Asked by rinu
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
Answers
Answered by
Reiny
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
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
Answered by
Diya K Vineeth
Thank you very much
Answered by
Sunami
You all r wrong
Answered by
Abhinav
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
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
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