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

Given that \((x+y)^2=(x-y)^2\) --> \(x^2+2xy+y^2=x^2-2xy+y^2\) --> \(4xy=0\) --> either \(x=0\) or \(y=0\).

(1) \(\sqrt{x}+\sqrt{y}>0\). It's possible that \(x=0\) and \(y\) is any positive number, as well as, it's possible that \(y=0\) and \(x\) is any positive number. Not sufficient.

(2) \(\sqrt{x}-\sqrt{y}>0\). \(x\) cannot be 0, since in this case we'd have that \(\sqrt{y}<0\) (which is wrong, since square root of a number is greater than or equal to 0), thus must be true that \(y=0\). Sufficient.

Square root function cannot give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{4}=2\) (not +2 and -2). In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.
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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]

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09 Jun 2015, 02:52

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This DS question is rooted in Classic Quadratics and Radical rules. There's a heavy "logic" component to this prompt, and you'll have to do a certain amount of math to get to the solution...

We're given a number of facts to work with: 1) A = X+Y 2) B = X-Y 3) A^2 = B^2

We're asked for the value of Y.

Before dealing with the two Facts, I'm going to take a moment to simplify the information in the prompt.

If we cancel out the terms, we're left with... 2XY = -2XY

There are only 3 ways for this equation to exist.... X = 0 OR Y = 0 OR BOTH = 0

This is an important restriction that will impact the answer to this question.

Fact 1: √X + √Y > 0

Since radicals are either POSITIVE or ZERO, either X or Y could = 0 (but not both). IF....X=1, Y=0 the answer to the question is 0 IF....X=0, Y=1 the answer to the question is 1 Fact 1 is INSUFFICIENT

Fact 2: √X − √Y > 0

Here, the DIFFERENCE is > 0, so the only way for that difference to be POSITIVE, given ALL of the restrictions we've already discussed, is for the X to be POSITIVE and the Y to be 0. Here, the answer to the question is ALWAYS 0. Fact 2 is SUFFICIENT.

Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]

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10 Jun 2015, 06:34

EMPOWERgmatRichC wrote:

Hi All,

This DS question is rooted in Classic Quadratics and Radical rules. There's a heavy "logic" component to this prompt, and you'll have to do a certain amount of math to get to the solution...

We're given a number of facts to work with: 1) A = X+Y 2) B = X-Y 3) A^2 = B^2

We're asked for the value of Y.

Before dealing with the two Facts, I'm going to take a moment to simplify the information in the prompt.

If we cancel out the terms, we're left with... 2XY = -2XY

There are only 3 ways for this equation to exist.... X = 0 OR Y = 0 OR BOTH = 0

This is an important restriction that will impact the answer to this question.

Fact 1: √X + √Y > 0

Since radicals are either POSITIVE or ZERO, either X or Y could = 0 (but not both). IF....X=1, Y=0 the answer to the question is 0 IF....X=0, Y=1 the answer to the question is 1 Fact 1 is INSUFFICIENT

Fact 2: √X − √Y > 0

Here, the DIFFERENCE is > 0, so the only way for that difference to be POSITIVE, given ALL of the restrictions we've already discussed, is for the X to be POSITIVE and the Y to be 0. Here, the answer to the question is ALWAYS 0. Fact 2 is SUFFICIENT.

Hi Rich, Nice explanation and powerful 'TESt It' methodology. I reached the same result: X = 0 OR Y = 0 OR BOTH = 0 but I was afraid that I'm mistaken because of zeros in the result so I applied 'Testing it' in the equation but it got so complicated with many variables and situations.

However, I want to ask about Fact 2. If I choose X=0 & Y= 9 for easy roots example. So The result will be the following: √X − √Y > 0 then Fact 2: √0 − √9 > 0------> .... √9 will be either 3 or -3. If it is 3, then 0-3>0 so not satisfying the condition but if Y=-3, then 0- (-3)= 3>0 so it is OK with condition. Therefore, Y will always any positive number. So Fact 2 should be insufficient as there many values for Y.

The equation X^2 = 9 has TWO solutions: +3 and -3.... but we have to work with what we were given and we were given RADICALS, so there is no negative option.

Second, knowing that the radical has just one solution, X = 0 is NOT an option in Fact 2, since we need the result of the calculation to be > 0. By extension, Y MUST = 0.

Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]

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10 Oct 2016, 21:42

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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