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

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a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  18 Jan 2013, 15:08
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$$a=x+y$$ and $$b=x-y$$. If $$a^2=b^2$$, what is the value of y?

(1) $$\sqrt{x}+\sqrt{y}>0$$

(2) $$\sqrt{x}-\sqrt{y}>0$$

M Advanced Quant, Chapter 9 (Workout Sets), Problem 73.
[Reveal] Spoiler: OA

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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  18 Jan 2013, 15:18
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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.

Hope it's clear.
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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  23 Jun 2013, 13:19
Hello Bunuel

But square root of 4 is +/-2 correct? And -2 < 0
Can u please explain why u said square root of integer cannot b less than 0?

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a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  23 Jun 2013, 13:33
Expert's post
Shashank1149 wrote:
Hello Bunuel

But square root of 4 is +/-2 correct? And -2 < 0
Can u please explain why u said square root of integer cannot b less than 0?

Posted from GMAT ToolKit

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]  09 Jun 2015, 01:52
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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  09 Jun 2015, 18:42
Expert's post
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.

A^2 = (X+Y)^2 = X^2 + 2XY + Y^2
B^2 = (X-Y)^2 = X^2 - 2XY + Y^2

A^2 = B^2.....so......

X^2 + 2XY + Y^2 = X^2 - 2XY + Y^2

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.

[Reveal] Spoiler:
B

GMAT assassins aren't born, they're made,
Rich
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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  10 Jun 2015, 05:34
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.

A^2 = (X+Y)^2 = X^2 + 2XY + Y^2
B^2 = (X-Y)^2 = X^2 - 2XY + Y^2

A^2 = B^2.....so......

X^2 + 2XY + Y^2 = X^2 - 2XY + Y^2

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.

[Reveal] Spoiler:
B

GMAT assassins aren't born, they're made,
Rich

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.

What is wrong in the above?
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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? [#permalink]  10 Jun 2015, 18:26
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Expert's post
Hi Mo2men,

There are a couple of issues here:

First, √9 has just ONE solution: +3

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.

GMAT assassins aren't born, they're made,
Rich
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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y?   [#permalink] 10 Jun 2015, 18:26
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