SOLUTION: Equations Quadratic in form X^2 + x + (sqrt x^2 + x)- 2 = 0

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Question 207916: Equations Quadratic in form
X^2 + x + (sqrt x^2 + x)- 2 = 0

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Quadratic equations are of the form:
ax%5E2+%2B+bx+%2B+c+=+0
At first glance
x%5E2+%2B+x+%2B+sqrt%28x%5E2+%2B+x%29-+2+=+0
does not appear to be of the quadratic form. But a deeper look reveals that it is a quadratic equation. The key is to notice that
  1. The expression x%5E2%2Bx appears twice; and
  2. x%5E2%2Bx+=+%28sqrt%28x%5E2%2Bx%29%29%5E2

With this in mind we can use a substitution to see and solve the quadratic equation.

Let's make y+=+sqrt%28x%5E2%2Bx%29. Then y%5E2+=+%28sqrt%28x%5E2%2Bx%29%29%5E2+=+x%5E2+%2B+x. Substituting these into the original equation we get:
y%5E2+%2B+y+-+2+=+0
We can solve this for y by factoring (or with the quadratic formula):
%28y+%2B+2%29%28y+-+1%29+=+0
The only way this product can be zero is if one of the products is zero:
y+%2B+2+=+0 or y+-+1+=+0
Solving each of these we get:
y+=+-2 or y+=+1
Substituting back in for the y's we get:
sqrt%28x%5E2%2Bx%29+=+-2 or sqrt%28x%5E2%2Bx%29+=+1
Since all square roots are positive it is impossible, in the first equation, for the square root to be a -2. So there are no solutions to the first equation. But there are solutions to the second equation. Square both sides to get:
%28sqrt%28x%5E2+%2B+x%29%29%5E2+=+1%5E2
x%5E2+%2B+x+=+1
Subtracting one from both sides we get:
x%5E2+%2B+x+-+1+=+0
This does not factor but we can use the quadratic formula on it:
x+=+%28-1+%2B-+sqrt%281%5E2+-+4%281%29%28-1%29%29%29%2F%282%281%29%29
Simplifying we get:
x+=+%28-1+%2B-+sqrt%281+-+4%281%29%28-1%29%29%29%2F%282%281%29%29
x+=+%28-1+%2B-+sqrt%281+-+%28-4%29%29%29%2F2
x+=+%28-1+%2B-+sqrt%285%29%29%2F2
Since we squared both sides while solving, we should check for extraneous solutions. First we'll check x+=+%28-1%2Bsqrt%285%29%29%2F2. If we square this we get:
x%5E2+=+%286+-+2sqrt%285%29%29%2F4
Add x to both sides:
x%5E2+%2B+x+=+%286+-+2sqrt%285%29%29%2F4+%2B+%28-1%2Bsqrt%285%29%29%2F2
Match the denominators so we can add:
x%5E2+%2B+x+=+%286+-+2sqrt%285%29%29%2F4+%2B+%28-2+%2B+2sqrt%285%29%29%2F4
x%5E2+%2B+x+=+4%2F4
x%5E2+%2B+x+=+1
Substituting into the original equation we get:
1+%2B+sqrt%281%29+-+2+=+0
which checks out.

Checking the other solution, x+=+%28-1+-+sqrt%285%29%29%2F2, in a similar way finds that it, too, works. So both x+=+%28-1+%2B+sqrt%285%29%29%2F2 and x+=+%28-1+-+sqrt%285%29%29%2F2 are solutions.