If alpha and beta are the zeros of polynomial 3 x^2-5x-9 find value of 2/ Alpha + 2/ beta.. Please solve this question fast, don’t

2 thoughts on “If alpha and beta are the zeros of polynomial 3 x^2-5x-9 find value of 2/ Alpha + 2/ beta.. Please solve this question fast, don’t”

1. Answer:

   -10/9

   Step-by-step explanation:

   I have already answered this question, you can check in my maths section recent answers.

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2. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that,

   [tex]\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {3x}^{2} – 5x – 9.[/tex]

   We know that,

   [tex]\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}[/tex]

   [tex]\bf\implies \: \alpha + \beta = – \dfrac{( – 5)}{3} = \dfrac{5}{3} – – – (1)[/tex]

   Also,

   [tex]\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}[/tex]

   [tex]\bf\implies \: \alpha \beta = \dfrac{ – 9}{ 3} = – 3 – – – (2)[/tex]

   Now,

   Consider,

   [tex]\rm :\longmapsto\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } [/tex]

   [tex]\rm \: \: = \: 2\bigg(\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \bigg) [/tex]

   [tex]\rm \: \: = \: 2\bigg(\dfrac{ \beta + \alpha }{ \alpha \beta } \bigg) [/tex]

   [tex]\rm \: \: = \: 2 \times \dfrac{( – 5)}{3} \times \dfrac{1}{3} [/tex]

   [tex]\rm \: \: = \: – \: \dfrac{10}{9} [/tex]

   [tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } = – \: \dfrac{10}{9} }}[/tex]

   Additional Information :-

   [tex]\rm :\longmapsto\: \alpha,\beta \: and \: \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d[/tex]

   then,

   [tex]\green{\boxed{ \tt \: \alpha + \beta + \gamma = – \: \dfrac{b}{a} }}[/tex]

   [tex]\green{\boxed{ \tt \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}[/tex]

   [tex]\green{\boxed{ \tt \: \alpha \beta \gamma = – \: \dfrac{d}{a} }}[/tex]

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