Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Oct 2014, 05:43

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Quadratics Factor problem from paper GMAT exam

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Manager
Manager
User avatar
Joined: 04 Aug 2005
Posts: 202
Followers: 1

Kudos [?]: 4 [1] , given: 0

From OG Quant. Review - Algebra, second-degree equations [#permalink] New post 23 Jul 2007, 08:41
1
This post received
KUDOS
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

77% (01:53) correct 23% (00:13) wrong based on 13 sessions
Basically, I'm looking for a clearer explanation on this one:

If (t-8) is a factor of t^2-kt-48, then k=

A. 16
B. 12
C. 2
D. 6
E. 14

OA is C

OE is:
If (t-8) is a factor of the expression t^2-kt-48, then the expression can be written as the product (t-8)(t+a). Where did the a come from? How do figure out to use a (or any other unkown)?
t^2-8t+at-8a multiply the binomials
t^2-(8-a)t-8a combine the t terms

Comparing this expression to the original expression sets -8a=-48 or a=6. Then, by substituting this value of a into the product of (t-8)(t+a):

(t-8)(t+6)
t^2-8t+6t-48
t^2-2t-48
Therefore, k=2

Thanks for any help!
[Reveal] Spoiler: OA
Director
Director
User avatar
Joined: 30 Nov 2006
Posts: 592
Location: Kuwait
Followers: 11

Kudos [?]: 154 [0], given: 0

 [#permalink] New post 23 Jul 2007, 14:29
If (t-8) is a factor of t^2-kt-48, then k=

A. 16
B. 12
C. 2
D. 6
E. 14

t^2-kt-48 = (t-8)(t+m) where m is any positive integer.

if 48/8 = 6, then we know as a matter of fact that:
m = +6
and thus k = 8-6 = 2


This is very basic algebra: for m and a are positive integers
t^2-kt-m = (t-a)(t+m) where a>m
t^2+kt-m = (t-a)(t+m) where a<m
t^2-kt+m = (t-a)(t-m)
t^2+kt+m = (t+a)(t+m)

GOT IT ?!
Manager
Manager
User avatar
Joined: 27 May 2007
Posts: 128
Followers: 1

Kudos [?]: 3 [0], given: 0

 [#permalink] New post 23 Jul 2007, 15:21
First, recognize t^2-kt-48 as a quadratic equation. That is, it is the product of 2 binomials (2 term polynomials) The first factor is given: (t-8). We know that the second factor will be (t+something) - the explanation uses "a" for the unknown. The rule for multiplying 2 binomials together is summarized by the acronym FOIL (first, outer, inner, last) So with the 2 terms (t-8) (t+a), the product would be t^2+at-8t-8a. If you look at the original equation, you see that the last term is -48, which correlates to -8a. So a=-48/-8 = 6. Plug 6 back into the place of a for "t^2+6t-8t-48". Simplify: t^2-2t-48. So k=2.
Senior Manager
Senior Manager
User avatar
Joined: 28 Jun 2007
Posts: 463
Followers: 2

Kudos [?]: 5 [0], given: 0

 [#permalink] New post 23 Jul 2007, 15:41
It should be noted that any equation of the second degree can be broken down to the form that the OE used.

Remember the formula [-b +/- sqrt(b^2 - 4ac)]/2a to find the two possible roots for an equation of the form ax^2+bx+c = 0.

Hope this helps.

Last edited by ioiio on 23 Jul 2007, 16:11, edited 1 time in total.
Senior Manager
Senior Manager
avatar
Joined: 14 Jun 2007
Posts: 399
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: From OG Quant. Review - Algebra, second-degree equations [#permalink] New post 23 Jul 2007, 16:04
ishcabibble wrote:
Basically, I'm looking for a clearer explanation on this one:

If (t-8) is a factor of t^2-kt-48, then k=

A. 16
B. 12
C. 2
D. 6
E. 14

OA is C

OE is:
If (t-8) is a factor of the expression t^2-kt-48, then the expression can be written as the product (t-8)(t+a). Where did the a come from? How do figure out to use a (or any other unkown)?
t^2-8t+at-8a multiply the binomials
t^2-(8-a)t-8a combine the t terms

Comparing this expression to the original expression sets -8a=-48 or a=6. Then, by substituting this value of a into the product of (t-8)(t+a):

(t-8)(t+6)
t^2-8t+6t-48
t^2-2t-48
Therefore, k=2

Thanks for any help!


iscabible,

don't worry about mishari's condescending banter... nerds get very upset when people in the real world can't solve math problems and don't obsess over the properties of numbers and other mathmetical nonsense that is of little use to 95% of tomorrow's managers and leaders. Also, recognize that many people on these boards come from parts of the world where where rote learning is the dominant system of education (i.e. memorization for little else besides recitation vs. true comprehension, creativity etc...)

anyway, like the others have mentioned recognize that this is a quadratic. while you need to memorize the quadratic forumla, it is not necessary in this problem; memorizing the basic forms of a quadratic that mishari listed can be a real time saver - but it is more important to be able to derive them and understand them first rather than learn them rote.

