Question

if alpha beta gamma are the zeros of the polynomial p x is equal to 6 x cube + 3 x square - 5 x + 1 find the value of one upon alpha plus one upon beta + one upon them

Answer 1

Heya Frnd.......☺

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

$= > \: \frac{ \alpha \beta + \beta \gamma + \gamma \alpha }{ \alpha \beta \gamma }$........(1)

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NOW,

We know that :-

$\alpha \beta \: + \: \beta \gamma \: + \: \gamma \alpha = \: \frac{ - b}{a} = \: \frac{ - 3}{6} \\ = > \: \frac{ - 1}{2}$
...............................( 2 )

Also,

$\alpha \beta \gamma = \: \frac{ - d}{a} = \: \frac{ - 1}{6}$
................................( 3 )

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NOW,

BY ( 1 ) , (2) & (3)

WE GET ==>>

$\frac{ \alpha \beta + \beta \gamma + \gamma \alpha }{ \alpha \beta \gamma } \\ \\ = > \: \frac{ \frac{ - 1}{2} }{ \frac{ - 1}{6} }$

$= > \: \frac{2}{6}$

$= > \: \frac{1}{3} ........ \infty \: ANS \: \infty$

________☺☺☺_______

$HOPE IT WILL HELP U..............^$☺

Answer 2

HELLO DEAR,

we know that in cubic Polynomial

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ = > \alpha + \beta + \gamma = \frac{ - 3}{6} = - \frac{1}{2} \\ = > \frac{1}{ \alpha } + \frac{1}{ \beta } + \frac{1}{ \gamma } = - 2$

I HOPE ITS HELP YOU DEAR,
THANKS