If alpha and beta αandβ are the roots of the equation x^2-p(x+1)-c=0x2p(x+1)c=0 then the numerical value of (alpha^2+2alpha+1)/(alpha^2+2alpha+c)+(beta^2+2beta+1)/(beta^2+2beta+c)α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=?

1 Answer
Jul 16, 2018

If alpha and beta αandβ are the roots of the equation x^2-p(x+1)-c=0x2p(x+1)c=0 or
x^2-px-(p+c)=0x2px(p+c)=0
then

alpha+beta=pand alphabeta=-(p+c)α+β=pandαβ=(p+c)

So c=-alpha-beta-alphabetac=αβαβ

Now the numerical value of (alpha^2+2alpha+1)/(alpha^2+2alpha+c)+(beta^2+2beta+1)/(beta^2+2beta+c)α2+2α+1α2+2α+c+β2+2β+1β2+2β+c

=(alpha^2+2alpha+1)/(alpha^2+2alpha-alpha-beta-alphabeta)+(beta^2+2beta+1)/(beta^2+2beta-alpha-beta-alphabeta)=α2+2α+1α2+2ααβαβ+β2+2β+1β2+2βαβαβ

=(alpha^2+2alpha+1)/(alpha^2+alpha-beta-alphabeta)+(beta^2+2beta+1)/(beta^2+beta-alpha-alphabeta)=α2+2α+1α2+αβαβ+β2+2β+1β2+βααβ

=(alpha+1)^2/(alpha(alpha+1)-beta(alpha+1))+(beta+1)^2/(beta(beta+1)-alpha(1+beta))=(α+1)2α(α+1)β(α+1)+(β+1)2β(β+1)α(1+β)
=(alpha+1)^2/((alpha-beta)(alpha+1))-(beta+1)^2/((beta+1)(alpha-beta))=(α+1)2(αβ)(α+1)(β+1)2(β+1)(αβ)

=(alpha+1)/(alpha-beta)-(beta+1)/(alpha-beta)=α+1αββ+1αβ

=(alpha+1-beta-1)/(alpha-beta)=α+1β1αβ

=1=1