Question

if zero's if the polynomial are 1/3x^2 + 12x + K -2 are reciprocal of each other find K class 10​

Answer 1

$\large \bf \clubs \:  Given :-$

Zeros of Polynomial $\bf \dfrac{1}{3} {x}^{2} + 12x + K - 2$

are Reciprocal of Each other.

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$\large \bf \clubs \:  To \: Find :-$

• The Value of K

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$\large \bf \clubs \:  Main \: Concept :-$

For A Quadratic Polynomial of the Form :

 ax² + bx + c

$\sf{Sum  \: of  \: Zeros  = - \dfrac{b}{a} } \\  \\ \sf{ Product \:  of \:  Zeros =  \frac{c}{a} }$

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$\large \bf \clubs \:  Solution :-$

Let,

• First Zero of the Polynomial be = α

• Second Zero must be = 1/α

Hence ,

$\sf Product \: of \: Zeros =\dfrac{K-2}{1/3}$

$\sf\large:\longmapsto \alpha \times \dfrac{1}{ \alpha } =\dfrac{K-2}{1/3}$

$:\longmapsto\dfrac{1}{3} = K - 2 \\ \\ :\longmapsto K = \dfrac{1}{3}+2=\dfrac{1+6}{3}\\\\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf K = \dfrac{7}{3}} }}}$

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Let Zeoros of the Given Polynomial be -

$\large \bf a \: \: and \: \: \dfrac{1}{a}$

We Know,

$\sf a \times \dfrac{1}{a} = \dfrac{k - 2}{ \dfrac{1}{3} } \\ \\ \frac{1}{3} = \sf k - 2 \\ \\ \sf k = 2 + \frac{1}{3} \\ \\ \huge :\implies \bf \red{k = \frac{7}{3} }$

Which Is The Answer