If #alpha and beta # are the roots of the equation #x^2-p(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)#=?

1 Answer
P dilip_k
Jul 16, 2018

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

#alpha+beta=pand alphabeta=-(p+c)#

So #c=-alpha-beta-alphabeta#

Now the numerical value of #(alpha^2+2alpha+1)/(alpha^2+2alpha+c)+(beta^2+2beta+1)/(beta^2+2beta+c)#

#=(alpha^2+2alpha+1)/(alpha^2+2alpha-alpha-beta-alphabeta)+(beta^2+2beta+1)/(beta^2+2beta-alpha-beta-alphabeta)#

#=(alpha^2+2alpha+1)/(alpha^2+alpha-beta-alphabeta)+(beta^2+2beta+1)/(beta^2+beta-alpha-alphabeta)#

#=(alpha+1)^2/(alpha(alpha+1)-beta(alpha+1))+(beta+1)^2/(beta(beta+1)-alpha(1+beta))#
#=(alpha+1)^2/((alpha-beta)(alpha+1))-(beta+1)^2/((beta+1)(alpha-beta))#

#=(alpha+1)/(alpha-beta)-(beta+1)/(alpha-beta)#

#=(alpha+1-beta-1)/(alpha-beta)#

#=1#

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