Question

the polynomial whose zeroes are underroot 2+1 and underoot 2 -1 is

Answer 1

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

Given zeroes of the polynomial are

$\rm :\longmapsto\: \sqrt{2} + 1$

and

$\rm :\longmapsto\: \sqrt{2} - 1$

Let assume that

$\rm :\longmapsto\: \alpha = \sqrt{2} + 1$

and

$\rm :\longmapsto\: \beta = \sqrt{2} - 1$

So, Consider

$\rm :\longmapsto\: \alpha + \beta$

$\rm \:  =  \: ( \sqrt{2} + 1) + ( \sqrt{2} - 1)$

$\rm \:  =  \: \sqrt{2} + 1+ \sqrt{2} - 1$

$\rm \:  =  \: 2 \sqrt{2}$

$\red{\rm\implies \: \alpha + \beta = 2 \sqrt{2}}$

Now, Consider

$\rm :\longmapsto\: \alpha \beta$

$\rm \:  =  \: ( \sqrt{2} + 1)( \sqrt{2} - 1)$

$\rm \:  =  \: {( \sqrt{2} )}^{2} - {1}^{2}$

$\rm \:  =  \: 2 - 1$

$\rm \:  =  \: 1$

$\red{\rm\implies \: \alpha \beta = 1}$

So, Required Quadratic polynomial is given by

$\red{\rm :\longmapsto\:f(x) = k\bigg[{x}^{2} - ( \alpha + \beta )x + \alpha \beta \bigg], \: k \: \ne \: 0}$

$\red{\rm :\longmapsto\:f(x) = k\bigg[{x}^{2} - 2 \sqrt{2} x + 1 \bigg], \: k \: \ne \: 0}$

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More to Know,

For Cubic Polynomial

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

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

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

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

For Quadratic polynomial

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

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

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