Question

If alpha and beta are the roots of x2-2x+3 then find the value of alpha/beta2 + beta/alpha2

Answer 1

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

Given that

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

We know,

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

$\bf\implies \: \alpha + \beta = - \: \dfrac{( - 2)}{1} = 2$

And

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

$\bf\implies \: \alpha \beta = \dfrac{3}{1} = 3$

Now, Consider

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

$\rm \: = \: \: \dfrac{ { \alpha }^{3} + { \beta }^{3} }{ { \alpha }^{2} { \beta }^{2} }$

$\rm \: = \: \: \dfrac{ {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta )}{ {( \alpha \beta )}^{2} }$

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

$\rm \: = \: \: \dfrac{8 - 18}{9}$

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

Additional Information

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

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

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

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

Some useful Identities :-

$\boxed{ \rm{ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy}}$

$\boxed{ \rm{ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}$

$\boxed{ \rm{ {(x - y)}^{2} = {(x + y)}^{2} - 4xy}}$