Question

if alpha and beta are zeroes of the polynomial p(x) =3x² -2x -6 then find 1/ alpha + 1/ beta​

Answer 1

Solution :

Given polynomial :

> p(x) = 3x^2 - 2x - 6 .

The roots of this polynomial are alpha and beta .

We need to find the value of 1/alpha + 1/beta.

One method to do this is to solve the expression and directly obtain the roots & substitute.

The other method is by comparing the coefficients.

We will be going with the later.

1/alpha + 1/beta

> ( alpha + beta )/( alpha beta )

For a polynomial of the form ax^2 + bx + c having the roots alpha and beta ;

alpha + beta = -b/a

alpha beta = c/a

> ( -b/a ) / ( c/a )

> ( -b/a ) * ( a/c)

> ( -b/c)

b = -2 and c = -6 in the original polynomial

Required value :

> ( 2/-6)

> -1/3 .

This is the answer.

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Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• if alpha and beta are zeroes of the polynomial p(x) =3x² -2x -6 then find 1/ alpha + 1/ beta

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$\huge{AηsωeR} ✍$

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$\large\underline\blue{\bold{Given :-  }}$

• A polynomial p(x) =3x² -2x -6 having zeroes alpha and beta.

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$\large\underline\blue{\bold{To \: Find :-  }}$

$\sf \:  \dfrac{1}{ \alpha } + \dfrac{1}{ \beta }$

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$\large\underline\blue{\bold{Formula \: Used :-  }}$

Let us consider a quadratic polynomial p(x) =ax² + bx + c having zeroes p and q,

$\begin{gathered}\begin{gathered}\bf then= \begin{cases} &\sf{p + q = - \dfrac{b}{a} } \\ &\sf{pq = \dfrac{c}{a} } \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

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$\sf \:  since \: \alpha \: and \: \beta \: are \: zeroes \: of \: p(x) =3x² -2x -6$

$\begin{gathered}\begin{gathered}\bf \begin{cases} &\sf{ \alpha + \beta = - \dfrac{ - 2}{3} = \dfrac{2}{3} } \\ &\sf{ \alpha \beta = \dfrac{ - 6}{3} = - 2} \end{cases}\end{gathered}\end{gathered}$

$\sf \:  Now, \: \dfrac{1}{ \alpha } + \dfrac{1}{ \beta }$

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

$\sf \:  = \dfrac{\dfrac{2}{3} }{ - 2}$

$\sf \:  = - \dfrac{1}{3}$

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$\large{\boxed{\boxed{\bf{Hence, \sf \:  \dfrac{1}{ \alpha } + \dfrac{1}{ \beta } = - \dfrac{1}{3} }}}}$