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]
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
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!
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.
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...