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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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