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
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Hi,

If x & y are negative then √x, √y would be out of scope of GMAT.

On serious note, supposedly x, y are negative. √x, √y would be imaginary quantities.
√x + √y would be a point on Argand Plane (where complex numbers are represented).
and you would be comparing two points (√x + √y > 0). Does this make any sense? No.

Thus, it is correctly mentioned in the book that x & y are non-negative.

Let me know if you need any more assistance on this topic.

Regards,

(x+y)^2 = (x-y)^2
x^2 + 2xy + y^2 = x^2 - 2xy + y^2

2xy = -2xy
4xy = 0

either x or y must be zero, or both

1) either one of x or y could be zero. NS
2)
√x > √y
√x > √y >= 0

y = 0

B)

Dear Bunuel,

your reasoning and answer does not match, could you please confirm that the answer is B and not D.

Typo edited. Thank you.
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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


a^2 = b^2 implies (a+b)(a-b) = 0

a + b = 2x
a - b = 2y

4xy = 0 or xy = 0 or either x = 0 or y = 0 or both = 0

(1) √x + √y > 0
if either x or y or both = 0 and √x + √y > 0, that means either x can be zero and y can be any positive number or y can be zero and x can be any positive number
Therefore, 1 is insufficient

(2) √x - √y > 0
In this case, x or y or both = 0 and √x - √y > 0. For √x - √y > 0, √x must be greater than √y. But one of x or y must be zero. Therefore either y = 0 and x is a positive number, or x is zero and y is an imaginary number. Why? If √x must be greater than √y and x is zero then √y must be negative. Since Imaginary numbers are out of scope for the GMAT exam, therefore y = 0 and x is a positive number -->> sufficient

if x = (-2)^3

√x = 4√-2

Is this correct?

If it is, is it negative number?

One general question:

If a^2 = b^2 (x+y)^2 = (x-y)^2
then a = b x+y = x-y

Obviously this does not work out. Where is my thinking flawed?

Hi,
you are missing on the point that if a=-b, a^2 will still be equal to b^2..
so you will have a case where (x+y)=-(x-y)
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