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

If α, β are roots of x2−px−(p+c)=0then α+β=p, αβ=−(p+c)Let (α+1)(β+1)= αβ+(α+β)+1=−(p+c)+p+1=1−cNow α2+2α+1α2+2α+c+β2+2β+1β2+2β+c= (α+1)2(α+1)2+c−1+(β+1)2(β+1)2+c−1=(α+1)2(α+1)2−(α+1)β+1+(β+1)2(β+1)2−(β+1)(α+1)= (α+1)α−β+β+1β−α= α+1−(β+1)α−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=

If α, β are roots of x2−px−(p+c)=0then α+β=p, αβ=−(p+c)Let (α+1)(β+1)= αβ+(α+β)+1=−(p+c)+p+1=1−cNow α2+2α+1α2+2α+c+β2+2β+1β2+2β+c= (α+1)2(α+1)2+c−1+(β+1)2(β+1)2+c−1=(α+1)2(α+1)2−(α+1)β+1+(β+1)2(β+1)2−(β+1)(α+1)= (α+1)α−β+β+1β−α= α+1−(β+1)α−β=1

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$I f α, β a r e t h e r o o t s o f x^{2} - p (x + 1) - c = 0, t h e n \frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} =$

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Detailed Solution

$\begin{array}{l} I f α, β a r e r o o t s o f x^{2} - p x - (p + c) = 0 \\ t h e n α + β = p, α β = - (p + c) \\ L e t (α + 1) (β + 1) = α β + (α + β) + 1 \\ = - (p + c) + p + 1 \\ = 1 - c \\ N o w \frac{α^{2} + 2 α + 1}{α^{2} + 2 α + c} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + c} = \frac{{(α + 1)}^{2}}{{(α + 1)}^{2} + c - 1} + \frac{{(β + 1)}^{2}}{{(β + 1)}^{2} + c - 1} \\ = \frac{{(α + 1)}^{2}}{{(α + 1)}^{2} - (α + 1) β + 1} + \frac{{(β + 1)}^{2}}{{(β + 1)}^{2} - (β + 1) (α + 1)} \\ = \frac{(α + 1)}{α - β} + \frac{β + 1}{β - α} \\ = \frac{α + 1 - (β + 1)}{α - β} = 1 \end{array}$