Question

IF a,ß are the zeroes of the polynomials f (x)=x^2+x+1 ,then 1/a+1/ß​

Answer 1

${\underline{\sf{Question}}}$

If alpha and beta are the zeroes of the polynomials f (x)=x^2+x+1 ,then 1/alpha+1/beta

${\underline{\sf{Theory}}}$

If $\sf \alpha \: and \: \beta$are zeroes of quadratic polynomial

$\sf \: f(x) = {x}^{2} + bx + c$

Then ,$\sf \: \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }$

$\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }$

${\underline{\sf{Solution}}}$

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

$\sf \: \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }$

$\implies \: \sf \: \alpha + \beta = \frac{ - 1}{1} = - 1..(1)$

and

$\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }$

$\implies \: \sf \: \alpha \beta = \frac{1}{1} = 1...(2)$

We have to find the value of

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

By taking L.C.M of dinominators

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

put the values of equation (1) and (2)

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