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# Tearing my hair out on this question!

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Intern
Joined: 11 Jul 2011
Posts: 3
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Tearing my hair out on this question! [#permalink]

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11 Jul 2011, 21:34
Hello,

I have been having difficulties with this question.

I originally got the question wrong so looked at the answer for help. It is on page 52 of GMAT review (red) book.
I'm trying to understand how the factoring works particularly the step from squaring both sides to the last line. I understand that I must use FOIL however I cannot work out how this is to be applied on the right-hand side.

If I may say, for a so-called "answer explanations" section this is the part which should be explained. The rest of the answer explanation has not given me a problem eventhough they are explained. So I am not so concerned about getting the answer right (creating 4x₂ on one side) as to understanding how we get from one step to another.

So basically can anyone help me to explain how (√2x+1)₂ becomes 2x + √2x+1 ?

Second degree equations

Work with the equation to create 4x₂ on one side.

√(〖3-2x〗^ ) = √2x+1
(√(〖3-2x〗^ ))₂ = (√2x+1)₂ Square both sides
3-2x = 2x + √2x+1

It may be that there is a piece of knowledge I am missing or I am applying FOIL wrong.

Any help would be appreciated!

Thank you,
Current Student
Joined: 08 Jan 2009
Posts: 326
GMAT 1: 770 Q50 V46
Followers: 24

Kudos [?]: 124 [0], given: 7

Re: Tearing my hair out on this question! [#permalink]

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11 Jul 2011, 22:44
Some strange formatting in your question makes it a little difficult to understand what is going on. Might want to use the formatting tags to make a little clearer. My expansion looks like:

$$(\sqrt{2x} + 1)^2$$

$$(\sqrt{2x} + 1)^2 = (\sqrt{2x} + 1)(\sqrt{2x} + 1)$$

$$(\sqrt{2x} + 1)^2 = \sqrt{2x}\sqrt{2x}+2\sqrt{2x} + 1$$

$$(\sqrt{2x} + 1)^2 = 2x + 2\sqrt{2x} + 1$$

$$(\sqrt{2x} + 1)^2 = 2(x + \sqrt{2x}) + 1$$
Intern
Joined: 03 Jul 2011
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Re: Tearing my hair out on this question! [#permalink]

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11 Jul 2011, 22:48
flynno37 wrote:
Hello,

I have been having difficulties with this question.

I originally got the question wrong so looked at the answer for help. It is on page 52 of GMAT review (red) book.
I'm trying to understand how the factoring works particularly the step from squaring both sides to the last line. I understand that I must use FOIL however I cannot work out how this is to be applied on the right-hand side.

If I may say, for a so-called "answer explanations" section this is the part which should be explained. The rest of the answer explanation has not given me a problem eventhough they are explained. So I am not so concerned about getting the answer right (creating 4x₂ on one side) as to understanding how we get from one step to another.

So basically can anyone help me to explain how (√2x+1)₂ becomes 2x + √2x+1 ?

Second degree equations

Work with the equation to create 4x₂ on one side.

√(〖3-2x〗^ ) = √2x+1
(√(〖3-2x〗^ ))₂ = (√2x+1)₂ Square both sides
3-2x = 2x + √2x+1

It may be that there is a piece of knowledge I am missing or I am applying FOIL wrong.

Any help would be appreciated!

Thank you,

First, Outside, Inside, Last
(√2x+1)*(√2x+1) = 2x + √2x + √2x + 1 = 2x+2√2x+1
Looks like you simply forgot to perform the operation on 1*√2x twice.
Hope this helped!
Intern
Joined: 11 Jul 2011
Posts: 3
Followers: 0

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

Re: Tearing my hair out on this question! [#permalink]

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12 Jul 2011, 09:15
Thank you for your responses so far. I think I am having difficulty in the application of the FOIL rule.
Here is how I am using FOIL to expand the equation. Can anyone pinpoint where I am going wrong?

(√(〖3-2x〗^ ))₂ = (√2x+1)₂ Square both sides
3-2x = 2x + √2x+1

So, I am doing this:
(√2x+1)₂ = (√2x+1) (√2x+1)
F (first): √2x √2x

O (outer): √2x +1

I (Inside): +1 √2x

L (last): + 1 + 1

Can anyone explain what I do next?
This may be a silly question but should I multiply each result of doing FOIL?
So the result of F (first) will be √4x₂ ?
And √2x multiplied by +1 will just be √2x?
Or is it more of a case of connecting the components? As I have done above.
Is it correct to be able to connect the √2x (in O) with the √2x (in I)? So that will be 2 √2x ?

I am trying not to get hung up on this but I have been told it’s pretty fundamental to get it!

Any general recommendations would be appreciated!

Thank you,

Sorry for the formatting. I'm pasting in from ms word.
Current Student
Joined: 08 Jan 2009
Posts: 326
GMAT 1: 770 Q50 V46
Followers: 24

Kudos [?]: 124 [0], given: 7

Re: Tearing my hair out on this question! [#permalink]

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12 Jul 2011, 15:47
Yep it is fundamental, so its good you are making sure you understand it!

Take each component and add them together (so add F + O + I + L). Then you have fully expanded the equation out of brackets.

What books are you using? I just did some quick google searches, there are tons and tons of guides and videos that may be helpful.

I can't vouch for any, but check out:

Intern
Joined: 11 Jul 2011
Posts: 3
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Kudos [?]: 0 [0], given: 0

Re: Tearing my hair out on this question! [#permalink]

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13 Jul 2011, 19:19
Thanks for your help. I think I have it now.

The books I'm using are:

Official guide for GMAT review, 12th edition
Kaplan GMAT math workbook
AS Level Mathematics (Text book from UK high school)

Do you recommend taking the Kaplan courses or others that are paid classroom / online packages?
Senior Manager
Joined: 08 Jun 2010
Posts: 397
Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32
Followers: 3

Kudos [?]: 77 [0], given: 13

Re: Tearing my hair out on this question! [#permalink]

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13 Jul 2011, 21:25
pike wrote:
Some strange formatting in your question makes it a little difficult to understand what is going on. Might want to use the formatting tags to make a little clearer. My expansion looks like:

$$(\sqrt{2x} + 1)^2$$

$$(\sqrt{2x} + 1)^2 = (\sqrt{2x} + 1)(\sqrt{2x} + 1)$$

$$(\sqrt{2x} + 1)^2 = \sqrt{2x}\sqrt{2x}+2\sqrt{2x} + 1$$

$$(\sqrt{2x} + 1)^2 = 2x + 2\sqrt{2x} + 1$$

$$(\sqrt{2x} + 1)^2 = 2(x + \sqrt{2x}) + 1$$

What software are you using to do the maths? It makes equations look good
Re: Tearing my hair out on this question!   [#permalink] 13 Jul 2011, 21:25
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# Tearing my hair out on this question!

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