2sqrt(r^2-x^2) - [2x^2] / [sqrt(r^2-x^2)] = [2(r^2-2x^2)] / [sqrt(r^2-x^2)]

How does 2sqrt(r^2-x^2) - [2x^2] / [sqrt(r^2-x^2)] = [2(r^2-2x^2)] / [sqrt(r^2-x^2)]?

Can anyone guide me through process of doing this? it should be easy because it was part of calculations of a problem from textbook and took place in only 1 step unexplained.

see : http://imageshack.us/photo/my-images/207…

assuming expression is

[2(√(r^2 - x^2)) -2x^2] / √(r^2-x^2) = 2(r^2-2x^2) / √(r^2-x^2)

square both sides

[(2(√(r^2-x^2)) - 2x^2) / √(r^2-x^2)]^2 = [2(r^2-2x^2) / √(r^2-x^2)]^2

[4√(r^2-x^2)^2 - 4x^2√(r^2-x^2) - 4x^2√(r^2-x^2) + 4x^4] / √(r^2-x^2)^2 EQUALS [4(r^2-2x^2)^2] / √(r^2-x^2)^2

(4(r^2-x^2) - 8x^2√(r^2-x^2) + 4x^4) / (r^2 - x^2) EQUALS 4(r^4 - 2r^2x^2 - 2r^2x^2 + 4x^4) / (r^2-x^2)

(4r^2 - 4x^2 - 8x^2√(r^2-x^2) + 4x^4) / (r^2-x^2) EQUALS 4(r^4 - 4r^2x^2 + 4x^4) / (r^2-x^2)

note that we have equal bases, so we can DROP them and solve using numerators

(4r^2 - 4x^2 - 8x^2√(r^2-x^2) + 4x^4) EQUALS 4(r^4 - 4r^2x^2 + 4x^4)

factor 4 out of the left

4(r^2 - x^2 - 2x^2√(r^2-x^2) + x^4) EQUALS 4(r^4 - 4r^2x^2 + 4x^4)

divide both sides by 4

r^2 - x^2 - 2x^2√(r^2-x^2) + x^4 EQUALS r^4 - 4r^2x^2 + 4x^4

2√(r^2-x^2) - 2x^2/√(r^2-x^2) = [2r^2 - 2x^2 - 2x^2]/√(r^2 -x^2) = 2r^2/√(r^2 -x^2)