Last visit was: 25 Nov 2024, 14:23 It is currently 25 Nov 2024, 14:23
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

(x + y)(x − y) = 0 xy ≠ 0

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4814
Own Kudos [?]: 11198 [2]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
(x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   Updated on: 31 Oct 2018, 06:21
1
Expert Reply
1
Bookmarks
00:00

Question Stats:

48% (01:26) correct 51% (01:26) wrong based on 27 sessions

Hide Show timer Statistics

\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(6\sqrt{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.
ShowHide Answer
Official Answer
C

Originally posted by sandy on 28 Oct 2018, 16:54.
Last edited by chetan2u on 31 Oct 2018, 06:21, edited 1 time in total.
avatar
ruposh6240
Intern
Intern
Joined: 22 Jul 2018
Posts: 39
Own Kudos [?]: 67 [1]
Given Kudos: 0
Send PM
Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   30 Oct 2018, 03:28
1
Bookmarks
B is answer
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [0]
Given Kudos: 0
Send PM
Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   31 Oct 2018, 03:56
Expert Reply
sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(\sqrt[6]{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C
avatar
MTEManik
Intern
Intern
Joined: 08 Sep 2018
Posts: 3
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   31 Oct 2018, 06:19
I have confusion about the answer C
avatar
Mercychee
Intern
Intern
Joined: 04 Nov 2018
Posts: 44
Own Kudos [?]: 22 [0]
Given Kudos: 0
Send PM
Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   10 Nov 2018, 10:50
chetan2u wrote:
sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(\sqrt[6]{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C


May I know how you derived the 36 that was included into the square root?
avatar
ShefaliSahu
Intern
Intern
Joined: 02 Oct 2018
Posts: 31
Own Kudos [?]: 11 [0]
Given Kudos: 0
Send PM
Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   11 Nov 2018, 21:11
The answer is C.
avatar
Salsanousi
Manager
Manager
Joined: 07 Aug 2016
Posts: 59
Own Kudos [?]: 68 [0]
Given Kudos: 0
GRE 1: Q166 V156
Send PM
Re: (x + y)(x − y) = 0 xy ≠ 0 [#permalink] New post   19 Nov 2018, 17:03
Mercychee wrote:
chetan2u wrote:
sandy wrote:
\((x + y)(x - y) = 0\)
\(xy \neq 0\)

Quantity A
Quantity B
\(\sqrt[6]{\frac{19}{2x^2}}\)
\(\sqrt{\frac{342}{y^2}}\)


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.


\((x + y)(x - y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C


May I know how you derived the 36 that was included into the square root?


I think what chetan tried to say is that there is probably a typo because it should be 6 * square root of (19/2x^2)

Square root of 36 is 6.

Because it was outside the root so he brought it inside as 6^2.
User avatar
B
BrDickinson
Manager
Manager
Joined: 03 Aug 2022
Posts: 54
Own Kudos [?]: 32 [1]
Given Kudos: 69
Send PM
Re: (x + y)(x y) = 0 xy 0 [#permalink] New post   15 Aug 2022, 07:39
1
One simple way to solve this is the following:
(x+y)(x−y)=0 --> x^2 - y^2 = 0 (difference of squares pattern)
So x^2 = y^2. We can take the square root of both sides --> x = y

Now, we can plug 1 for both and square both square roots.
We will be left with
A: 18(19/2) = 342
B: 342

Forgive me if I made any mistake.
Prep Club for GRE Bot
gmatclubot
Re: (x + y)(x y) = 0 xy 0 [#permalink]
15 Aug 2022, 07:39
Moderators:
adewale223
GRE Instructor
84 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1111 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact