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Second Degree Equations

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by m&m » Wed May 13, 2009 7:32 am
if t-8 is a factor than it must be that

(t-8)(at+b) = t^2-kt-48

so finding the coefficients for a and b we get
b=6 we know because -8*a = -48
and a=1 we know because t*at = t^2

so (t-8)(t+6) = t^2 -2t - 48 so k=2 -->C
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Re: Second Degree Equations

by sureshbala » Wed May 13, 2009 12:40 pm
moonstone1012001 wrote:Hi everyone,

I don't get how to factor this out, please help! :(

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

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

Thank you so much!
Let f(t) = t^2 - kt - 48

Since t-8 is a factor of f(t), f(8) = 0

i.e 8^2 - 8k -48 = 0

i.e. k = 2
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