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

Put p=−2, x2+2(x+1)−c=0x2+2x+1=c−1 and x2+2x+c=2(c−1)α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=(c−1)2(c−1)+(c−1)2(c−1)=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=

Put p=−2, x2+2(x+1)−c=0x2+2x+1=c−1 and x2+2x+c=2(c−1)α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=(c−1)2(c−1)+(c−1)2(c−1)=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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Put $p = - 2, x^{2} + 2 (x + 1) - c = 0 x^{2} + 2 x + 1 = c - 1$ and $x^{2} + 2 x + c = 2 (c - 1)$

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

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

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