eybrj2 wrote:
a = x + y and b = x - y. If a^2 = b^2, what is the value of y?
(1) √x + √y > 0
(2) √x - √y > 0
The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be non-negative, in order for their square roots to be real values.
I don't understand this explantion.
Why can't x and y be negative?
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: D.?
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 \(\sqrt{-25}=undefined\).
eybrj2 wrote:
if x = (-2)^3
√x = 4√-2
Is this correct?
If it is, is it negative number?
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.htmlyour reasoning and answer does not match, could you please confirm that the answer is B and not D.