SOLUTION: Solve algebraically? Solve algebraically and check your potential solutions: squareroot(x+2) -x=0 I know that the answers are x=2, x=How do I know for sure they are correct aft

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Question 122042: Solve algebraically?
Solve algebraically and check your potential solutions:
squareroot(x+2) -x=0
I know that the answers are x=2, x=How do I know for sure they are correct after checking them. X=2 works out but x=-1 is funny. Thanks

Answer by solver91311(16897) About Me  (Show Source):
You can put this solution on YOUR website!
I presume you solved this in the following fashion:

sqrt%28x%2B2%29-x=0
sqrt%28x%2B2%29=x
Square both sides:
x%2B2=x%5E2
x%5E2-x-2=0
%28x-2%29%28x%2B1%29=0

x=2 or x=-1

The reason that x=-1 doesn't check (sqrt%28-1%2B2%29-%28-1%29=2%3C%3E0) is that you introduced an extraneous root when you squared both sides of the equation in the process of solving it. 2 is the only element of the solution set of the original equation.

The following is a trivial example of what happens when you square both sides of an equation:

Let x=1

Square both sides:

x%5E2=1%5E2=1

x%5E2-1=0

%28x-1%29%28x%2B1%29=0

x=1 or x=-1

x=1 is fine, but x=-1, by substitution in the original equation, leads us to the absurdity that -1=1. Therefore x=-1 is an extraneous root introduced in the process of squaring both sides of the equation and it must be excluded from the solution set.