Question

$\huge \fbox \blue{question } : -$


$\dashrightarrow \rm \: if \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the$
$\rm \: quadratic \: polynomial \: f(x) = {ax}^{2} +$
$\rm + bx + c \: \: them \: evalute : -$

$\rm \: (1) \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$
$\rm \: (2) \: \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha }$


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Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha , \: \beta \: are \: zeroes \: of \: {ax}^{2} + bx + c$

We know,

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

$\bf\implies \: \alpha \beta = \dfrac{c}{a}$

And

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

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

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

$\rm \: = \: \bigg[ {\bigg( - \dfrac{b}{a} \bigg) }^{2} - \dfrac{2c}{a}\bigg] \div \dfrac{c}{a}$

$\rm \: = \: \bigg[\dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{2c}{a}\bigg] \div \dfrac{c}{a}$

$\rm \: = \: \bigg[\dfrac{ {b}^{2} - 2ac}{ {a}^{2} }\bigg] \times \dfrac{a}{c}$

$\rm \: = \: \dfrac{ {b}^{2} - 2ac}{ac}$

Consider,

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

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

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

On substituting the values, we get

$\rm \: = \: \bigg[ {\bigg( - \dfrac{b}{a} \bigg) }^{3} + \dfrac{3c}{a} \times \dfrac{b}{a} \bigg] \div \dfrac{c}{a}$

$\rm \: = \: \bigg[ - \dfrac{ {b}^{3} }{ {a}^{3} } + \dfrac{3bc}{ {a}^{2} }\bigg] \div \dfrac{c}{a}$

$\rm \: = \: \bigg[\dfrac{ - {b}^{3} + 3abc}{ {a}^{3} }\bigg] \times \dfrac{a}{c}$

$\rm \: = \: \dfrac{ - {b}^{3} + 3abc}{c {a}^{2} }$

Answer 2

⇝ Given :-

$\rm \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the\: quadratic$

$\rm \: polynomial \: f(x) = {ax}^{2} + bx + c$

⇝ To Find :-

Values of

• $\rm \dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }$

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

⇝ Main Concept :-

For A Quadratic Equation of the Form

ax² + bx + c =0.

$\purple{ \text{Sum of Zeros} = \dfrac{ - \text b}{ \: \: \text a} }$

$\purple{ \text{Product of Zeros} = \dfrac{ \text c}{\text a} }$

⇝ Solution :-

Here,

$\text{Sum of Zeros} = \frac{ -\text b}{ \: \: \: \text a}$

$\large:\longmapsto \green{\pmb{ \alpha + \beta = \frac{ -\text b}{ \: \: \:\text a }}} \: - - - (1)$

And

$\text{Product of Zeros} = \frac{\text c}{\text b}$

$\large:\longmapsto \green{\pmb{ \alpha \beta = \frac{ \text c}{\text a }}} \: - - - (2)$

❒ Finding Value of $\pmb{\dfrac{ \alpha }{ \beta } + \dfrac{ \beta }{ \alpha }}$

$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$

$= \frac{ \alpha^2 + \beta^2 }{ \alpha \beta }$

⏩ Adding and Subtracting $\pmb{2\alpha \beta }$ in Numerator :

$= \frac{( \alpha {}^{2} + { \beta }^{2} + 2 \alpha \beta ) - 2 \alpha \beta }{ \alpha \beta }$

$= \frac{( \alpha + \beta {)}^{2} - 2 \alpha \beta }{ \alpha \beta }$

✈ Purtting Values from (1) and (2) :

$= \frac{ \bigg(\dfrac{ - \text b}{ \: \: \text a} \bigg)^{2} - 2 \times \dfrac{\text c}{\text a} }{ \dfrac{\text c}{\text a} }$

$= \frac{ \dfrac{ {\text b}^{2} }{ {\text a}^{2}} - {\dfrac{2\text c}{\text a} } }{ \dfrac{\text c}{\text a} }$

$= \dfrac{ \dfrac{ {\text b}^{2} }{\text a} - 2\text c }{\text c}$

$= \frac{ {\text b}^{2} }{\text a\text c} - 2$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{ {b}^{2} }{ac} - 2}}}}}$

--------------------------------

❒ Finding Value of $\pmb{\dfrac{ { \alpha }^{2} }{ \beta } + \dfrac{ { \beta }^{2} }{ \alpha }}$

$\frac{ \alpha {}^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha }$

$= \frac{ { \alpha }^{3} + { \beta }^{3} }{ \alpha \beta }$

⏩ Adding and Subtracting $\pmb{3\alpha \beta(\alpha+\beta) }$ in Numerator :

$= \frac{ { \alpha }^{3} + { \beta }^{3} + 3 \alpha \beta ( \alpha + \beta) - 3 \alpha \beta ( \alpha + \beta ) }{ \alpha \beta }$

$= \frac{ { (\alpha + \beta) }^{3} - 3 \alpha \beta ( \alpha + \beta )}{ \alpha \beta }$

✈ Purtting Values from (1) and (2) :

$= \frac{ \bigg(\dfrac{ - \text b}{ \: \: \: \text a} \bigg) ^{3} - 3 \times \dfrac{\text c}{\text a} \times \bigg( \dfrac{ - \text b}{\text a} \bigg) }{ \dfrac{\text c}{\text a} }$

$= \frac{ - \dfrac{ {\text b}^{3} }{\text a {}^{3} } + \dfrac{3\text b\text c}{ {\text a}^{2} } }{ \dfrac{\text c}{\text a} }$

$= \frac{ - {\text b}^{3} }{ {\text a}^{2}\text c } + \frac{3\text b}{\text a}$

Hence,

$\large\underline{\pink{\underline{\frak{\pmb{ \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } = \frac{ - { b}^{3} }{ { a}^{2} c } + \frac{3 b}{a} }}}}}$