Question

if alpha beta and gamma are the zeros of the polynomial f(x)=x³-3px²+qx-r such that 2 beta=alpha+gamma then show that 2p³=pq-r​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: \alpha, \: \beta , \: \gamma \: are \: zeroes \: of \: {x}^{3} - {3px}^{2} + qx - r$

and

$\rm :\longmapsto\: 2\beta = \gamma + \alpha$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\: \alpha + \beta + \gamma = - \dfrac{( - 3p)}{1}$

$\rm :\longmapsto\:2 \beta + \beta = 3p$

$\rm :\longmapsto\:3\beta = 3p$

$\rm :\longmapsto\: \beta = p$

Since,

$\rm :\longmapsto\: \: \beta\: is\: zero\: of \: {x}^{3} - {3px}^{2} + qx - r$

$\rm :\longmapsto\: \: p\: is\: zero\: of \: {x}^{3} - {3px}^{2} + qx - r$

$\rm :\longmapsto\: \: {p}^{3} - {3p(p)}^{2} + qp - r = 0$

$\rm :\longmapsto\: \: {p}^{3} - {3p}^{3} + qp - r = 0$

$\rm :\longmapsto\: \:- {2p}^{3} + qp - r = 0$

$\rm :\longmapsto\: \:{2p}^{3} = qp - r$

Hence, Proved

Additional Information :-

$\rm :\longmapsto\: \alpha, \: \beta , \: \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d \: then$

$\boxed{ \sf \: \alpha + \beta + \gamma = - \: \dfrac{b}{a}}$

$\boxed{ \sf \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a}}$

$\boxed{ \sf \: \alpha\beta \gamma = - \: \dfrac{d}{a}}$