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

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New post Updated on: 05 Jun 2013, 02:37
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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

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?

Originally posted by eybrj2 on 12 Jun 2012, 22:35.
Last edited by Bunuel on 05 Jun 2013, 02:37, edited 3 times in total.
EDITED THE QUESTION AND MOVED TO DS FORUM.
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a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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New post 13 Jun 2012, 02:55
5
6
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.

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 \(\sqrt{-25}=undefined\).

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.

P.S. Please read and follow: 11-rules-for-posting-133935.html
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Re: From Advanced GMAT Quant  [#permalink]

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New post 12 Jun 2012, 22:39
if x = (-2)^3

√x = 4√-2

Is this correct?

If it is, is it negative number?
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Re: From Advanced GMAT Quant  [#permalink]

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New post 12 Jun 2012, 22:51
1
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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Re: From Advanced GMAT Quant  [#permalink]

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New post 13 Jun 2012, 00:34
1
(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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Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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New post 28 Nov 2014, 13:46
1
Bunuel wrote:
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.

Answer: D.?

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 \(\sqrt{-25}=undefined\).

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.

P.S. Please read and follow: 11-rules-for-posting-133935.html


Dear Bunuel,

your reasoning and answer does not match, could you please confirm that the answer is B and not D.
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Joined: 02 Sep 2009
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Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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New post 29 Nov 2014, 04:16
santorasantu wrote:
Bunuel wrote:
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.

Answer: D.?

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 \(\sqrt{-25}=undefined\).

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.

P.S. Please read and follow: 11-rules-for-posting-133935.html


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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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

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New post 21 Mar 2015, 06:54
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?
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a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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New post 21 Mar 2015, 07:10
lucky1829 wrote:
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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Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of &nbs [#permalink] 13 Mar 2018, 09:47
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