the OG explanations at times see counterintuitive. They are not Kaplan etc... they DO NOT CARE if you do well on the test. keep this in mind when reading their explanations. they are not going to necessarily show you the best way to solve a problem

in this problem they are giving you one of the roots as (t-8) you need to find t + or - something... when you use the foil method the product of the last two terms must equal -48. -8*6=-48. so we have that the other root is t+6. now with FOIL the sum of the products of the inner and outer terms = the middle term of the quadratic. 6t-8t=-2t. a value of 2 for k gives you -2t
1 KUDOS received
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5095
Location: Singapore
Followers: 19

Kudos [?]: 153 [1] , given: 0

 [#permalink] New post 23 Jul 2007, 16:09
1
This post received
KUDOS
If t-8 is a factor of the equation, then when t = 8, the equation must become zero.

So

64-8k-48 = 0
k = 2
Manager
Manager
User avatar
Joined: 27 May 2007
Posts: 128
Followers: 1

Kudos [?]: 3 [0], given: 0

 [#permalink] New post 23 Jul 2007, 16:14
ywilfred wrote:
If t-8 is a factor of the equation, then when t = 8, the equation must become zero.

So

64-8k-48 = 0
k = 2

Good point! That's what I love about this board - seeing the different ways people tackle a problem to come up with the same answer. Learning to use these methods will put alot of tools in our toolboxes when we go in to take the test.
Senior Manager
Senior Manager
User avatar
Joined: 28 Jun 2007
Posts: 463
Followers: 2

Kudos [?]: 5 [0], given: 0

 [#permalink] New post 23 Jul 2007, 16:15
ywilfred wrote:
If t-8 is a factor of the equation, then when t = 8, the equation must become zero.

So

64-8k-48 = 0
k = 2


Why didnt I think of this? Definitely easier than finding roots and imagining quadratic equations.

Sometimes we all over-complicate things thanks to the ever-increasing number of formulae we tend to remember.
Manager
Manager
User avatar
Joined: 04 Aug 2005
Posts: 202
Followers: 1

Kudos [?]: 4 [0], given: 0

 [#permalink] New post 24 Jul 2007, 05:18
Thanks everyone for your help! I did see that it is a quadratic equation, but for some reason adding that extra unknown just totally threw me off. Unfortunately for me, my brain just has a really hard time thinking that way. At least I rock on the verbal section!

Robin in NC, thanks for your very clear explanation - that's exactly what I was looking for. And ywilfred, that is a genius way of thinking about it! I'm sure that you're going to save a lot of us a lot of time on the test now! And thanks for the words of encouragement, anonymousegmat! Trust me, it is probably way more frustrating for me not being able to wrap my mind around certain concepts than it is for anyone else. :-D
Intern
Intern
avatar
Joined: 20 Jul 2007
Posts: 13
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: From OG Quant. Review - Algebra, second-degree equations [#permalink] New post 24 Jul 2007, 06:29
ishcabibble wrote:
Basically, I'm looking for a clearer explanation on this one:

If (t-8) is a factor of t^2-kt-48, then k=

A. 16
B. 12
C. 2
D. 6
E. 14

OA is C

OE is:
If (t-8) is a factor of the expression t^2-kt-48, then the expression can be written as the product (t-8)(t+a). Where did the a come from? How do figure out to use a (or any other unkown)?
t^2-8t+at-8a multiply the binomials
t^2-(8-a)t-8a combine the t terms

Comparing this expression to the original expression sets -8a=-48 or a=6. Then, by substituting this value of a into the product of (t-8)(t+a):

(t-8)(t+6)
t^2-8t+6t-48
t^2-2t-48
Therefore, k=2

Thanks for any help!



Hi

t-8 is one of the factors of this quadratic equation. Hence t =8.

Subsititute t=8 in the equation and you get 64-8k-48.

Eqaute it and you get k = 2.

Regards

Nikhil
Manager
Manager
avatar
Joined: 30 May 2010
Posts: 191
Followers: 3

Kudos [?]: 46 [0], given: 32

Equation problem from QR 2nd PS 57 [#permalink] New post 26 Jul 2010, 10:51
If (t-8) is a factor of t^2-kt-48, then k =

A. -6
B. -2
C. 2
D. 6
E. 14
Manager
Manager
avatar
Joined: 30 May 2010
Posts: 191
Followers: 3

Kudos [?]: 46 [0], given: 32

Re: Equation problem from QR 2nd PS 57 [#permalink] New post 26 Jul 2010, 10:55
Almost every time I get stumped on a problem, I come here to post it and then figure out what I did wrong as I post or right after!

If (t-8) is one factor, then the second factor would have to be (t+6) to satisfy the -48 part.

