Originally Posted by
HallsofIvy I would have done that slightly differently. From $\displaystyle \sqrt{x+ 8}- \sqrt{x- 4}= 2$, add $\displaystyle \sqrt{x- 4}$ to both sides to get $\displaystyle \sqrt{x+ 8}= \sqrt{x- 4}+ 2$. Squaring both sides of that you get $\displaystyle x+ 8= x- 4+ 4\sqrt{x- 4}+ 4$. You still have the square root form the "cross term" but since you don't have both square roots on the same side, it's a little simpler. That can be reduced, by canceling the "x" terms on both sides and canceling the "-4" and "4", to $\displaystyle 8= 4\sqrt{x- 4}$ which reduces to $\displaystyle \sqrt{x- 4}= 2$. Now square both sides again to get x- 4= 4.
Notice that while e^(i*pi)'s equation $\displaystyle (x+ 8)(x- 4)= x^2$ looks like a quadratic, it is not be cause the [itex]x^2[/itex] terms on each side will cancel.