Question

Form a quadratic polynomial whose zeroes are : (2+1/√2) and (2-1/√2)

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Answer 1

Step-by-step explanation:

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Answer 2

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

Given that

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

and

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

Now, in order to find a quadratic polynomial, we have to first find sum of zeroes and product of zeroes.

Consider,

Sum of zeroes,

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

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

$\rm \: = \: \: 4$

So,

$\bf\implies \: \alpha + \beta = 4 - - - (1)$

Consider,

Product of zeroes,

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

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

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

$\rm \: = \: \: 4 - \dfrac{1}{2}$

$\rm \: = \: \: \dfrac{8 - 1}{2}$

$\rm \: = \: \: \dfrac{7}{2}$

$\bf\implies \: \alpha \beta = \: \: \dfrac{7}{2}$

Now,

Let assume that the required Quadratic polynomial be f(x).

So,

The required Quadratic polynomial is given by

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

On substituting the values, we get

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - 4x + \dfrac{7}{2} \bigg) \: where \: k \ne \: 0$

$\bf :\longmapsto\:f(x) = \dfrac{k}{2} \bigg( {2x}^{2} - 8x + 7 \bigg) \: where \: k \ne \: 0$

Additional Information :-

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

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

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

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

$\green{ \boxed{ \rm{\rm :\longmapsto\: \: \alpha + \beta + \gamma = - \: \dfrac{b}{a} }}}$

$\green{ \boxed{ \rm{\rm :\longmapsto\: \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}}$

$\green{ \boxed{ \rm{\rm :\longmapsto\: \: \alpha \beta \gamma = - \: \dfrac{d}{a} }}}$