A. is a trap answer, because -6 is the root, but that is not what the question is asking for!

-8 + 6 = -2

B. is another trap answer, because the negative sign is already given in the original equation.

Therefore, the correct answer is k = 2.
Intern
Intern
avatar
Joined: 26 Jul 2010
Posts: 5
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Equation problem from QR 2nd PS 57 [#permalink] New post 26 Jul 2010, 23:42
useful information
:-D
Manager
Manager
avatar
Joined: 16 Apr 2010
Posts: 223
Followers: 3

Kudos [?]: 46 [0], given: 12

Re: Equation problem from QR 2nd PS 57 [#permalink] New post 27 Jul 2010, 03:30
Hi,

For quadratic equations in the form ax^2+bx+c=0, you can make use of the following:
Sum of the roots: x1+x2 = - b/a
Product of the roots: x1*x2 = c/a


regards,
Jack
CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2794
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 182

Kudos [?]: 985 [0], given: 235

Reviews Badge
Re: Equation problem from QR 2nd PS 57 [#permalink] New post 27 Jul 2010, 05:19
jpr200012 wrote:
If (t-8) is a factor of t^2-kt-48, then k =

A. -6
B. -2
C. 2
D. 6
E. 14


If (t-8) is a factor of t^2-kt-48

then t =8 must satisfy t^2-kt-48=0

=> 8^2 -8k - 48 = 0 => 8 -k -6 = 0 => k =2
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Current Student
User avatar
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 129
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
Followers: 8

Kudos [?]: 68 [0], given: 210

Quadratics Factor problem from paper GMAT exam [#permalink] New post 11 Oct 2010, 13:20
Can you help me break down the steps to arriving at the answer for this problem. Thanks!

If (t - 8) is a factor of t^2-kt-48, then k =

a. -6
b. -2
c. 2
d. 6
e. 14
1 KUDOS received
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 500 [1] , given: 25

GMAT ToolKit User Reviews Badge
Re: Quadratics Factor problem from paper GMAT exam [#permalink] New post 11 Oct 2010, 13:23
1
This post received
KUDOS
Manager
Manager
User avatar
Joined: 04 Aug 2005
Posts: 202
Followers: 1

Kudos [?]: 4 [0], given: 0

Re: Quadratics Factor problem from paper GMAT exam [#permalink] New post 12 Oct 2010, 05:00
I have to say that it is pretty cool and kind of funny that I'm still getting responses from a post I wrote 3 years ago. Especially since I'm now done with my MBA!

I'm glad that this thread is still helping folks and good luck on the GMAT!
Manager
Manager
avatar
Joined: 10 Sep 2010
Posts: 133
Followers: 2

Kudos [?]: 12 [0], given: 7

Re: Quadratics Factor problem from paper GMAT exam [#permalink] New post 15 Nov 2010, 19:49
I am having problem with this question.

The quadratic equations has the form ax^2+bx+c=0

The question stem has: t^2-kt-48. One root of the equation is 8 (as given), which means the other one is 6.
So then the equation start looking like this: t^2-2t-48.

Going back to a generic formula for quadratic equation: a=1, b=-2, c=-48.
So the question asks what is "k", meaning that they ask for the absolute value of "b"?
Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4876
Location: Pune, India
Followers: 1150

Kudos [?]: 5348 [0], given: 165

Re: Quadratics Factor problem from paper GMAT exam [#permalink] New post 15 Nov 2010, 20:04
Expert's post
Fijisurf wrote:
I am having problem with this question.

The quadratic equations has the form ax^2+bx+c=0

The question stem has: t^2-kt-48. One root of the equation is 8 (as given), which means the other one is 6.
So then the equation start looking like this: t^2-2t-48.

Going back to a generic formula for quadratic equation: a=1, b=-2, c=-48.
So the question asks what is "k", meaning that they ask for the absolute value of "b"?


To answer your question:
General form: ax^2+bx+c=0
Here the equation is t^2-kt-48 = 0
So -k = b or we can say k = -b
When you get the equation t^2-2t-48 = 0, you get b = -2. So k = - (-2) = 2

Something to note: Given t^2-kt-48 = 0
You got the equation is t^2-2t-48 = 0
Just compare the two right here and you see that k = 2

Another approach: When we say 8 is a root of the equation, we mean it satisfies the equation.
So if I put 8 in place of t, it should satisfy my equation.
t^2-kt-48 = 0 will become 8^2 -8k - 48 = 0 or k = 2
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Re: Quadratics Factor problem from paper GMAT exam   [#permalink] 15 Nov 2010, 20:04
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic Problem factoring difficult quadratic equation Samwong 3 30 Aug 2013, 10:18
3 Experts publish their posts in the topic Factoring quadratic equations jfk 7 02 Apr 2012, 18:06
1 GMAT Real Exam paper akbism 5 30 Aug 2011, 05:28
1 Tough SC problem from ETS paper GMAT tonebeeze 5 11 Oct 2010, 18:06
1 Experts publish their posts in the topic Got this problem from old GMAT papers - There are no legal cumic 8 13 Jun 2008, 23:58
Display posts from previous: Sort by

Quadratics Factor problem from paper GMAT exam

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 25 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.