SOLUTION: (((Square root of (x+2)))=x I don't know where to start! Thanks for the help! :)

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Question 697143: (((Square root of (x+2)))=x
I don't know where to start! Thanks for the help! :)
Found 2 solutions by Stitch, god2012:
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
Given sqrt%28X%2B2%29+=+X
I would start by squaring both sides of teh equation to remove the radical sign.
X+%2B+2+=+X%5E2
Now subtract X from both sides
2+=+X%5E2+-+X
Subtract 2 from both sides
0+=+X%5E2+-+X+-+2
Now you can use the quadratic equation.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aX%5E2%2BbX%2Bc=0 (in our case 1X%5E2%2B-1X%2B-2+=+0) has the following solutons:

X%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-2=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+9+%29%29%2F2%5Ca.

X%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+9+%29%29%2F2%5C1+=+2
X%5B2%5D+=+%28-%28-1%29-sqrt%28+9+%29%29%2F2%5C1+=+-1

Quadratic expression 1X%5E2%2B-1X%2B-2 can be factored:
1X%5E2%2B-1X%2B-2+=+1%28X-2%29%2A%28X--1%29
Again, the answer is: 2, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-2+%29

Answer by god2012(113) About Me  (Show Source):
You can put this solution on YOUR website!
Given :
sqrt%28x%2B2%29 = x
Solution :
Taking square on both sides, we get
x%2B2+=+x%5E2
x%5E2-x-2=0
x%5E2-2x%2Bx-2=0
x%28x-2%29%2B1%28x-2%29=0
%28x-2%29%28x%2B1%29=0
x=2 or x=-1