Question

If
$\alpha \: and \: \beta \: are \: the \: zeros \: of \: the \: polynomial \:f(x) = {x }^{2} + px + q \: then \: \: a \: polynomial \: having \: \frac{1}{ \alpha } \: and \: \frac{1}{ \beta } \: is \: its \: zeros \: is \\ a) {x}^{2} + qx + p \\ \\ b) {x}^{2} - px + q \\ c)q {x}^{2} + px + 1 \\ d)p {x}^{2} + qx + 1$
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Answer 1

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Answer 2

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

$\: \: \: \: \: \: \: \bull \: \: \sf \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) = {x}^{2} + px + q$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \: \bull \: \: \sf \: a \: polynomial \: having \: zeroes \: \dfrac{1}{ \alpha } \: and \: \dfrac{1}{ \beta }$

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

• Given that,

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

We know that,

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

Or

$\boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

And

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

Or

$\boxed{\purple{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}$

So,

$\rm :\longmapsto\: \alpha + \beta = \dfrac{-b}{a} = \: - \: p$

And

$\rm :\longmapsto\: \alpha \beta = \: \dfrac{c}{a} = \: q$

Now,

Consider,

$\rm :\longmapsto\:S = \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \dfrac{ \alpha + \beta }{ \alpha \beta } = \dfrac{ - p}{q}$

And

$\rm :\longmapsto\:P = \dfrac{1}{ \alpha } \times \dfrac{1}{ \beta } = \dfrac{1}{ \alpha \beta } = \dfrac{1}{q}$

Now,

• We know that, a quadratic polynomial whose sum of zeroes is S and Product of zeroes is P, is given by

$\rm :\longmapsto\:p(x) =k({x}^{2} - Sx + P) \: where \: k \ne \: 0$

$\sf \: Hence, \sf \: a \: polynomial \: having \: zeroes \: \dfrac{1}{ \alpha } \: and \: \dfrac{1}{ \beta } is \: given \: by$

So, on substituting the values of S and P, we get

$\rm :\longmapsto\:p(x) = k \bigg({x}^{2} - (\dfrac{ - p}{q} )x + \dfrac{1}{q} \bigg) \: where \: k \ne \: 0$

$\rm :\longmapsto\:p(x) = \dfrac{k}{q} \bigg( {qx}^{2} + px + 1 \bigg) \:where \: k \ne \: 0$

$\overbrace{ \underline { \boxed { \rm \therefore \: The \: required \: polynomial \: = \: {qx}^{2} + px + 1 }}}$

$\bf\implies \:option \: (c) \: is \: correct$

Additional Information

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

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

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