If `alpha, beta` are roots of the equation`x^(2)-p(x+1)-c=0` show that `(alpha+1)(beta+1)=1-c` Hence prove that `(alpha^(2)+2alpha+1)/(alpha^(2)+2alpha+c)+(beta^(2)+2beta+1)/(beta^(2)+2beta+c)=1`

Answer

A

Explanation

`alpha+beta=p` and `alphabeta=-(p+q)`
Now, `(alpha + 1) (beta+1) =(alpha +beta)+alphabeta+1=1-q`
The given expression =`((alpha+1)^2)/((alpha+1)^2+(q-1))+((beta+1)^2)/((beta+1)^2+(q-1))`
`=((alpha+1)^2)/((alpha+1)^2 -(alpha +1) (beta+1))+((beta+1)^2)/((beta+1)^2-(alpha +1)(beta+1))`
`=((alpha +1))/((alpha -beta))+((beta+1))/((beta-alpha ))=((alpha -beta))/((alpha -beta))=1`