What is the answer to this problem?

fiona_1
Tue, 03 Jun 2008 02:24:37 +0000

(x+y)^2 is 4^2 which is 16. Squared not times 2.

Anyway, you can’t set x and y to be anything, vashthestampede0. The question itself gives you the information that (x+y)^2 = x^2 + y^2. As Rhapsody said, if you expand it, you will know that xy = 0, so x or y (or both) must be 0.