Question

If alpha, beta, gamma are zeroes of x3+3x2-x-3 then find 1/alpha+1/beta+1/gamma​

Answer 1

Answer:

1/alpha+1/beta+1/gamma=-1/3

Answer 2

Given : $\alpha, \beta, \gamma$ are roots of the equation,

$x^3+3x^2-x-3=0\hfill (1)$

To find : the value of $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$

Step-by-step explanation: We know if the general equation of cubic be,

$ax^3+bx^2+cx+d=0$ and $\alpha, \beta, \gamma$ are roots then,

$\alpha+\beta+\gamma=(-1)^1\times\frac{b}{a}$

$\alpha\beta+\beta\gamma+\gamma\alpha=(-1)^2\frac{c}{a}$

$\alpha\beta\gamma=(-1)^3\frac{d}{a}$

Hence,

$\alpga+\beta+\gamma=(-1)^1\times \frac{3}{1}=-3$

$\alpha\beta+\beta\gamma+\gamma\alpha=(-1)^2\frac{-1}{1}=1$

$\alpha\beta\gamma=(-1)^3\frac{-3}{1}=3$

Therefore,

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

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

$=-\frac{1}{3}$