Question

If alpha, beta and gamma are the zeroes of the polynomial 2x3 – 2x2 + 3x – 4, then evaluate alpha/beta gamma + beta / gamma alpha + gamma/alpha beta

Answer 1

Answer:

on simplifying, you get

(alpha² + beta² + gamma²) / alpha.beta.gamma

= (1)²- 2(3/2) / 4 = -1/2

Answer 2

▪Given :-

α , β and γ Are Zeros of the Cubic Polynomial :

$\bf 2 {x}^{3} - 2 {x}^{2} + 3x - 4$

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▪To Find :-

We have to Find the value of :

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

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▪Key Concept :-

☆ Relationship b/w Zeros And Coefficients of a Cubic Polynomial

If α , β and γ are the Zeros of a Cubic Polynomial

of the Form:

$\bf a{x}^{3} + b {x}^{2} + cx + d$

Then,

$\bf \alpha + \beta + \gamma = - \frac { b}{a} \\ \\ \bf\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} \\ \\ \bf \alpha \beta \gamma = - \frac{d}{a}$

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▪Solution :-

Here we have,

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

$\bf \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{3}{2} \\ \\ \bf \alpha \beta \gamma = - \frac{ ( - 4)}{2} = 2$

Now ,

$\frac{ \alpha }{ \beta \gamma } + \frac{ \beta }{ \gamma \alpha } + \frac{ \gamma }{ \alpha \beta } \\ \\ = \frac{ { \alpha }^{2} + { \beta }^{2} + { \gamma }^{2} }{ \alpha \beta \gamma } \\ \\ \bf \bigg \{ \frac{ \large Add \: and \: Subtract}{ \underline{2( \alpha \beta + \beta \gamma + \gamma \alpha ) \: in \: Numerator } }\bigg \}$

$= \frac{( { \alpha + \beta + \gamma )}^{2} - 2( \alpha \beta + \beta \gamma + \gamma \alpha) }{ \alpha \beta \gamma } \\ \\ \bf \Large\{ Putting \: \: values \}$

$= \dfrac{ {(1)}^{2} - 2( \dfrac{3}{2} )}{2} \\ \\ = \frac{1 - 3}{2} \\ \\ = \frac{ - 2}{ \: \: \: 2} \\ \\ = \huge\purple{{ {\bf \: - 1}}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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