Question

The roots of the polynomial $f(x) = x^3 - 12x^2 + 39x + k$ are in AP. Find the value of k.

Answer 1

Firstly, we will assume the roots as a-d, a, a+d since it is given that they are in AP. Then, we are given a cubic equation, so we must know the relationship between their coefficients and roots. The formulas that will be used to verify this relationship are-
$\mathrm\alpha+\mathrm\beta+\mathrm\gamma=\dfrac{-\mathrm b}{\mathrm a}\\\\\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{\mathrm\beta\mathrm\gamma}+\mathrm{\mathrm\gamma\mathrm\alpha}=\dfrac{\mathrm c}{\mathrm a}\\\\\mathrm{\mathrm\alpha\mathrm\beta\mathrm\gamma}=\dfrac{-\mathrm d}{\mathrm a}$

Complete step-by-step answer :
To begin with the solution, first we have to assume the roots of the cubic equation. We know that a cubic equation has three roots. Also, it is given that roots are in AP. The terms in an AP are such that they have a common difference between them.
So, let us assume the roots as a-d, a, a+d. From the given polynomial the sum of roots can be written as-
$a - d + a + a + d = 12$
$3a = 12$
$a = 4$

Next, let us write the sum of product of roots, which is given by-
$a(a - d) + a(a + d) + (a - d)(a + d) = 39$
$4(4 - d) + 4(4 + d) + 16 - d^2 = 39$
$d^2 = 48 - 39 = 9$
$d = 3$

From the values of a and d we can see the roots as 1, 4 and 7.

The product of roots is given by-
$1 \times 4 \times 7 = k = 28$

Hence, the value of k is 28.

Note : In such types of questions be careful in what we assume the roots. In this question, we assumed the roots as $a-d$, $a$ and $a+d$ so that d gets cancelled and we get the value of a in the first equation itself. This makes the calculation a lot easier.