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If the zeroes of the polynomial $f x = x 3 − 12 x 2 + 39 x + k$ are in AP, then the value of k is

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-28

answer is D.

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Detailed Solution

The correct answer is: -28
We know that for a cubic polynomial,

S u m o f r o o t s = α + β + γ = − B A

Sum of the product of roots = α β + β γ + γ α = C A

Product of roots = α β γ = D A

On comparing the given equation with

A x 3 + B x 2 + C x + D

A = 1, B = – 12, C = 39, D = k
On comparing the roots,

α = a, β = a + d, γ = a + 2 d

Now,

α + β + γ = − B A a + a + d + a + 2 d = − − 12 1 3 a + 3 d = 12 a + d = 4

α β + β γ + γ α = C A a a + d + a + d a + 2 d + a + 2 d a = 391 3 a 2 + 6 a d + 2 d 2 = 39

On substituting,

3 a 2 + 6 a 4 − a + 2 4 − a 2 = 39 3 a 2 + 24 a − 6 a 2 + 2 16 + a 2 − 8 a = 39 3 a 2 + 24 a − 6 a 2 + 32 + 2 a 2 − 16 a = 39 − a 2 + 8 a − 7 = 0 a 2 − 8 a + 7 = 0 a 2 − 7 a − a + 7 = 0 (a − 7) (a − 1) = 0 a = 7; a = 1

When a=7

a + d = 4 7 + d = 4 d = − 3

Using

α β γ = − D A

a × (a + d) × (a + 2 d) = − k 1 7 × 4 × 7 + 2 × − 3 = − k 28 × 1 = − k k = − 28

When a=1

a + d = 4 d = 3

Using

α β γ = − D A

a × a + d × a + 2 d = − k 1 1 × 4 × 1 + 2 × 3 = − k 1 × 4 × 7 = − k k = − 28

Hence, option 2 is correct.

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