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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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24 Jan 2014, 04:55

Factoring the expression: (t - 8) (t - ?), since (t - 8) is a factor, the other bracket has to be (t + 6). The net result is -2. So (B)? Not sure of the answer because I got lost with the signs a bit.

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

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24 Jan 2014, 23:46

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

(A) - 6 (8) - 2 (C) 2 (0) 6 (E) 14

It is given that \((t - 8)\) is a factor of the quadratic expression \(t^2 - kt - 48\) Hence, we need to find the other factor of \(-48\) such that the sum of factors is\(-(\frac{-k}{1})=k\).

Thus, \((8) * x = -48\)

Or, \(x = -6\)

So, \(k = 8 + (- 6) = 2\)

Answer: (C)

Last edited by arunspanda on 01 Nov 2014, 02:02, edited 1 time in total.

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

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22 May 2014, 23:38

Vieta's formulas applied to quadratic: x1+x2= -b/a & x1*x2=c/a From what is given in the question: -8+x2=-k and -8*x2=-48 so, x2=6 and thus -8+6=-k i.e. k=2
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Last edited by NoHalfMeasures on 11 Nov 2015, 20:45, edited 1 time in total.

Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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30 Jun 2015, 15:40

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Re: If (t- 8) is a factor of t^2 - kt - 48, then k= [#permalink]

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15 May 2016, 10:04

In this question, I didn't understand this step:

(t−8)(t−8) is a factor of t2−kt−48t2−kt−48 means that t=8t=8 is a solution of the equation t2−kt−48=0t2−kt−48=0.

Why do we assume both equations are equal to 0? What is the rationale of this step? It would be great if someone could explain with examples, or with videos.

gmatclubot

Re: If (t- 8) is a factor of t^2 - kt - 48, then k=
[#permalink]
15 May 2016, 10:04

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