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Source: — Problem Solving |
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by vivek.kapoor83 » Fri Oct 10, 2008 8:30 am
SHorter method would be put the values of K back in the eqn and solve the quad eqn and check for which value of K, t= 8. It will not take more that 60sec.
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by 4meonly » Sat Oct 11, 2008 1:25 am
(t^2)- kt-48 = (t+6)(t-8 )
because we know that (t-8 ) is a factor of (t^2)- kt-48
solving (t+6)(t-8 ) = (t^2)-8t+6t-48 = t^2-2t-48

k=2
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by mental » Sat Oct 11, 2008 7:35 am
simplest way i can think might be.........

t-8 is a factor of quadratic equation t^2-kt-48

when we say this it means
t^2-kt-48 = 0
and t-8 =0

put t=8 in the equation

64-8k-48=0.............k=2
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