Question

Answer the questions with steps ​

Answer 1

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

Given that

$\rm :\longmapsto\: \alpha, \: \beta \: are \: zeroes \: of \: f(x) = {x}^{2} - px + q$

We know that,

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

$\rm :\implies\: \alpha + \beta = - \: \dfrac{( - p)}{1} = p$

Also,

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

$\rm :\implies\: \alpha \beta = \dfrac{q}{1} = q$

Now,

Consider,

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

$\rm \:  =  \:  \: \dfrac{ { \alpha }^{4} + { \beta }^{4} }{ { \alpha }^{2} { \beta }^{2} }$

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

We know,

$\underbrace{\boxed{ \tt{ {x}^{2} + {y}^{2} \: = \: {(x + y)}^{2} - 2xy}}}$

So, using this identity, we get

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

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

So, on substituting the values, we get

$\rm \:  =  \:  \: \dfrac{ {( {p}^{2} - 2q) }^{2} - 2 {q}^{2} }{ {q}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {p}^{4} + {4q}^{2} - 4 {p}^{2} q - {2q}^{2} }{ {q}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {p}^{4} + {2q}^{2} - 4 {p}^{2} q}{ {q}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {p}^{4} }{ {q}^{2} } + \dfrac{ {2q}^{2} }{ {q}^{2} } - \dfrac{ {4p}^{2} q}{ {q}^{2} }$

$\rm \:  =  \:  \: \dfrac{ {p}^{4} }{ {q}^{2} } + 2 - \dfrac{ {4p}^{2}}{ {q}}$

$\rm \:  =  \:  \: \dfrac{ {p}^{4} }{ {q}^{2} } - \dfrac{ {4p}^{2}}{ {q}} + 2$

Hence, Proved

Additional Information :-

$\rm :\longmapsto\: \alpha,\beta, \gamma \: are \: zeroes \: of \: f(x) = {ax}^{3} + {bx}^{2} + cx + d, \: then$

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

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

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