Question

please answer quickly m

Answer 1

Given p(x)= x^2+x+1---(1)
zeroes of the p(x) are p and q ( here i'm taking p, q instead of alfa and beta b'coz i don't have that symbols)
compare p(x) with ax^2+bx+c=0
a=1,b=1,c=1
1) sum of the zeroes=p+q=-b/a=-1/1=-1----(2)
2) product of the zeroes=pq=c/a=1---(3)
3)1/p+1/q=(q+p)/pq=(1/1)=1 [from (2)and (3)]
4) p^2+q^2=(p+q)^2-2pq
=1-2=-1

Answer 2

HI friend
as p(x)=x²+x+1
let α and β are two zeros of quadratic equation
and we know that α+β=$\frac{-b}{a}$
                                   =$\frac{-x}{ x^{2} }$
                                   = -$\frac{1}{x}$
also,αβ=$\frac{c}{a}$
           =$\frac{1}{ x^{2} }$
and, $\frac{1}{ \alpha } + \frac{1}{ \beta }$=$\frac{ \beta + \alpha }{ \alpha \beta }$
=$\frac{- x^{2} }{x}$
= -x
and α²+β² = (α+β)²-2αβ
                =[tex] \frac{1}{ x^{2} } - \frac{2}{ x^{2} } = \frac{-1}{ x^{2} } [/tex]