SOLUTION: x^2+x+(√x^2+x)-2=0 I'm thinking this can be re-written as 2x^2+(√2x^2)-2=0 but I'm completely confused in all of this. At some point, don't I need to balance the rig

Algebra ->  Radicals -> SOLUTION: x^2+x+(√x^2+x)-2=0 I'm thinking this can be re-written as 2x^2+(√2x^2)-2=0 but I'm completely confused in all of this. At some point, don't I need to balance the rig      Log On


   

Question 422167: x^2+x+(√x^2+x)-2=0
I'm thinking this can be re-written as 2x^2+(√2x^2)-2=0 but I'm completely confused in all of this. At some point, don't I need to balance the right and the left side of the equation?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
No, you can't do that, this is a toughie
x%5E2%2Bx%2B%28sqrt%28x%5E2%2Bx%29%29-+2+=+0
:
x%5E2+%2B+x-+2 = -sqrt%28x%5E2%2Bx%29
Square both sides
%28x%5E2+%2B+x-+2%29%5E2 = x%5E2+%2B+x
FOIL the left
x%5E4%2B2x%5E3-3x%5E2-4x%2B4 = x%5E2+%2B+x
combine on the left
x%5E4%2B2x%5E3-3x%5E2-x%5E2-4x-x%2B4 = 0
x%5E4%2B2x%5E3-4x%5E2-5x%2B4 = 0
What can you do with this? it won't factor, graph the original and the above equation, see what we have

find the x intercepts, -1.6, and + .62 (not exact) Correction!!
:
Check both values in the original equation, see if it works