Finding the Value of k Using A.P. Condition on Zeroes

Video Explanation

Question

If the zeroes of the polynomial

\[ f(x) = x^3 – 12x^2 + 39x + k \]

are in arithmetic progression, find the value of \(k\).

Solution

Step 1: Assume the Zeroes in A.P.

Let the three zeroes be

\[ a-d,\; a,\; a+d \]

where \(a\) is the middle term and \(d\) is the common difference.

Step 2: Use Relations Between Zeroes and Coefficients

For the given polynomial \[ x^3 – 12x^2 + 39x + k, \]

we have

\[ a = 1,\quad b = -12,\quad c = 39,\quad d = k. \]

(i) Sum of the zeroes

\[ (a-d) + a + (a+d) = 3a \]

But,

\[ \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-12}{1} = 12 \]

So,

\[ 3a = 12 \Rightarrow a = 4 \]

(ii) Sum of the products of zeroes taken two at a time

\[ (a-d)a + a(a+d) + (a-d)(a+d) \]

\[ = a^2 – ad + a^2 + ad + (a^2 – d^2) = 3a^2 – d^2 \]

But,

\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = 39 \]

So,

\[ 3a^2 – d^2 = 39 \]

Substituting \(a = 4\),

\[ 3(4)^2 – d^2 = 39 \Rightarrow 48 – d^2 = 39 \]

\[ \Rightarrow d^2 = 9 \Rightarrow d = 3 \]

(iii) Product of the zeroes

\[ (a-d)a(a+d) = a(a^2 – d^2) \]

But,

\[ \alpha\beta\gamma = -k \]

So,

\[ 4(16 – 9) = -k \]

\[ 28 = -k \Rightarrow k = -28 \]

Conclusion

The value of \(k\) for which the zeroes of the polynomial \[ x^3 – 12x^2 + 39x + k \] are in arithmetic progression is

\[ \boxed{k = -28} \]