Answers

2014-06-08T22:33:51+05:30
Given α , β are the roots of x²-p(x+1)-q=0
Now we have to find the value of 
(α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q)
But from equation it is clear that 
Sum of the roots = 
α+β = +p
Product of the roots = α*β = -p-q
(α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) = (α²+β²)+4(α+β)+(1/α²+1/β²)+2q
(α+β)² = α²+β²+2αβ
⇒α²+β² = (α+β)² - 2αβ = p² - 2*(-p-q) = p²+2p+2q
(1/α²+1/β²) = (α²+β²)/(αβ)² = (p²+2p+2q)/(-p-q)² =  (p²+2p+2q)/(p²+q²+2pq)
(α²+2α+1/α²+2α+q)+(β²+2β+1/β²+2β+q) = (α²+β²)+4(α+β)+(1/α²+1/β²)+2q =
p²+2p+2q+4p+ (p²+2p+2q)/(p²+q²+2pq)+2q = p²+6p+4q+ (p²+2p+2q)/(p²+q²+2pq)
 = {(p²+6p+4q)(p²+q²+2pq)+(p²+2p+2q)}/(p²+q²+2pq)
= {p⁴+6p³+4qp²+p²q²+6pq²+4q³+2p³q+12p²q+8pq²+p²+2p+2q}/(p²+q²+2pq)
This is the answer.

0