SOLUTION: √x +2 x =0 I squared both sides to get (√x+2)^2 – x^2 = 0 then simplifying it I got, -x^2 + x+2 = 0 setting up the quadratic equation I got (-x +2)(x+1)=0 s

Algebra ->  Functions -> SOLUTION: √x +2 x =0 I squared both sides to get (√x+2)^2 – x^2 = 0 then simplifying it I got, -x^2 + x+2 = 0 setting up the quadratic equation I got (-x +2)(x+1)=0 s      Log On


   

Question 205613: √x +2 x =0
I squared both sides to get
(√x+2)^2 – x^2 = 0 then simplifying it I got,
-x^2 + x+2 = 0 setting up the quadratic equation I got
(-x +2)(x+1)=0 setting each side to zero,
-x+2 =0 and x+1=0 solving for x in each equation
-x+2-2=0-2 =
X=-2 and for the other one
X+1-1=0-1=
X=-1 so the solution set is {-2,-1) is this correct???
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%29+%2B+2x+=+0
Let sqrt%28x%29 be represented by y. So the original equation then becomes
sqrt%28x%29+%2B+2x+=+0
y+%2B+2y%5E2+=+0
y+%281%2B2y%29+=+0
So y = 0 or y = -1/2
Now convert back to x
sqrt%28x%29+=+0 --> x =0
sqrt%28x%29+=+-1%2F2 --> x is imaginary at best, and non-existent at worst
The only real solution is x = 0
See this for a plot --> http://www82.wolframalpha.com/input/?i=x^(1%2F2)+%2B+2x+%3D+0