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.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: OG Diag #16 PS [#permalink]
18 Jan 2011, 01:04

2

This post received KUDOS

tonebeeze wrote:

This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?

Hey, below mentioned is my strategy to solve this problem,

Sqrt both side you will get, 3-2x= 2x+1+2sqrt{2x} 3-2x-2x-1= 2sqrt{2x} 2-4x= 2sqrt{2x} devide by 2 both side 1-2x = sqrt{2x} Again sqrt of both side 1+4x^2-4x= 2x 1+4x^2-6x=0 4x^2= 6x-1

Re: OG Diag #16 PS [#permalink]
18 Jan 2011, 14:32

SoniaSaini wrote:

tonebeeze wrote:

This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?

Hey, below mentioned is my strategy to solve this problem,

Sqrt both side you will get, 3-2x= 2x+1+2sqrt{2x} 3-2x-2x-1= 2sqrt{2x} 2-4x= 2sqrt{2x} devide by 2 both side 1-2x = sqrt{2x} Again sqrt of both side 1+4x^2-4x= 2x 1+4x^2-6x=0 4x^2= 6x-1

\sqrt{3-2x} = \sqrt{2x} +1 --> square both sides: (\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2 --> 3-2x=2x+2*\sqrt{2x}+1 --> rearrange so that to have root at one side: 2-4x=2*\sqrt{2x} --> reduce by 2: 1-2x=\sqrt{2x} --> square again: (1-2x)^2=(\sqrt{2x})^2 --> 1-4x+4x^2=2x --> rearrange again: 4x^2=6x-1.

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
26 Jun 2012, 02:37

1

This post received KUDOS

Expert's post

metallicafan wrote:

Solving the equation, I arrive to this point: 2 - 2\sqrt{2x}=4x

How can we know how to manipulate the equation in order to arrive to one of the choices and not another valid solution?

After this point, you see that none of the options has a square root sign. You need to get rid of it to get to the answer. Hence, it is apparent that you need to square it again. You then take the square root term to one side and the rest of the terms to the other side and square both sides of the equation. This helps you get rid of the square root sign. _________________

It’s been a long time, since I posted. A busy schedule at office and the GMAT preparation, fully tied up with all my free hours. Anyways, now I’m back...

Burritos. Great, engaging session about how to network properly. How better can it get? Hosted jointly by Human Capital Club and Engineers in Management, we had a chance to...