If α,β are the roots of x2−p(x+1)−c=0 , then α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=

x2−p(x+1)−c=0 x2−px−p−c=0 α+β=p αβ=−p−c (1+α)(1+β)=1+(α+β)+αβ =1+p−p−c =1−c α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=(α+1)2(α+1)2−1+c+(β+1)2(β+1)2−1+c =(α+1)2(α+1)2−(1+α)(1+β)+(β+1)2(β+1)2−(1+α)(1+β) =α+1α+1−1−β+β+1β+1−1−α =α+1α−β+β+1β−α =α−βα−β =1

If α,β are the roots of x2−p(x+1)−c=0 , then α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=

x2−p(x+1)−c=0 x2−px−p−c=0 α+β=p αβ=−p−c (1+α)(1+β)=1+(α+β)+αβ =1+p−p−c =1−c α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=(α+1)2(α+1)2−1+c+(β+1)2(β+1)2−1+c =(α+1)2(α+1)2−(1+α)(1+β)+(β+1)2(β+1)2−(1+α)(1+β) =α+1α+1−1−β+β+1β+1−1−α =α+1α−β+β+1β−α =α−βα−β =1

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If $α, β$ are the roots of $x^{2} - p (x + 1) - c = 0$ , then $\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} =$

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Detailed Solution

x^{2} - p (x + 1) - c = 0

$x^{2} - p x - p - c = 0$

$α + β = p$

$α β = - p - c$

$(1 + α) (1 + β) = 1 + (α + β) + α β$

$= 1 + p - p - c$

$= 1 - c$

$\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} = \frac{{(α + 1)}^{2}}{{(α + 1)}^{2} - 1 + c} + \frac{{(β + 1)}^{2}}{{(β + 1)}^{2} - 1 + c}$