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
61
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]
13 Apr 2013, 08:01
Expert's post
Hi Anshuman,
For this you need to go thru some strategy books of GMAT Quant. In My Opinion, two important sources are, 1) Manhattan GMAT Quant strategy guides 4th or 5th Ed 2) GMAT Club Math Book - This is free and you can download it from this website free of cost (Check BB's profile) Furthermore you can check my signature. There i have attached my articles prepared on some peculiar topics. You may find them useful.
Re: New Algebra Set!!! [#permalink]
22 May 2013, 21:19
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.
Re: New Algebra Set!!! [#permalink]
23 May 2013, 01:00
Expert's post
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? _________________
Re: New Algebra Set!!! [#permalink]
23 May 2013, 04:03
2. The equation x^2 + ax - b = 0 has equal roots, and one of the roots of the equation x^2 + ax + 15 = 0 is 3. What is the value of b?
A. -64 B. -16 C. -15 D. -1/16 E. -1/64
Solution:
x^2 + ax + 15 = 0 equation has one root as 3. only possible values of getting product of the roots 15 , having one root 3 is 5*3. x^2-5x-3x+15=0 take common values out => x(x-5)-3(x-5) =0 => x=5 and x=3(what was given) , from this we get a= -8.
x^2 -8x - b =0 => if roots are equal then b^2-4ac=0, 64+4b=0 from this b=-16
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 _____________________________________________________________________________
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=\(3\)
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...