Question

if the zeros of the polynomial f(x)=2x^3-15x^2+37x-30 are in A.P;find them​

Answer 1

       $\underline{\bold{Solution:}}$

       $\text{The given equation is }f(x) = 2x^{3} - 15x^{2} + 37x - 30 = 0$

       $\text{It is also given that the roots of }f(x) \text{ are in A.P}$

       $\text{Let the roots of the equation be }(a-d),a,(a+d)$

       $\text{Applying the relationship between roots and coefficients}$

       $\text{Sum of roots} = \dfrac{-b}{a}$

$\implies a-d+a+a+d = \dfrac{-(-15)}{2}$

$\implies 3a = \dfrac{15}{2}$

$\implies a = \dfrac{5}{2}$

       $\text{Product of roots} = \dfrac{-d}{a}$

$\implies (a-d)(a)(a+d) = \dfrac{-(-30)}{2}$

       $\text{We know that }(y+z)(y-z) = y^{2} - z^{2}$

$\implies (a^{2} - d^{2})(a) = \dfrac{30}{2}$

$\implies (a^{2}-d^{2})(a) = 15$

        $\text{We know that }a = \dfrac52$

$\implies \{(\dfrac52)^{2} - d^{2}\}(\dfrac52) = 15$

$\implies \dfrac{25}{4} - d^{2} = 6$

$\implies -d^{2} = 6 - \dfrac{25}{4}$

$\implies -d^{2} = \dfrac{24-25}{4}$

$\implies d^{2} = \dfrac{1}{4}$

$\implies d = +\dfrac12,-\dfrac12$

       $\text{We can take either value of }d\text{ as we will get the same roots}$

$\implies\text{The roots are:}$

       $(a-d) = \dfrac52 - \dfrac12 = 2$

       $a = \dfrac52 = 2.5$

       $(a+d) = \dfrac52 + \dfrac12 = 3$

$\therefore \text{ }\text{ }\boxed{\text{The roots of }f(x) \text{ are }2,2.5,3}$