Question 5.

It is given that $x^2 + bx + c = 0$ has two real roots. Let the roots of the equation be $\alpha, \beta$. ($\alpha$ > $\beta$)

Then, we can say that $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{1}{3}$ .... Eq(1)

Similarly, $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{5}{9}$ .... Eq(2)

Eq(2) can be written as $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)^2 + 2 \cdot \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{5}{9}$

=> $\left(\frac{1}{3}\right)^2 + 2 \cdot \frac{1}{\alpha \cdot \beta} = \frac{5}{9}$

=> $\frac{2}{\alpha \cdot \beta} = \frac{4}{9} \Rightarrow \frac{1}{\alpha \cdot \beta} = \frac{2}{9}$

=> $\alpha \cdot \beta = \frac{9}{2}$

We know that the product of the roots is equal to c, which implies $c = \frac{9}{2}$

We also know that the sum of the roots is equal to -b.

=> $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)^2 - \frac{2}{\alpha \beta} = \frac{5}{9}$

=> $\left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 - \frac{4}{9} = \frac{5}{9}$

=> $\left(\frac{\alpha + \beta}{\alpha \beta}\right)^2 = (1)^2$

=> $\alpha + \beta = \pm \alpha \beta$

Hence, the maximum value of b is $\frac{9}{2}$.

Hence, the maximum value of (b+c) is 9