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Manager  Joined: 31 Oct 2011
Posts: 219
a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 50% (02:09) correct 50% (02:23) wrong based on 367 sessions

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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, 23:35.
Last edited by Bunuel on 05 Jun 2013, 03:37, edited 3 times in total.
EDITED THE QUESTION AND MOVED TO DS FORUM.
##### Most Helpful Expert Reply
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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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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Manager  Joined: 31 Oct 2011
Posts: 219
Re: From Advanced GMAT Quant  [#permalink]

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if x = (-2)^3

√x = 4√-2

Is this correct?

If it is, is it negative number?
Current Student B
Joined: 29 Mar 2012
Posts: 301
Location: India
GMAT 1: 640 Q50 V26 GMAT 2: 660 Q50 V28 GMAT 3: 730 Q50 V38 Re: From Advanced GMAT Quant  [#permalink]

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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,
Current Student Joined: 08 Jan 2009
Posts: 297
GMAT 1: 770 Q50 V46 Re: From Advanced GMAT Quant  [#permalink]

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2
(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)
Senior Manager  G
Joined: 27 Aug 2014
Posts: 294
Location: Netherlands
Concentration: Finance, Strategy
Schools: LBS '22, ISB '21
GPA: 3.9
WE: Analyst (Energy and Utilities)
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 57025
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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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.
_________________
Intern  Joined: 27 Jan 2015
Posts: 5
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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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?
Math Expert V
Joined: 02 Aug 2009
Posts: 7756
a = x + y and b = x - y. If a^2 = b^2, what is the value of  [#permalink]

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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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Non-Human User Joined: 09 Sep 2013
Posts: 12007
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   [#permalink] 02 Aug 2019, 02:55
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