Question

If the zeroes of the polynomial f(x) = x3 12x2 + 39x + a are in ap, find the value of

a.​

Answer 1

Step-by-step explanation:

Since 'a' is a zero of the polynomial f(x). Hence, the value of k is -28.

Answer 2

Step-by-step explanation:

f(x) = x^3 - 12x^2 + 39x + a

Since, roots of this equation are in A.P.

Let a−d,a,a+d are roots.

Now, sum of roots = -b/a

a−d+a+a+d= 12/1

3a=12

a=4

Sum of products of two consecutive roots = c/a

(a−d)a+a(a+d)+(a−d)(a+d)= 39/1

a^2 - ad + a^2 + ad + a^2 - d^2 = 39

3a^2 - d^2 = 39

3 × 16 -d^2 = 39

48 - d^2 = 39

d^2 =

d^2 = ±3

Therefore, the roots are 1, 4, 7 or 7, 4, 1

Now, product of roots=(a−d)a(a+d) = -d/a

1×4×7=−a

a = -28

Hence, the value of a is −28.

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