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by neelgandham » Sat Aug 18, 2012 12:45 pm
If (t - 8) is a factor of t^2 -kt - 48, then
k =
(A) -6
(B) -2
(C) 2
(D) 6
(E) 14

Let (t-8)*(t+a) = t^2 -kt - 48
t^2 +(a-8)t - 8a = t^2 -kt - 48
Comparing the constants, -8a = -48. i.e. a = 6
Comparing the coefficients of t, a-8 = -k, i.e 6-8 = -k, k = 2
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by theCEO » Sat Aug 18, 2012 12:59 pm
grandh01 wrote:If (t - 8) is a factor of t^2 -kt - 48, then
k =
(A) -6
(B) -2
(C) 2
(D) 6
(E) 14
(t-8) * (t+x) ---- x = 48/8 = 6
(t-8) * (t+6)
t^2 + 6t - 8t -48
t^2 - 2t - 48 = t^2 - kt - 48
2t = kt
k = 2
ans = c
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by pemdas » Sat Aug 18, 2012 2:12 pm
grandh01 wrote:If (t - 8) is a factor of t^2 -kt - 48, then
k =
(A) -6
(B) -2
(C) 2
(D) 6
(E) 14
I don't know ...
why we should equate quadratic function to zero; some taste of gmatters ;)
f(x)=t^2 -kt -48 ===> t^2 -kt -48=-48, t^2 -kt=0 and t(t-k)=0. We get t=0 and t=k and the two coordinates for parabola (0,-48) and (k,-48). Since the quadratic function has positive coefficient for t^2 our parabola opens upwards, and the vertex of parabola will be placed at (x,y) where y<-48. So f(x)<-48 and t^2 -kt -48 < -48, t(t-k)<0, t<0 and t<k. If you noticed answer choices C, D, E will be correct for the function given.
let's review the options
C) k=2, 2>0>t and the vertex of parabola is set at x=1 (mid-point of [2-0]/2=1). y=1-2*1-48=-49. Hence the vertex coordinate is (1,-49)
D) k=6, x=3 and y=9-6*3-48=-57. The vertex coordinate is (3,-57)
E) k=14, x=7 and y=49-14*7-48=-97. The vertex coordinate is (7,-97)

I believe this question is very badly designed by a non-mathematician
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