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From OG Quant. Review - Algebra, second-degree equations [#permalink]
23 Jul 2007, 08:41

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This post received KUDOS

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Difficulty:

5% (low)

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^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)

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.

Re: From OG Quant. Review - Algebra, second-degree equations [#permalink]
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

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.

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.

Re: From OG Quant. Review - Algebra, second-degree equations [#permalink]
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.

Re: Quadratics Factor problem from paper GMAT exam [#permalink]
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!

Re: Quadratics Factor problem from paper GMAT exam [#permalink]
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"?

Re: Quadratics Factor problem from paper GMAT exam [#permalink]
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 _________________