Question

find the value alpha upon beta + beta upon alfa for 3x²+2x+5​

Answer 1

Answer:

YZ and XZ

Step-by-step explanation:

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Answer 2

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: 3{x}^{2} + 2x + 5$

We know that,

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

$\bf\implies \: \alpha \beta = \dfrac{5}{3}$

and

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

$\bf\implies \: \alpha + \beta = - \: \dfrac{2}{3}$

Now, Consider

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

$\rm \: = \:\dfrac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta }$

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

On substituting the values, we get

$\rm \: = \:\bigg[ {\bigg( - \dfrac{2}{3} \bigg) }^{2} - 2 \times \dfrac{5}{3} \bigg] \div \dfrac{5}{3}$

$\rm \: = \:\bigg[\dfrac{4}{9} - \dfrac{10}{3} \bigg] \times \dfrac{3}{5}$

$\rm \: = \:\bigg[\dfrac{4 - 30}{9}\bigg] \times \dfrac{3}{5}$

$\rm \: = \:\bigg[\dfrac{- 26}{9}\bigg] \times \dfrac{3}{5}$

$\rm \: = \: - \: \dfrac{26}{15}$

Hence,

$\rm :\longmapsto\: \boxed{ \bf{ \: \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha } = - \frac{26}{15}}}$

Additional Information :-

$\boxed{ \bf{ \: {( \alpha + \beta )}^{2} - 2 \alpha \beta = { \alpha }^{2} + { \beta }^{2} }}$

$\boxed{ \bf{ \: {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) = { \alpha }^{3} + { \beta }^{3} }}$

$\boxed{ \bf{ \: {( \alpha - \beta )}^{2} = {( \alpha + \beta) }^{2} - 4 \alpha \beta }}$