Properties of Roots of Quadratic Equation

Let the roots of $x^2 + bx + c = 0$ be $\alpha$ and $\beta$. We know $\alpha + \beta = -b$ and \alpha \beta = c.

Given:

1. $\left|\frac{1}{\alpha} - \frac{1}{\beta}\right| = \frac{1}{3} \implies \left(\frac{1}{\alpha} - \frac{1}{\beta}\right)^2 = \frac{1}{9}$
2. $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5}{9}$

Expanding $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)^2$: $\frac{1}{\alpha^2} + \frac{1}{\beta^2} - \frac{2}{\alpha \beta} = \frac{1}{9}$ $\frac{5}{9} - \frac{2}{c} = \frac{1}{9} \implies \frac{2}{c} = \frac{4}{9} \implies c = \frac{9}{2}$

Now, $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{b^2 - 2c}{c^2} = \frac{5}{9}$: $\frac{b^2 - 2(9/2)}{(9/2)^2} = \frac{5}{9} \implies \frac{b^2 - 9}{81/4} = \frac{5}{9}$ $b^2 - 9 = \frac{45}{4} \implies b^2 = \frac{81}{4} \implies b = \pm \frac{9}{2}$

To maximize $b + c$, take $b = \frac{9}{2}$: $b + c = \frac{9}{2} + \frac{9}{2} = 9$