Question

If alpha and beta are the roots of polynomials x square + x + 1 ,then value of 1 upon alpha + 1 upon beta is

Answer 1

Given :

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

To Find :

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

Solution :

$\implies\sf \alpha +\beta = - \frac{b}{a}\\ \\\sf \implies\alpha + \beta = - \frac{1}{1} \\ \\\implies\boxed{ \alpha + \beta = -1}$

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$\implies \sf \alpha \beta = \frac{c}{a} \\ \\\implies\sf \alpha \beta = \frac{1}{1} \\ \\\implies\boxed{\sf \alpha \beta = 1}$

Now

$\implies \frac{1}{ \alpha} + \frac{1}{ \beta} = \frac{ \beta + \alpha}{ \alpha \beta} \\ \\\implies\frac{1}{ \alpha} + \frac{1}{ \beta} = \frac{ - 1}{1} \\ \\ \implies \boxed{ \boxed{\frac{1}{ \alpha} + \frac{1}{ \beta} = - 1}}$

Answer 2

Given that ,

The quadratic eq is (x)² + x + 1 = 0

As we know that ,

The sum of roots i.e α + β is

$\star \: \: α + β = - \frac{b}{a}$

The product of roots i.e α × β is

$\star \: \: α \times β = \frac{c}{a}$

Thus ,

$\sf \Rightarrow α + β = - 1$

and

$\sf \Rightarrow α \times β =1$

Now , we have to find the value of

$\star \: \: \mathtt{ \underline{ \fbox{ \frac{1}{α} + \frac{1}{β} }}}$

It can be written as ,

$\star \: \: \mathtt{ \underline{ \fbox{ \frac{ β \: + \: α}{αβ} }}}$

Substitute the known values in above equation , we get

$\sf \Rightarrow \frac{ - 1}{1} \\ \\\sf \Rightarrow - 1$

$\therefore \underline{\sf \bold{The \: required \: value \: is \: -1}}$