Question

If `alpha` and `beta` are the roots of `x^2 - p (x+1) - c = 0`, then the value of `(alpha^2 + 2alpha+1)/(alpha^2 +2 alpha + c) + (beta^2 + 2beta + 1)/(beta^2 + 2beta + c)`

Answer 1

Since `alpha` and `beta` are the roots of the equation
`x^(2)-px-p-c=0`
`:.alpha+beta=p`
and `alpha beta=-p-c`
Now `(alpha+1)(beta+1)=alpha beta+alpha +beta+1`
`=-p-c+p+q=1-c`
Hence `(alpha+1)(beta+1)=1-c`……….i
Second part LHS`=(alpha^(2)+2alpha+1)/(alpha^(2)+2alpha+c)+(beta^(2)+2beta+1)/(beta^(2)+2beta+c)`
`=((alpha+1)^(2))/((alpha+1)^(2)-(1-c))+((beta+1)^(2))/((beta+1)^(2)-(1-c))`
`=(alpha+1)^(2))/((alpha+1)^(2)-(alpha+1)(beta+1))`
`+((beta+1)^(2))/((beta+1)^(2)-(alpha+1)(beta+1))` [from eq i ]
`=(alpha+1)/(alpha-beta)+(beta+1)/(beta-alpha)=(alpha-beta)/(alpha-beta)=-1=RHS`
Hence RHS=LHS`