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 350,000 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

29

This post received KUDOS

Expert's post

47

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]
18 Mar 2013, 07:06

1

This post received KUDOS

Edited

1. x^4 = x^3 + 6x^2 => x^2 (x^2 - x - 6) = 0 The roots of x^2 - x - 6 are -2 and 3, but -2 cannot be the value of x. So 3 and 0 are the only possible roots. => Sum of all possible solutions = 3.

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:11

1

This post received KUDOS

Expert's post

GyanOne wrote:

1. x^4 = x^3 + 6x^2 => x^2 (x^2 - x - 6) = 0 Sum of roots of x^2 - x - 6 = 0 is 1 and the only other solution is x=0 => Sum of all possible solutions = 1.

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:17

1

This post received KUDOS

2. Let one of the roots be x, then other root is also x. Sum of roots = 2x = -a Product of roots = x^2 = - b

For other equation one root is 3 and product of the roots is 15, so other root is 5. Sum of roots = 8 = -a or a = -8. From above 2x = 8 or x = 4, b = - x^2 = -16. Answer is B.

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 07:22

4

This post received KUDOS

3. a^2 + b^ 2 = m a^2 - b^2 = n Solving both the equations( adding them, and then subtracting them ): 2a^2 = m + n 2b^2 = m - n. a = ((m+n)/2)^(1/2) b = ((m-n)/2)^(1/2)

Now -x^2 +4x - 13 has its maximum value at x = -4/-2 = 2 and -y^2 - 6y has its maximum value at y=6/-2 = -3 Therefore max value of the expression = 3(-4 + 8 -13) + 2 (-9+18) = -27 + 18 = -9

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:04

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 \(x=\sqrt[4]{x^3+6x^2}\) =>\(x^4=x^3+6x^2\) =>\(x^4-x^3-6x^2=0\)

Sum of roots for \(ax^n+bx^(n-1)+cx^(n-2)+... + constant\) = \(-b/a\) In the given prob : b= -1, a=1

Sum of roots = \(-b/a = -(-1)/1 = 1\) C _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Product of all n = product of roots of the above equation. For ax^2 + bx + c = 0 => Product of roors = c/a Product of roots = -145/1 = -145 Ans = A _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

None of -50, 25 or 100 is possible with product of two possible values of x. what I am missing ? _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:33

3

This post received KUDOS

1)\(x^4=x^3+6x^2\) \(x^4-x^3-6x^2=0\) \(x^2(x^2-x-6)=0\) \(x^2=0 (1) x = 0\) \(x^2-x-6=0 (2) x = 3 (3) x = -2\) \(0+3-2=1\) C

2)The equation x^2 + ax - b = 0 has equal roots \(a^2+4b=0\) one of the roots of the equation x^2 + ax + 15 = 0 is 3 \((-a+-\sqrt{a^2-4*15})/2=3\) \(+-\sqrt{a^2-60}=6+a\) \((\sqrt{a^2-60})^2=(6+a)^2\) \(-96=12a\) \(a=-8\)

\(a^2+4b=0\) \(64+4b=0\) \(b=-16\) B

3)We can use some numbers \(2^2+1^2=5=m 2^2-1^2=5=n\) \(ab=2\)

\(\sqrt{m^2-n^2}/2 = \sqrt{25-9}/2 = 2 = ab\) C

4) -3x^2 + 12x -2y^2 - 12y - 39 MAX \(-3x^2 + 12x\) has max in (2,12) \(-2y^2 - 12y - 39\) has max in (-6,-3) -3x^2 + 12x -2y^2 - 12y - 39 MAX = 12 - 3 =9 D _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:36

1

This post received KUDOS

Expert's post

Zarrolou wrote:

1)\(x^4=x^3+6x^2\) \(x^4-x^3-6x^2=0\) \(x^2(x^2-x-6)=0\) \(x^2=0 (1) x = 0\) \(x^2-x-6=0 (2) x = 3 (3) x = -2\) \(0+3-2=1\) C

2)The equation x^2 + ax - b = 0 has equal roots \(a^2+4b=0\) one of the roots of the equation x^2 + ax + 15 = 0 is 3 \((-a+-\sqrt{a^2-4*15})/2=3\) \(+-\sqrt{a^2-60}=6+a\) \((\sqrt{a^2-60})^2=(6+a)^2\) \(-96=12a\) \(a=-8\)

\(a^2+4b=0\) \(64+4b=0\) \(b=-16\) B

3)We can use some numbers \(2^2+1^2=5=m 2^2-1^2=5=n\) \(ab=2\)

\(\sqrt{m^2-n^2}/2 = \sqrt{25-9}/2 = 2 = ab\) C

4) -3x^2 + 12x -2y^2 - 12y - 39 MAX \(-3x^2 + 12x\) has max in (2,12) \(-2y^2 - 12y - 39\) has max in (-6,-3) -3x^2 + 12x -2y^2 - 12y - 39 MAX = 12 - 3 =9 D

Better to post each solution as a separate post, since I cannot give more than 1 kudos point for one post. _________________

Re: New Algebra Set!!! [#permalink]
18 Mar 2013, 08:37

5. I have taken a longer approach, there should be a better solution. x^2 + 2x -15 = -m x^2 + 5x - 3x - 15 = -m (x+5)(x-3) = -m x will have integer values, since m is also an integer. Putting positive values for x such as ( 1, 2....) we will get values for m which lies between -10 and 10, they are -9, 0 and 7. We will get same result for negative values of x, the equation is a parabola, it will be symmetric Hence the probability is 1/3 Answer is B.

safe.txmblr Can business make a difference in the great problems that we face? My own view is nuanced. I think business potentially has a significant role to play...

Still 7 months to go to be at Lausanne. But, as Lausanne has a vacancy rate of 0.1% for rental properties, I booked my rental apartment yesterday for...