Question

For what value of k , the roots of eq x³-6x²+kx+64=0 are in geometric progression?

Answer 1

Answer:

-24

Step-by-step explanation:

The eqn. is cubic. Therefore, it has three roots.

Let them be a/r, a and ar, with common ratio = r and first term = a/r

Product of roots = a/r*a*ar = a^3 = -d/a = -64

So, a = -4

Now, sum of roots = a/r + a + ar = a(1 + r + 1/r) = -4(1 + r + 1/r) = 6

r^2 + r + 1 = -3r/2

2r^2 + 5r + 2 = 0

(r + 2)(2r + 1) = 0

r = -2 or -1/2

So, the GP is 2, -4, 8 or 8, -4, 2, and these three are the roots.

Therefore, Sum of roots taken two at a time = 2(-4)+(-4)8+2(8) = -8 - 32 + 16              

                                                                                                      = -24.

- 24 = k/1  => k = -24.