SOLUTION: Please help solve the equation (20-x)^2+(20+x)^2=8 including the smaller and larger values of x

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Question 621853: Please help solve the equation (20-x)^2+(20+x)^2=8 including the smaller and larger values of x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%2820-x%29%5E2%2B%2820%2Bx%29%5E2=8
Solving equations often starts with simplifying each side of the equation. We can use FOIL or patterns to square the binomials on the left side. (I like using the patterns.)

which simplifies as follows:
400-40x%2Bx%5E2+%2B+400%2B40x%2Bx%5E2+=+8
2x%5E2+%2B+800+=+8

This is a quadratic equation. It has no "x term" in it so probably the easiest way to solve this is to isolate the squared term and find a square root. Subtracting 800 from each side:
2x%5E2+=+-792
and then dividing by 2:
x%5E2+=+-396

Normally we would now find the square root of each side. But if we look at what we have we will realize that there is no solution to this equation because x%5E2 cannot turn out to be a negative number, like -796 (unless x is a complex number).

If you know about complex numbers and if you are supposed to find complex solutions to this equation, we can continue:
sqrt%28x%5E2%29+=+sqrt%28-796%29
x+=+sqrt%28-796%29 or x+=+-sqrt%28-796%29
With a negative number in the square root, we start by factoring out -1:
x+=+sqrt%28-1%2A796%29 or x+=+-sqrt%28-1%2A796%29
x+=+sqrt%28-1%29%2Asqrt%28796%29 or x+=+-sqrt%28-1%29%2Asqrt%28796%29
x+=+i%2Asqrt%28796%29 or x+=+-i%2Asqrt%28796%29
Now we simplify the remaining square root by finding perfect square factors:
x+=+i%2Asqrt%284%2A199%29 or x+=+-i%2Asqrt%284%2A199%29
x+=+i%2Asqrt%284%29%2Asqrt%28199%29 or x+=+-i%2Asqrt%284%29%2Asqrt%28199%29
x+=+i%2A2%2Asqrt%28199%29 or x+=+-i%2A2%2Asqrt%28199%29
or
x+=+2i%2Asqrt%28199%29 or x+=+-2i%2Asqrt%28199%29
These are the only (complex) solutions to your equation.