a = x + y and b = x - y. If a^2 = b^2, what is the value of

a = x + y and b = x - y. If a^2 = b^2, what is the value of y?

Since \(a = x + y\) and \(b = x - y\) then from \(a^2 = b^2\) we have that \((x + y)^2=(x-y)^2\) --> \(x^2+2xy+y^2=x^2-2xy+y^2\) --> \(4xy=0\) --> \(xy=0\) --> either \(x\) or \(y\) equals to zero (or both).

(1) √x + √y > 0. Two cases are possible: \(x=0\) and \(y\) is ANY positive number OR \(y=0\) and \(x\) is ANY positive number. Not sufficient.

(2) √x - √y > 0 --> \(\sqrt{x}>\sqrt{y}\). Now, since square root function can not give negative result (\(\sqrt{some \ expression}\geq{0}\)), then \(\sqrt{x}>\sqrt{y}\geq{0}\). So, \(x>0\) and \(y=0\). Sufficient.

Or another way: square \(\sqrt{x}>\sqrt{y}\) (we can safely do that since both parts of the inequality are non-negative): \(x^2>y^2\) --> \(y^2\) (square of a number) is always non-negative, so \(x^2\) is more than some non-negative number, which makes \(x^2\) a positive value which excludes the possibility of \(x=0\), so \(y=0\). Sufficient.

Answer: B.

As for your question:

The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers so even roots from negative number is undefined on the GMAT: \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{-25}=undefined\).


No, that's not correct. If \(x=(-2)^3=-8\) then \(\sqrt{x}=\sqrt{-8}=undefined\).

Hope it's clear.

P.S. Please read and follow: 11-rules-for-posting-133935.html