SOLUTION: (square root of x + 2)- x = 0 (x+2)^2 -x = 0 (x+2)^1/2^2/1 - x + x = 0 + x^2/1 x + 2 - 2 = x^2/1 - 2 x = x^2/1^1/2 - 2 x - x = x - 2 0 + 2 = x - 2 + 2 2 = x

Algebra ->  Radicals -> SOLUTION: (square root of x + 2)- x = 0 (x+2)^2 -x = 0 (x+2)^1/2^2/1 - x + x = 0 + x^2/1 x + 2 - 2 = x^2/1 - 2 x = x^2/1^1/2 - 2 x - x = x - 2 0 + 2 = x - 2 + 2 2 = x      Log On


   

Question 156543: (square root of x + 2)- x = 0
(x+2)^2 -x = 0
(x+2)^1/2^2/1 - x + x = 0 + x^2/1
x + 2 - 2 = x^2/1 - 2
x = x^2/1^1/2 - 2
x - x = x - 2
0 + 2 = x - 2 + 2
2 = x
Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B2%29-x=0
then
sqrt%28x%2B2%29=x
then
x%2B2=x%5E2
then
x%5E2-x-2=0
then
x+=+%281+%2B-+sqrt%28+1%5E2-4%2A1%2A%28-2%29+%29%29%2F%282%29+ =
x+=+%281+%2B-+3+%29%2F2+
then x=-1 or x=2
-1 is not a solution of the first eq because sqrt%28-1%2B2%29-%28-1%29=2
2 is a solution of the first eq because sqrt%282%2B2%29-%282%29=0
Answer: x=2