Find the value of $\frac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+b}$+$\frac{\beta^2+2\beta+1}{\beta^2+2\beta+b}$

Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-a(x+1)-b=0$

Find the value of $\frac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+b}$+$\frac{\beta^2+2\beta+1}{\beta^2+2\beta+b}$

I tried to multiply by $\beta^2$ to first fractional part but equation got complicated. I tried to substitute $\alpha+\beta =a$ and $\alpha\beta=-(a+b)$ also but not getting the desired result

Answer 1

Notice that: $$\left( \alpha +1 \right)\left( \beta +1 \right)=\alpha +\beta +\alpha \beta +1=1-b$$ Now $$\frac{{{\alpha }^{2}}+2\alpha +1}{{{\alpha }^{2}}+2\alpha +b}+\frac{{{\beta }^{2}}+2\beta +1}{{{\beta }^{2}}+2\beta +b}=\frac{{{\left( \alpha +1 \right)}^{2}}}{{{\left( \alpha +1 \right)}^{2}}-\left( 1-b \right)}+\frac{{{\left( \beta +1 \right)}^{2}}}{{{\left( \beta +1 \right)}^{2}}-\left( 1-b \right)}$$ Substituting the value of $1-b$

$\begin{align} & =\frac{{{\left( \alpha +1 \right)}^{2}}}{{{\left( \alpha +1 \right)}^{2}}-\left( \alpha +1 \right)\left( \beta +1 \right)}+\frac{{{\left( \beta +1 \right)}^{2}}}{{{\left( \beta +1 \right)}^{2}}-\left( \alpha +1 \right)\left( \beta +1 \right)} \\ & =\frac{\left( \alpha +1 \right)}{\left( \alpha +1 \right)-\left( \beta +1 \right)}+\frac{\left( \beta +1 \right)}{\left( \beta +1 \right)-\left( \alpha +1 \right)} \\ & =\frac{\alpha +1}{\alpha -\beta }+\frac{\beta +1}{\beta -\alpha } \\ & =1 \\ \end{align}$

Answer 2

\begin{align} \alpha^2 &= a(\alpha+1)+b \\ \beta^2 &= a(\beta+1)+b \\ \alpha^2+2\alpha+b &= a(\alpha+1)+b+2\alpha+b \\ &= \alpha(a+2)+a+2b \\ \beta^2+2\beta+b &= \beta(a+2)+a+2b \\ (\alpha^2+2\alpha+b)(\beta^2+2\beta+b) &= \color{blue}{\alpha \beta}(a+2)^2+ \color{red}{(\alpha+\beta)}(a+2)(a+2b)+(a+2b)^2 \\ &= \color{blue}{-(a+b)}(a+2)^2+\color{red}{a}(a+2)(a+2b)+(a+2b)^2 \\ &= (b-1)(a^2+4a+4b) \\ \frac{\alpha^2+2\alpha+1}{\alpha^2+2\alpha+b}+ \frac{\beta^2+2\beta+1}{\beta^2+2\beta+b} &= 2+\frac{1-b}{\alpha^2+2\alpha+b}+ \frac{1-b}{\beta^2+2\beta+b} \\ &= 2-\frac{\color{red}{(\alpha+\beta)}(a+2)+2(a+2b)}{a^2+4a+4b} \\ &= 2-\frac{\color{red}{a}(a+2)+2(a+2b)}{a^2+4a+4b} \\ &= 1 \end{align}

Answer 3

HINT.-We have $$\frac{A+1-b}{A}+\frac{B+1-b}{B}=2+\frac{(A+B)(1-b)}{AB}$$ $$AB=(\alpha\beta)^2+2\alpha^2\beta+\alpha^2b+2\alpha\beta^2+4\alpha\beta+2\alpha b+\beta^2b+2\beta b+b^2$$ Using $$\alpha\beta=-(a+b)\\\alpha+\beta= a\\\alpha^2+\beta^2=a^2+2(a+b)$$ it is easy to verify that $$\frac{(A+B)(1-b)}{AB}=-1$$ Thus the result equal to $1$.