Question

If the zeros of the polynomial f(x) = x³ –12x² + 39x + k are in A.P., find the value of k.

Answer 1

k = -28

Step-by-step explanation:

f(x) = x³ –12x² + 39x + k

Let the zeroes be α, β and γ.

Let α = a-d, β = a, and γ = a+d as the zeroes are in an AP.

Sum of the zeroes = -b/a

α + β + γ = 12

a-d + a +a +d = 12

3a =12

Therefore a = 4

Product of the zeroes = c/a

αβ + βγ + αγ = 39

a(a-d) + a(a+d) +(a+d)(a-d) = 39

a² -ad +a² +ad +a² - d² = 39

3a² - d² = 39

d² = 3a² - 39

   = 3(16) - 39

   = 48 - 39

   = 9

Therefore d = +-3

So the zeroes are:  

α = a-d = 4-3 = 1  or 7

β = a = 4

γ = a+d = 7  or 1

So the AP is 1, 4, 7.

-k = α * β * γ  

-k  = 1 * 4 * 7

Therefore k = -28