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

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Kudos [?]: 98 [1] , given: 18

a = x + y and b = x - y. If a^2 = b^2, what is the value of [#permalink]  12 Jun 2012, 22:35
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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

[Reveal] Spoiler:
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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Jun 2013, 02:37, edited 3 times in total.
EDITED THE QUESTION AND MOVED TO DS FORUM.
Senior Manager
Joined: 31 Oct 2011
Posts: 331
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Kudos [?]: 98 [0], given: 18

if x = (-2)^3

√x = 4√-2

Is this correct?

If it is, is it negative number?
Senior Manager
Joined: 28 Mar 2012
Posts: 285
Concentration: Entrepreneurship
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
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Kudos [?]: 139 [0], given: 23

eybrj2 wrote:
In chapter 9,

Q 73. 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?

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,

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My posts: Solving Inequalities, Solving Simultaneous equations, Divisibility Rules

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Kudos [?]: 76 [0], given: 7

(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)

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Kudos [?]: 22197 [1] , given: 2601

Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of [#permalink]  13 Jun 2012, 02:55
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Expert's post
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.

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.

_________________
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of   [#permalink] 13 Jun 2012, 02:55
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