If x is an integer, is 16^x + 16^-x = b
(1) 4^x + 4^-x = √(b+2)
(2) x>0
I know that the answer is A. But my question is, why, when you square on both sides doesn't √(b+2) end up being |b+2|?
(1) 4^x + 4^-x = √(b+2)
(2) x>0
I know that the answer is A. But my question is, why, when you square on both sides doesn't √(b+2) end up being |b+2|?
















