Question

If $\alpha ,\beta$ are the roots of $x^2-a(x-1)+b=0,$ then $\frac{1}{{\alpha }^2-a\alpha }+\frac{1}{{\beta }^2-a\beta }+\frac{2}{a+b}$ is equal to

Answer 1

$x^2-a(x-1)+b=0$

$\alpha ,\beta$ are the roots of the equation,

$\therefore$${\alpha }^2-a\alpha +a+b=0{\alpha }^2-a\alpha =-a-b$

Likewise,

${\beta }^2-a\beta +a+b=0{\beta }^2-a\beta =-a-b$

$\therefore$ $\frac{1}{{\alpha }^2-a\alpha }+\frac{1}{{\beta }^2-a\beta }+\frac{2}{a+b}$

$⟹$$-\frac{1}{a+b}-\frac{1}{a+b}+\frac{2}{a+b}$

$⟹0$