NIMCET 2009 Question Paper with solutions

Question : 31 of 120

Marks: +1, -0

Find the value of

k

in the equation

x^{3}-6 x^{2}+k x+64=0

, if it is known that the roots of the equation are in geometric progression.

Solution:

Let

\;{a}/{r}, a

and

a r

be the three roots. (product of roots)
Then,

\;{a}/{r} × a × a r=-64

⇒ \;\; a^{3}=-64 ⇒ a=-4

Sum of the roots

⇒ \;\; -4(1+r+\;{1}/{r})=6

⇒ \;\; r+\;{1}/{r}=-\;{5}/{2} ⇒ r=-2

Now, sum of the roots taken two at a time

2(-4)+(+16)(-2)+8(2)=k

⇒ \;\; k=-24

Go to Question:

Prev Question

Next Question