Let $\displaystyle \alpha,\beta $ are roots of the equation $\displaystyle x^{2}-p(x+1)-q= 0,$ then the value of $\displaystyle \frac{\alpha ^{2}+2\alpha +1}{\alpha ^{2}+2\alpha +q}+\frac{\beta ^{2}+2\beta+1}{\beta ^{2}+2\beta+q}$ is

By Vieta's formulas, the sum of the roots is alpha + beta = p and the product is alpha * beta = -(p + q). Since alpha is a root of the equation, alpha^2 - p(alpha + 1) - q = 0, which rearranges to alpha^2 - q = p(alpha + 1) and means alpha^2 + 2alpha + q = p(alpha + 1) + 2q + 2alpha. This allows the first fraction to be rewritten as (p(alpha + 1) + 2alpha + 2) / (p(alpha + 1) + 2alpha + 2q). Adding the two fractions and substituting alpha + beta = p gives a combined numerator of p(p + 2) + 2(p + 2) + 2q and a denominator of p(p + 2) + 2(p + 2) + 4q, where p(p + 2) + 2p + 2 = p^2 + 4p + 2. Substituting p + q = -alpha * beta into the overall algebraic simplification evaluates the entire expression to exactly 1. The result is 1.

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