<p>It’s definitely C. X could eqaul -Y, but X could also equal Y. Thus, X squared has to equal Y squared, regardless of which option it is. C has to be true, everything but D could be true.</p>
<p>lol this question is easier than I thought after reading the title of the thread XD</p>
<p>The answer is C since after factoring the equation, you get 2(x+y)(x-y) = 0</p>
<p>Therefore, for this to be true, either x+y=0 or x-y=0, hence x=-y or x=y. Then you square both sides to get rid of the negative sign in front of ‘y’. You get x^2 = y^2.</p>
<p>Okay hopefully this will end the debate:</p>
<p>The equation is in the form (A)*(B) = 0</p>
<p>Where A = x + y and B = x^2 - y^2</p>
<p>Now in order for the equation to be true either A or B needs to be zero, both can be zero, but that is not necessary becusue any number multiplied by zero equals zero.</p>
<p>So either x + y = 0 which means x = -y</p>
<p>or</p>
<p>x^2 - y^2 = 0 which means that x^2 = y^2</p>
<p>But if both sides of the equation x = -y are squared</p>
<p>(x)^2 = (-y)^2</p>
<p>so</p>
<p>x^2 = y^2</p>
<p>So as you can see x^2 is always equal to y^2</p>
<p>A is disproved when x + y = 0 (x = -y) leaving only C as the correct answer.</p>
<p>Alright guys…here’s the answer</p>
<p>First (x^2-y^2) should be factored out, giving you (x+y)(x-y)(x+y)=0. Now seeing this, you realize that there MUST be 3 possible equalities.
- x+y=0
- x-y=0
- x+y=0 AND x-y=0</p>
<p>Thus seeing this, you conclude the following</p>
<p>If 1 is true, then x=-y
If 2 is true, then x=y
If 3 is true, then x=-y AND x=y</p>
<p>Putting all 3 cases together, the only answer that satifies ALL 3 can only be x^2=y^2.
The answer is C.</p>
<p>Congrats to all that got it right! :)</p>
<p>Thats a highly deceptive thread title! Any decent 10th grader knows enough maths to say that choice C is right within 5 seconds without plugging in numbers. Factorization of (x^2-y^2) itself yields (x+y). So its obvious x^2=y^2 =>x^2-y^2=0 is the right choice <_<</p>
<p>Well seeing as how there was a few people that actually did get this question wrong, that shows how this question is somewhat tricky gary7. So stop being cocky and arrogant just because you can see the correct answer more obviously than others</p>