What is the answer to this problem?

<p>If (x + y)2 = x2 + y2, which of the following statements must also be true?</p>

<p>x = 0
(x - y)2 = x2 + y2
xy = 0
None</p>

<p>I only</p>

<p>II only</p>

<p>III only</p>

<p>II and III </p>

<p>iii only right? (all the 2's are squares)</p>

<p>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.</p>

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

<p>It's easy to find a counterexample for 1. Try setting y=0 for any value of x.</p>

<p>It's 2am, so maybe I'm not making any sense this late but isn't the answer none...? let's make both x and y equal to 2.
so...(x+y)2 and x2+y2 both equal 8.
therefore x doesn't equal 0,(2-2)2=4+4 is also wrong, and 2*2 doesn't equal 0as well.</p>

<p>(x+y)^2 is 4^2 which is 16. Squared not times 2.</p>

<p>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.</p>

<p>SAT Questions didn't type in (x+y)^2, he/she typed (x + y)2, which doesn't mean squared so, unless he/she meant squared and not times two,which is what is written, then I'm right, pretty sure SAT meant times two since that is umm...what SAT wrote....<em>confusion grows</em></p>

<p>Didn't read the bottom part where they said all the 2's are squares, lazy bum couldn't just put a ^ every now and then eh...hehe oh well.</p>

<p>'s cool bb. I think this qn is either from Barron's or Blue Book.</p>

<p>to Rhap. Green
um.. (x-y)^2=x^2-2xy+y^2, and (x-y)(x+y)=X^2-y^2, and x^2+y^2 doesnt have a factor.</p>

<p>Oh eff, you're completely right, I wasn't even paying attention. Well I feel retarded now. That's what I get for posting before I think.</p>