If alpha and beta are the zeros of the polynomial p(x)=x^2+x+1 then find the value of 1÷alpha+1÷beta 2)alpha^2+beta^2

Answer 1

I will use a for alpha and b for beta

then from the properties of zeros of a quadratic
a+b = -1
ab = 1

then 1/a + 1/b
= (b+a)/(ab) = -1/1 = -1

a^2 + b^2
= (a+b)^2 - 2ab
= (-1)^2 - 2(1)
= 1 - 2 = -1

Answer 2

Thank you very much

Answer 3

You all r wrong

Answer 4

Answer
Given polynomial is x2−a(x+1)−b=0
x
2
−ax−a−b=0
x
2
−ax−(a+b)=0

α+β=a,αβ=−a−b.......... given
(α+1)(β+1)=αβ+α+β+1
=αβ+(α+β)+1
=−a−b+a+1=0⇒b=1