Question

hey guys pls ans my ques fast​

Answer 1

Answer:

$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = -1$

Step-by-step explanation:

The given function-

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

Comparing with $ax^{2} +bx+c$

We have, a = 1, b = 1 and c = 1

Given that $\alpha  \ and  \  \beta$ are the roots of given function.

Then, sum of roots = $-\dfrac{b}{a}$

$\alpha+\beta = -\dfrac{1}{1}$

So, $\alpha + \beta = -1$

Product of roots = $\dfrac{c}{a}$

$\alpha\beta=\dfrac{1}{1}$

$\alpha \beta =1$

Now, $\dfrac{1}{\alpha } + \dfrac{1}{\beta }$ = ?

$\dfrac{1}{\alpha } +\dfrac{1}{\beta } = \dfrac{\alpha+\beta  }{\alpha\beta  }$

                                      = $\dfrac{-1}{1}$

$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = -1$