Question

Hello!

if ,
$\alpha \: and \: \beta$
are the zeroes of the polynomial p(x)=
${x}^{2} + x + 1$
then, find the sum of their reciprocals
i.e,
$( \frac{1}{ \alpha } + \frac{1}{ \beta } )$


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Answer 1

${\huge {\mathfrak {Answer :-}}}$

We have,

${} p(x) = x^2 + x + 1$

✨Comparing it with the general form of quadratic equation,

${} p(x) = ax^2 + bx + c$

◾We get that,

a = 1, b = 1, c = 1.

◾Now,

${} \alpha \: and \: \beta$

are the roots of the equation.

◾So, we have

${} \alpha + \beta$

= - b / a = - 1 / 1 = - 1.

◾And,

${} \alpha × \beta$

= c / a = 1 / 1 = 1.

✨Therefore,

${} \frac {1}{ \alpha} + \frac {1}{\beta}$

${} = \frac { \alpha + \beta }{ \alpha × \beta}$

◾Now, putting the values

= - 1 / 1 = - 1.

✨Hence, (-1) is the answer.

Answer 2

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