SOLUTION: Could you help me with this? {{{8sqrt(x-1)+1}}} = {{{x}}} Thank-you

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Question 93556: Could you help me with this?
8sqrt%28x-1%29%2B1 = x
Thank-you
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

Could you help me with this?

8sqrt%28x-1%29%2B1 = x

Thank-you

8sqrt%28x-1%29%2B1 = x

Add -1 to both sides to get the radical 
term alone on the left:

  8sqrt%28x-1%29 = x-1

Square both sides of the equation:

 +%28+8sqrt%28x-1%29+%29%5E2 = +%28x-1%29%5E2+

8%5E2%28sqrt%28x-1%29%29%5E2 = %28x-1%29%28x-1%29

64%28x-1%29 = x%5E2-x-x%2B1

64x-64 = x%5E2-2x%2B1

Switch left and right sides for convenience:

x%5E2-2x%2B1 = 64x-64

Get 0 on the right 

x%5E2-66x%2B65 = 0

Factor left side by thinking of two whole numbers
which have product 65 and sum 66, whch are 65 and 1:

%28x-65%29%28x-1%29 = 0

Set each factor = 0

x-65=0 gives x=65

x-1=0 gives x=1

However we must check these answers because radical
equations often have extraneous answers, which are
not solutions.  

Checking x=65

8sqrt%28x-1%29%2B1 = x

8sqrt%2865-1%29%2B1 = 65

8sqrt%2864%29%2B1 = 65

8%288%29%2B1 = 65

64%2B1 = 65

65 = 65

So we know x=65 is a valid solution.

Checking x=1

8sqrt%28x-1%29%2B1 = x

8sqrt%281-1%29%2B1 = 1

8sqrt%280%29%2B1 = 1

8%280%29%2B1 = 1

0%2B1 = 1

1 = 1

So as it turns out, both x=1 and x=65
are valid solutions.

But be aware that this is not always the case.
Sometimes when you get two answers, one of them
will be a solution and the other will not check,
and thus must be discarded; sometimes neither one 
is a solution and the equation has no solution.
Just be sure to ALWAYS CHECK a RADICAL EQUATION.

Edwin