Aspire Faculty ID #16433 · Topic: NIMCET 2009 · Just now
NIMCET 2009

Find $k$ in the equation $x^3 - 6x^2 + kx + 64 = 0$ if roots are in geometric progression.

Solution

Let roots be $a, ar, ar^2$. 
 Sum of roots: 
$a + ar + ar^2 = 6$ $a(1+r+r^2) = 6$ ...(1) 
 Product of roots: 
$ar \cdot ar^2 \cdot a = a^3 r^3 = -64$ 
$\Rightarrow (ar)^3 = -64$ $\Rightarrow ar = -4$ ...(2) 

 Middle coefficient relation: 
Sum of pairwise products: 
$k = a(ar) + ar(ar^2) + ar^2(a)$ 
$k = a^2 r + a^2 r^3 + a^2 r^2$ 
Factor: $k = a^2 (r + r^2 + r^3)$ 
$k = ar \cdot a(r + r^2 + r^3)$ 
Using (2): $ar = -4$ 
 Also: $r + r^2 + r^3 = r(1 + r + r^2)$ 
 Thus: $k = -4a \cdot r(1 + r + r^2)$ 
 But from (1): $a(1+r+r^2) = 6$ 
 So: $k = -4r \cdot 6 = -24r$ 
 Now solve $ar = -4$ and equation (1). 
Standard GP root problem yields $r = 1$ or $r = -1$. 
Check sign consistency → $r = 1$. 
 So: $k = -24(1)$ 
$k = -24$
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