SOLUTION: x^2+x+ the square root of (x^2+x) -2 =0. I know to iscolate the radical one one side of the equation and then square both sides. I'm supposed to get a quadratic equation, set it

Algebra ->  Radicals -> SOLUTION: x^2+x+ the square root of (x^2+x) -2 =0. I know to iscolate the radical one one side of the equation and then square both sides. I'm supposed to get a quadratic equation, set it       Log On


   

Question 70575: x^2+x+ the square root of (x^2+x) -2 =0. I know to iscolate the radical one one side of the equation and then square both sides. I'm supposed to get a quadratic equation, set it equal to zero and solve. But instead I'm getting x^4+2x^3-4x^2-5x+4=0. As this is not in standard ax^2+bx+c form and the degree is 4, I have no idea now to solve it for x. Please help.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
One way you can simplify the solving of this problem is by substituting y+=+sqrt%28x%5E2%2Bx%29 so y%5E2+=+x%5E2%2Bx. Let's see what we get:
y%5E2%2By-2+=+0 Solve by factoring.
%28y-1%29%28y%2B2%29+=+0 Apply the zero product principle.
y-1+=+0 and/or y%2B2+=+0
If y-1+=+0 then y+=+1
If y%2B2+=+0 then y+=+-2
Now re-substitute y+=+sqrt%28x%5E2%2Bx%29
sqrt%28x%5E2%2Bx%29+=+1 and
sqrt%28x%5E2%2Bx%29+=+-2
Now square both sides of these two equations.
x%5E2%2Bx+=+1 or x%5E2%2Bx-1+=+0
x%5E2%2Bx+=+4 or x%5E2%2Bx-4+=+0
So now you have two quadratic equations that can be solved using the quadratic formula: x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
Taking the first equation:
x+=+%28-1%2B-sqrt%281%5E2-4%281%29%28-1%29%29%29%2F2%281%29
x+=+%28-1%2B-sqrt%285%29%29%2F2
x%5B1%5D+=+%28-1%2Bsqrt%285%29%29%2F2 and x%5B2%5D+=+%28-1-sqrt%285%29%29%2F2
And the second equation:
x+=+%28-1%2B-sqrt%281%5E2-4%281%29%28-4%29%29%29%2F2%281%29
x+=+%28-1%2B-sqrt%2817%29%29%2F2
x%5B1%5D+=+%28-1%2Bsqrt%2817%29%29%2F2 and x%5B2%5D+=+%28-1-sqrt%2817%29%29%2F2