SOLUTION: How do I solve the square root of x+6 - 4 ( -4 being on the outside of the square) = x what are the proposed solutions

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Question 709682: How do I solve the square root of x+6 - 4 ( -4 being on the outside of the square) = x what are the proposed solutions
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
First we isolate the square root:
sqrt%28x%2B6%29-4=x --> sqrt%28x%2B6%29=x%2B4
Now, we square both sides of the equal sign,
knowing that we may be adding extraneous solutions,
but not worrying about them at this point.
sqrt%28x%2B6%29=x%2B4 --> x%2B6=%28x%2B4%29%5E2 --> x%2B6=x%5E2%2B8x%2B16
Now, we solve, still not worrying about extraneous solutions.
x%2B6=x%5E2%2B8x%2B16 --> 0=x%5E2%2B7x%2B10
If we are good at factoring, we factor
x%5E2%2B7x%2B10=0 --> %28x%2B2%29%28x%2B5%29=0
and find the solutions to that equation as
x=-2 and x=-5
(If we are not good at factoring,
we may solve the equation a different way,
such as "completing the square"
or using the quadratic formula).

At this point we worry about extraneous solutions,
but not to much.
We just check to see if the solutions we found for x%5E2%2B7x%2B10=0
are solutions to
sqrt%28x%2B6%29-4=x
Substituting x=-2
sqrt%28x%2B6%29-4=sqrt%28-2%2B6%29-4=sqrt%284%29-4=2-4=-2
so highlight%28x=-2%29 is a solution of the original equation.
Substituting x=-5
sqrt%28x%2B6%29-4=sqrt%28-5%2B6%29-4=sqrt%281%29-4=1-4=-3
so highlight%28x=-5%29 is an extraneous solution.
It is not a solution of the original equation.