Last visit was: 10 Jul 2025, 07:55 It is currently 10 Jul 2025, 07:55
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
eybrj2
Joined: 31 Oct 2011
Last visit: 02 Oct 2013
Posts: 200
Own Kudos:
8,515
 [56]
Given Kudos: 18
Posts: 200
Kudos: 8,515
 [56]
8
Kudos
Add Kudos
48
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,615
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,615
Kudos: 740,073
 [17]
8
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
General Discussion
User avatar
eybrj2
Joined: 31 Oct 2011
Last visit: 02 Oct 2013
Posts: 200
Own Kudos:
Given Kudos: 18
Posts: 200
Kudos: 8,515
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
cyberjadugar
Joined: 29 Mar 2012
Last visit: 28 May 2024
Posts: 266
Own Kudos:
1,672
 [2]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
GMAT 3: 730 Q50 V38
Posts: 266
Kudos: 1,672
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
eybrj2
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,
User avatar
pike
User avatar
Current Student
Joined: 08 Jan 2009
Last visit: 27 Dec 2020
Posts: 245
Own Kudos:
476
 [3]
Given Kudos: 7
GMAT 1: 770 Q50 V46
GMAT 1: 770 Q50 V46
Posts: 245
Kudos: 476
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
(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)
User avatar
santorasantu
Joined: 27 Aug 2014
Last visit: 06 Apr 2023
Posts: 242
Own Kudos:
453
 [1]
Given Kudos: 76
Location: Netherlands
Concentration: Finance, Strategy
Schools: ISB '21 LBS '22
GPA: 3.9
WE:Analyst (Energy)
Schools: ISB '21 LBS '22
Posts: 242
Kudos: 453
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
eybrj2
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
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,615
Own Kudos:
740,073
 [1]
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,615
Kudos: 740,073
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
santorasantu
Bunuel
eybrj2
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
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.
avatar
lucky1829
Joined: 27 Jan 2015
Last visit: 15 Jun 2015
Posts: 5
Own Kudos:
Given Kudos: 1
Posts: 5
Kudos: 6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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?
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 09 Jul 2025
Posts: 11,295
Own Kudos:
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,295
Kudos: 41,652
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lucky1829
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)
User avatar
TheBipedalHorse
Joined: 16 Jun 2021
Last visit: 12 Dec 2023
Posts: 109
Own Kudos:
36
 [1]
Given Kudos: 98
Posts: 109
Kudos: 36
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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


a^2 = b^2 implies (a+b)(a-b) = 0

a + b = 2x
a - b = 2y

4xy = 0 or xy = 0 or either x = 0 or y = 0 or both = 0

(1) √x + √y > 0
if either x or y or both = 0 and √x + √y > 0, that means either x can be zero and y can be any positive number or y can be zero and x can be any positive number
Therefore, 1 is insufficient

(2) √x - √y > 0
In this case, x or y or both = 0 and √x - √y > 0. For √x - √y > 0, √x must be greater than √y. But one of x or y must be zero. Therefore either y = 0 and x is a positive number, or x is zero and y is an imaginary number. Why? If √x must be greater than √y and x is zero then √y must be negative. Since Imaginary numbers are out of scope for the GMAT exam, therefore y = 0 and x is a positive number -->> sufficient
User avatar
Yeetyeti38
Joined: 16 May 2023
Last visit: 19 Apr 2025
Posts: 13
Own Kudos:
Given Kudos: 3
Posts: 13
Kudos: 11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I'm following the explanation but not sure why √y being negative makes it an imaginary number. If y=16, then √y could be -4 but that wouldn't be an imaginary number would it?

Is the rule for the GMAT that the square root of a number must be positive unless otherwise indicated?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,615
Own Kudos:
740,073
 [2]
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,615
Kudos: 740,073
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Yeetyeti38
I'm following the explanation but not sure why √y being negative makes it an imaginary number. If y=16, then √y could be -4 but that wouldn't be an imaginary number would it?

Is the rule for the GMAT that the square root of a number must be positive unless otherwise indicated?
­
That's not specific for the GMAT.

Mathematically, \(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

√9 = 3, NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3. Because x^2 = 9 means that x =-√9 =-3 or x = √9 = 3.

Hope it helps.
 
User avatar
Yeetyeti38
Joined: 16 May 2023
Last visit: 19 Apr 2025
Posts: 13
Own Kudos:
Given Kudos: 3
Posts: 13
Kudos: 11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Yeetyeti38
I'm following the explanation but not sure why √y being negative makes it an imaginary number. If y=16, then √y could be -4 but that wouldn't be an imaginary number would it?

Is the rule for the GMAT that the square root of a number must be positive unless otherwise indicated?
­
That's not specific for the GMAT.

Mathematically, \(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

√9 = 3, NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3. Because x^2 = 9 means that x =-√9 =-3 or x = √9 = 3.

Hope it helps.

 
­that is very helpful, thank you! the gaps in my understanding of high school math continue to reveal themselves...
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,375
Own Kudos:
Posts: 37,375
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
102615 posts