# 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)=?

Jul 16, 2018

If $\alpha \mathmr{and} \beta$ are the roots of the equation ${x}^{2} - p \left(x + 1\right) - c = 0$ or
${x}^{2} - p x - \left(p + c\right) = 0$
then

$\alpha + \beta = p \mathmr{and} \alpha \beta = - \left(p + c\right)$

So $c = - \alpha - \beta - \alpha \beta$

Now the numerical value of $\frac{{\alpha}^{2} + 2 \alpha + 1}{{\alpha}^{2} + 2 \alpha + c} + \frac{{\beta}^{2} + 2 \beta + 1}{{\beta}^{2} + 2 \beta + c}$

$= \frac{{\alpha}^{2} + 2 \alpha + 1}{{\alpha}^{2} + 2 \alpha - \alpha - \beta - \alpha \beta} + \frac{{\beta}^{2} + 2 \beta + 1}{{\beta}^{2} + 2 \beta - \alpha - \beta - \alpha \beta}$

$= \frac{{\alpha}^{2} + 2 \alpha + 1}{{\alpha}^{2} + \alpha - \beta - \alpha \beta} + \frac{{\beta}^{2} + 2 \beta + 1}{{\beta}^{2} + \beta - \alpha - \alpha \beta}$

$= {\left(\alpha + 1\right)}^{2} / \left(\alpha \left(\alpha + 1\right) - \beta \left(\alpha + 1\right)\right) + {\left(\beta + 1\right)}^{2} / \left(\beta \left(\beta + 1\right) - \alpha \left(1 + \beta\right)\right)$
$= {\left(\alpha + 1\right)}^{2} / \left(\left(\alpha - \beta\right) \left(\alpha + 1\right)\right) - {\left(\beta + 1\right)}^{2} / \left(\left(\beta + 1\right) \left(\alpha - \beta\right)\right)$

$= \frac{\alpha + 1}{\alpha - \beta} - \frac{\beta + 1}{\alpha - \beta}$

$= \frac{\alpha + 1 - \beta - 1}{\alpha - \beta}$

$= 1$