Question

If a polynomial p(x) has zeroes as 2 and 3, Write a polynomial whose zeros
are reciprocal of zeroes of p(x). ( With Explaintion please)

Answer 1

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

$\sf \: \: \: \: \: \: \: \: \: \bull \: \dfrac{1}{2} \: and \: \dfrac{1}{3} \: are \: zeroes \: of \: polynomial$

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

$\sf \: \: \: \: \: \: \: \: \: \bull \: the \: polynomial \: having \: zeroes \: \dfrac{1}{2} \: and \: \dfrac{1}{3}$

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

$\sf \: \: \: \: \: \: \: \: \: \bull \: Let \: f(x) \: be \: the \: required \: polynomial.$

$\sf \: \: \: \: \: \: \: \: \: \bull \: Let \: \alpha = \dfrac{1}{2} \: \: and \: \: \beta = \dfrac{1}{3}$

Now,

• Sum of zeroes is given by

$\rm :\longmapsto\: \alpha + \beta = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3 + 2}{6} = \dfrac{5}{6}$

and

• Product of zeroes is given by

$\rm :\longmapsto\: \alpha \beta = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$

Hence,

• The required polynomial f(x) is given by

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

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - \dfrac{5}{6} x + \dfrac{1}{6} \bigg) \: where \: k \ne \: 0$

$\rm :\longmapsto\:f(x) = \dfrac{k}{6} ( {6x}^{2} - 5x + 1) \: where \: k \ne \: 0$

Additional Information :-

$\sf \: Let \: f(x) = {ax}^{2} + bx + c \: be \: a \: quadratic \: polynomial \: then$

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

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

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

Answer 2

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