Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 14 Sep 2014, 18:43

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Help with simple maths problem

Author Message
TAGS:
Manager
Joined: 29 Sep 2006
Posts: 57
Followers: 0

Kudos [?]: 1 [0], given: 0

Help with simple maths problem [#permalink]  11 Jan 2007, 07:26
Help needed with the following:

(x-y)^2 = 0.

I understand that this translates to (x-y)(x-y) = 0. But kaplan says that this is the same as (x-y) = 0, why is this.

Thanks
 Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes Manhattan GMAT Discount Codes
Director
Affiliations: FRM Charter holder
Joined: 02 Dec 2006
Posts: 736
Schools: Stanford, Chicago Booth, Babson College
Followers: 6

Kudos [?]: 17 [0], given: 4

Re: Help with simple maths problem [#permalink]  11 Jan 2007, 07:31
GMAT? wrote:
Help needed with the following:

(x-y)^2 = 0.

I understand that this translates to (x-y)(x-y) = 0. But kaplan says that this is the same as (x-y) = 0, why is this.

Thanks

can we give any other value to (x - y), and ensure that (x-y)(x-y) = 0 ?
Manager
Joined: 09 Nov 2006
Posts: 129
Followers: 1

Kudos [?]: 1 [0], given: 0

Re: Help with simple maths problem [#permalink]  11 Jan 2007, 07:47
GMAT? wrote:
(x-y)^2 = 0.

I understand that this translates to (x-y)(x-y) = 0.

Actually, (x-y)^2 = x^2 - 2xy + y^2
Manager
Joined: 29 Sep 2006
Posts: 57
Followers: 0

Kudos [?]: 1 [0], given: 0

This is what I thought it translated to as well:

(x-y)^2 = x^2 - 2xy + y^2

The whole question is:

Does x = y

(1) x^2 - y^2 = 0
(2) (x-y)^2 = 0

I know that the answer if B.

With (1): x^2 = y^2 but that does not mean x = y as 2^2 and -2^2 both = 16.

With (2): (x-y)^2 = 0. So x^2 - 2xy + y^2 = 0, hence x^2 -2xy = -y^2 hence x=y.

Just was unsure how kaplan got that (x-y)(x-y) = 0 is the same as (x-y) = 0.

Thanks
Manager
Joined: 04 Oct 2006
Posts: 190
Followers: 1

Kudos [?]: 1 [0], given: 0

In order to have (x-y)(x-y) equal 0, one of the two things you are multiplying must be equal to 0.

So either (x-y) or (x-y) has to be equal to 0.

And if one of them is, then the other one has to be as well.

Try working the logic using X * X = 0 (substitute X for x-y)

Hope that helps
_________________

wall street...bulls, bears, people from connecticut

Similar topics Replies Last post
Similar
Topics:
simple math question (not actual problem) 2 23 Aug 2012, 16:17
simple ps problems...can anyone help 1 10 Feb 2011, 05:04
simple word problem help 2 19 Aug 2010, 12:34
I need help with a simple math question 2 18 Feb 2009, 06:21
Need Help - Explain Math Problem 4 04 Dec 2005, 10:01
Display posts from previous: Sort by