Question

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 post any irrelevant answer.. Class 10 maths ​

Answer 1

Answer:

-10/9

Step-by-step explanation:

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

Answer 2

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

Given that,

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {3x}^{2} - 5x - 9.$

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha + \beta = - \dfrac{( - 5)}{3} = \dfrac{5}{3} - - - (1)$

Also,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha \beta = \dfrac{ - 9}{ 3} = - 3 - - - (2)$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta }$

$\rm \: \: = \: 2\bigg(\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \bigg)$

$\rm \: \: = \: 2\bigg(\dfrac{ \beta + \alpha }{ \alpha \beta } \bigg)$

$\rm \: \: = \: 2 \times \dfrac{( - 5)}{3} \times \dfrac{1}{3}$

$\rm \: \: = \: - \: \dfrac{10}{9}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:\:\dfrac{2}{ \alpha } + \dfrac{2}{ \beta } = - \: \dfrac{10}{9} }}$

Additional Information :-

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

then,

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

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

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