Question

If alpha,beta,sigma are zeroes of polynomial 2X^3 - 4X^2 +6X^ + 9 = 0, then find the value of alpha^-1 + beta^-1 + sigma^-1

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\underline{\text{Formulation :}}$

$\text{Let us consider the following polynomial of degree n}$

$\mathrm{f(x)=a_{0}x^{n}+a_{1}x^{n-1}+...+a_{n}}$

$\mathrm{where\:a_{0},\:a_{1},...\:a_{n}\:are\:given\:numbers\:with\:a_{0}\neq 0}$

$\mathrm{If\:x_{1},\:x_{2},\:...,\:x_{n}\:be\:the\:zeroes,}$

$\mathrm{\sum x_{1}=-\frac{a_{1}}{a_{0}}}$

$\mathrm{\sum x_{1}x_{2}=\frac{a_{2}}{a_{0}}}$

$\mathrm{\sum x_{1}x_{2}x_{3}=-\frac{a_{3}}{a_{0}}}$

$\mathrm{\;\;\;... ... ...}$

$\mathrm{x_{1}x_{2}...x_{n}=(-1)^{n}\frac{a_{n}}{a_{0}}}$

$\underline{\text{Solution :}}$

$\text{The given polynomial is}$

$\mathrm{P(x)=2x^{3}-4x^{2}+6x+9}$

$\mathrm{If\:\alpha,\:\beta,\:\gamma\:be\:the\:zeroes}$

$\mathrm{\sum \alpha=-\frac{-4}{2}}$

$\to \boxed{\mathrm{\alpha+\beta+\gamma=2}}$

$\mathrm{\sum \alpha \beta=\frac{6}{2}}$

$\to \boxed{\mathrm{\alpha\beta+\beta\gamma+\gamma\alpha=3}}$

$\boxed{\mathrm{\alpha\beta\gamma=-\frac{9}{2}}}$

$\mathrm{Now,\:{\alpha}^{-1}+{\beta}^{-1}+{\gamma}^{-1}}$

$\mathrm{=\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}$

$\mathrm{=\frac{\beta\gamma+\gamma\alpha+\alpha\beta}{\alpha\beta\gamma}}$

$\mathrm{=\frac{3}{-\frac{9}{2}}}$

$\mathrm{=-3\times \frac{2}{9}}$

$\mathrm{=-\frac{2}{3}}$

$\to \boxed{\mathrm{{\alpha}^{-1}+{\beta}^{-1}+{\gamma}^{-1}=-\frac{2}{3}}}$

$\text{Hence, proved.}$