Question

If the zeros of the polynomial f(x) = $2x^3 - 15x^2 + 37x - 30$ are in AP, find them.

Answer 1

Hint: To solve this problem, one should have theoretical knowledge of cubic equations and 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:
 Let the roots of the polynomial be a - d, a, a + d where d is the common difference of the AP. Applying the above three formulas-
$$\left(\mathrm a-\mathrm d\right)+\mathrm a+\left(\mathrm a+\mathrm d\right)=\dfrac{15}2\\3\mathrm a=\dfrac{15}2\\\mathrm a=\dfrac52$$
Now the product of roots is given by-
$$a\left(a-d\right)\left(a+d\right)=\dfrac{30}2=15\\\dfrac52\left(\dfrac52-d\right)\left(\dfrac52+d\right)=15\\\dfrac{25}4-d^2=6\\d^2=\dfrac14\\d=\pm\dfrac12$$
Applying the values of a and d we get the roots of a the equation as-
$$2,\dfrac{\;5}2,\;3$$

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.