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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]

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18 Jan 2013, 16:08
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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.

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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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18 Jan 2013, 16:18
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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]

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23 Jun 2013, 14:19
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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]

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23 Jun 2013, 14:33
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]

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09 Jun 2015, 19:42
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.

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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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21 Mar 2018, 21:11
HumptyDumpty wrote:
$$a=x+y$$ 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.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

$$a^2 = b^2$$
$$⇔ (x+y)^2 = (x-y)^2$$
$$⇔ x^2 + 2xy + y^2 = x^2 - 2xy + y^2$$
$$⇔ 2xy = -2xy$$
$$⇔ 4xy = 0$$
$$⇔ x = 0$$ or $$y = 0$$

Condition 1)
$$x = 1, y = 0$$ or $$x = 0, y = 1$$
Since we don't have a unique solution, condition 1) is not sufficient.

Condition 2)
$$\sqrt{x} - \sqrt{y} > 0$$
$$⇔ \sqrt{x} > \sqrt{y}$$
$$⇔ x > y ≥ 0$$
With the original condition, we have $$y = 0$$.
Condition 2) is sufficient.

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Re: a=x+y and b=x-y. If a^2=b^2, what is the value of y? &nbs [#permalink] 21 Mar 2018, 21:11
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