Question

$if \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \: \\ polynomials \: ax {}^{2} + bx \: + c. \: then \: \: \\ find \: the \: other \: polynomial \: \\ whose \: zeroes \: are \: \frac{ \alpha {}^{2} }{ \beta } \: and \: \frac{ \beta {}^{2} }{ \alpha } .$
$note \: = > \\ answer \: of \: these \: question \: = \\ x {}^{2} + \frac{b}{a {}^{2} c} (b {}^{2} - 3ac)x + \frac{c}{a}$
please answer step by step, briefly..
for your help I provided you the answer.​

Answer 1

$\bf\large \underbrace \orange{Solution:}$

$\bf \: \: since \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \:$

$\bf \:polynomials \: \: ax {}^{2} + bx \: + c \: \: then$

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

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

★ Now, Consider

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

$\bf\implies \:S = \dfrac{ { \alpha }^{3} + { \beta }^{3} }{ \alpha \beta }$

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

$\bf\implies \:S = \dfrac{ { - (\dfrac{b}{a} })^{3} + 3\dfrac{c}{a} \times \dfrac{b}{a} }{\dfrac{c}{a} }$

$\bf\implies \:S = \dfrac{ (- {b}^{3 } + 3abc) }{ {a}^{3} } \times \dfrac{a}{c}$

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

$\bf\implies \:S = \dfrac{ - b( {b}^{2} - 3ac) }{c {a}^{2} }$

★ Now, Consider

$\bf \:P = \dfrac{ { \alpha }^{2} }{ \beta } \times \dfrac{ { \beta }^{2} }{ \alpha }$

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

★ So, required quadratic polynomial is

$\bf\implies f(x) = \: {x}^{2} - Sx + P$

$\bf\implies \: {x}^{2} - \dfrac{ - b( {b}^{2} - 3ac) }{c {a}^{2} } x + \dfrac{c}{a}$

$\bf\implies \: {x}^{2} + \dfrac{ b( {b}^{2} - 3ac) }{c {a}^{2} } x + \dfrac{c}{a}$

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