What is the answer to this problem?

Rhapsody_in_Green
Tue, 03 Jun 2008 01:08:41 +0000

2 and 3. For 3, expand out (x+y)^2 = x^2 + 2xy + y^2. Thus, if, that is equal to x^2 +y^2, 2xy=0.

2 is always true for any quadratic. Again, expand it out: (x - y)^2 = x^2 + xy - xy + y^2 = x^2 + y^2.

It’s easy to find a counterexample for 1. Try setting y=0 for any value of x.