SOLUTION: (the square root of X+2) = x? How do I solve this? ALSO: (the square root of 2-x) = 2-x How do I solve this?

Algebra ->  Radicals -> SOLUTION: (the square root of X+2) = x? How do I solve this? ALSO: (the square root of 2-x) = 2-x How do I solve this?      Log On


   

Question 288676: (the square root of X+2) = x? How do I solve this?
ALSO: (the square root of 2-x) = 2-x How do I solve this?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1
sqrt%28x%2B2%29+=+x Start with the given equation.


x%2B2+=+x%5E2 Square both sides.


0+=+x%5E2-x-2 Get everything to one side.


x%5E2-x-2=0 Rearrange the equation.


%28x-2%29%28x%2B1%29=0 Factor.


x%2B2=0 or x%2B1=0 Use the zero product property.


x=2 or x=-1 Solve for 'x' in each case.


So the possible solutions for sqrt%28x%2B2%29+=+x are x=2 or x=-1. However, if you plug x=-1 into sqrt%28x%2B2%29+=+x, you'll get 1=-1 which is NOT true. So x=-1 is NOT a solution.


So the only solution is x=2

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# 2

sqrt%282-x%29=2-x Start with the given equation.


2-x=%282-x%29%5E2 Square both sides.


2-x=4-4x%2Bx%5E2 FOIL


0=4-4x%2Bx%5E2%2Bx-2 Get everything to one side.


0=x%5E2-3x%2B2 Combine like terms.


0=%28x-2%29%28x-1%29 Factor.


x-2=0 or x-1=0 Use the zero product property.


x=2 or x=1 Solve for 'x' in each case.


So the possible solutions are x=2 or x=1. If you plug in each solution to check, you'll find that they both work.


So the solutions are x=2 or x=1