On


   

Question 54534: √(6x+1)= x-1 Solve.
I got 6x+1=x^2-2x+1
Subtracted 6x and 1 and got 0=x^2-8x+0
x(x-8)
x=0
x=8
What did I do wrong?
Thanks
Found 2 solutions by stanbon, Nate:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
√(6x+1)= x-1 Solve.
I got 6x+1=x^2-2x+1
Subtracted 6x and 1 and got 0=x^2-8x+0
x(x-8)
x=0
x=8
Now you have to check these answers to see if one
or both are extraneous.
Checking x=0 in the original problem you get:
sqrt(1)=-1. So x=0 is extraneous.
Checking x=8 in the original problem you get:
sqrt(49)=8-1=7
That is a good result. x=8 is your solution.
You might ask why you got an extraneous answer.
It happened when you squared both sides of the
equation. When you did that you introduced a
root that was not part of the original equation.
Cheers,
Stan H.

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
You need to check your answers with square root values.
sqrt(6 * 0 + 1) = 0 - 1
sqrt(1) = -1 Wrong *even though sqrt(1) = +- 1
sqrt(6 * 8 + 1) = 8 - 1
sqrt(49) = 7 True