Question

If alpha and bita are the zeroes of the polynomial 5x^2 - 4x + 10, find the quadratic polynomial whose zeroes are 1/alpha and 1/bita....
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Answer 1

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

Given that

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {5x}^{2} - 4x + 10}$

We know that

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

So,

$\rm :\longmapsto\: \alpha \beta = \dfrac{10}{5} = 2$

Also,

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

So,

$\rm :\longmapsto\: \alpha + \beta = - \dfrac{( - 4)}{5} = \dfrac{4}{5}$

Now,

We have to form a quadratic polynomial whom zeroes are

$\red{\rm :\longmapsto\:\dfrac{1}{ \alpha }, \dfrac{1}{ \beta }}$

Let, we first find the sum of zeroes.

Consider,

$\red{\rm :\longmapsto\:\dfrac{1}{ \alpha } + \dfrac{1}{ \beta }}$

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

$\rm \: = \: \: \dfrac{4}{5} \times \dfrac{1}{2}$

$\rm \: = \: \: \dfrac{2}{5}$

Now, we find product of zeroes.

Consider,

$\red{\rm :\longmapsto\:\dfrac{1}{ \alpha } \times \dfrac{1}{ \beta }}$

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

$\rm \: = \: \: \dfrac{1}{2}$

So,

Required quadratic polynomial is given by

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

On substituting the values, we get

$\rm :\longmapsto\:p(x) = k\bigg( { {x}^{2} - \dfrac{2}{5 }x} + \dfrac{1}{2} \bigg), \: where \: k \ne \: 0$

$\rm :\longmapsto\:p(x) = \dfrac{k}{10} \bigg( {10 {x}^{2} - 4x} + 5 \bigg), \: where \: k \ne \: 0$

Additional Information :-

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

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

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

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