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:

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:

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.

[ 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

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.

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=\(3\)

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, 01:35

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...