Question

if alpha and beta are zeroes of the polynomial x^2-4x+3 the form a quadratic polynomial whose zeroes are 1/alpha and 1/beta​

Answer 1

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

Given that

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {x}^{2} - 4x + 3$

We know,

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

$\rm :\implies\: \alpha + \beta = - \: \dfrac{( - 4)}{1} = 4$

And

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

$\rm :\implies\: \alpha \beta = \dfrac{3}{1} = 3$

Now,

Consider,

$\rm :\longmapsto\:Sum \: of \: zeroes$

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

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

$\rm \: \: = \: \: \dfrac{4}{3}$

$\bf\implies \:\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = \dfrac{4}{3}$

Consider,

$\rm :\longmapsto\:Product \: of \: zeroes$

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

$\rm \: \: = \: \: \dfrac{1}{ 3 }$

$\bf\implies \:\dfrac{1}{ \alpha \beta } = \dfrac{1}{3}$

Hence,

The required Quadratic polynomial is

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

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - \dfrac{4}{3}x + \dfrac{1}{3}\bigg) \: where \: k \: \ne \: 0$

$\rm :\longmapsto\:f(x) = \dfrac{k}{3} \bigg( {3x}^{2} - 4x + 1\bigg) \: where \: k \: \ne \: 0$

Additional Information :-

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

$\boxed{ \sf{ \: \alpha + \beta + \gamma = - \: \frac{b}{a}}}$

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \frac{c}{a}}}$

$\boxed{ \sf{ \: \alpha \beta \gamma = - \: \frac{d}{a}}}$