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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
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:
New Algebra Set!!! [#permalink]
18 Mar 2013, 06:56
32
This post received KUDOS
Expert's post
62
This post was BOOKMARKED
The next set of medium/hard PS algebra questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers.
1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
Re: New Algebra Set!!! [#permalink]
23 Jun 2013, 00:37
Expert's post
aquax wrote:
Hi,
For the first question, what do you mean by "But X cannot be negative as it equals to the even (4th) root of some expression"?
\(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt[even]{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).
When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or -5. Even roots have only non-negative value on the GMAT.
In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\).
So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).
Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).
Answer: B.
Hi Bunuel,
Just as a matter of fact can you please explain that how you reached to factorization. In other words how a candidate will decide how to split -39 between (X,Y).
Your reply is appreciated !!
Rgds, TGC !! _________________
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________
So, we need to maximize the value of \(-3(x-2)^2-2(y+3)^2-9\).
Since, the maximum value of \(-3(x-2)^2\) and \(-2(y+3)^2\) is zero, then the maximum value of the whole expression is \(0+0-9=-9\).
Answer: B.
Hi Bunuel,
Just as a matter of fact can you please explain that how you reached to factorization. In other words how a candidate will decide how to split -39 between (X,Y).
Your reply is appreciated !!
Rgds, TGC !!
I completed the squares for -3x^2 + 12x - ... and for -2y^2 - 12y-... So, I asked myself what do I need there in order to have (a+b)^2.
Re: New Algebra Set!!! [#permalink]
23 Oct 2013, 22:54
5
This post received KUDOS
Bunuel wrote:
imhimanshu wrote:
Bunuel wrote:
[ Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);
Factorize: \(x^2(x-3)(x+2)=0\);
So, the roots are \(x=0\), \(x=3\) and \(x=-2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).
The sum of all possible solutions for x is 0+3=3.
Answer: D.
Hi Bunuel, Request you to please let me know where I'm making a mistake.
Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\)
i.e \((x-0)^2(x-(-6))>=0\) This implies that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) Please let me know where I am going wrong.
Thanks
Plug x=-2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true?
Hi Brunel if say x=16. x^1/2 has two values +4 & -4. So why x^1/4 cannot have +2 & -2 as as values ? Please explain
Re: New Algebra Set!!! [#permalink]
24 Oct 2013, 00:01
Expert's post
NeetiGupta wrote:
Bunuel wrote:
Bunuel wrote:
Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);
Factorize: \(x^2(x-3)(x+2)=0\);
So, the roots are \(x=0\), \(x=3\) and \(x=-2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).
The sum of all possible solutions for x is 0+3=3.
Answer: D.
Hi Bunuel, Request you to please let me know where I'm making a mistake.
Since \(x^3+6*x^2 >= 0\) Then, \(x^2(x+6)>=0\)
i.e \((x-0)^2(x-(-6))>=0\) This implies that\(x>=-6\) Hence, x=-2 is a valid root, and sum of all roots should be\(x=3+(-2) = 1\) Please let me know where I am going wrong.
Thanks
Plug x=-2 into \(x=\sqrt[4]{x^3+6x^2}\). Does the equation hold true?
Hi Brunel if say x=16. x^1/2 has two values +4 & -4. So why x^1/4 cannot have +2 & -2 as as values ? Please explain
First of all it's 4th root not 2nd root. Next, \(\sqrt[4]{16}=2\), not +2 or -2.
When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or -5.
In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.
Re: New Algebra Set!!! [#permalink]
05 Apr 2014, 02:42
Q1. Option D. Raising both LHS and RHS to fourth power,and making RHS=0 we get \(x^4-x^3-6x^2=0\) Taking \(x^2\) common and factorizing, \(x^2(x+2)(x-3)=0\) Three possible values of \(x\) are \(-2,3\) and \(0\) But \(x\) can't be \(-ve\). So sum of possible values=\(3\)
Re: New Algebra Set!!! [#permalink]
21 Jun 2014, 00:35
Bunuel wrote:
SOLUTIONs:
1. If \(x=\sqrt[4]{x^3+6x^2}\), then the sum of all possible solutions for x is:
A. -2 B. 0 C. 1 D. 3 E. 5
Take the given expression to the 4th power: \(x^4=x^3+6x^2\);
Re-arrange and factor out x^2: \(x^2(x^2-x-6)=0\);
Factorize: \(x^2(x-3)(x+2)=0\);
So, the roots are \(x=0\), \(x=3\) and \(x=-2\). But \(x\) cannot be negative as it equals to the even (4th) root of some expression (\(\sqrt{expression}\geq{0}\)), thus only two solution are valid \(x=0\) and \(x=3\).
The sum of all possible solutions for x is 0+3=3.
Answer: D.
Bunuel, how do we know that both sides of the equation are positive? if x is a negativve then we cannot raise even power. _________________
If my post was helpful, press Kudos. If not, then just press Kudos !!!
gmatclubot
Re: New Algebra Set!!!
[#permalink]
21 Jun 2014, 00:35
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...