A quadratic equation $x^{2}+b x+c=0$ has two real roots. It the difference between the reciprocals of the roots is $1 / 3$ and the sum of the reciprocals of the squares of the roots is $5 / 9$, then the largest possible value of ( $b+c$ ) is

Given quadratic equation: $x^2 + bx + c = 0$ with roots $m$ and $n$.

From Vieta's formulas:

Sum of roots: $m + n = -b$

Product of roots: $mn = c$

Given conditions:

Difference of reciprocals: $\tfrac{1}{n} - \tfrac{1}{m} = \tfrac{1}{3}$

Sum of reciprocals of squares: $\tfrac{1}{n^2} + \tfrac{1}{m^2} = \tfrac{5}{9}$

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Using the algebraic identity $(a - b)^2 = a^2 + b^2 - 2ab$:

$\left(\tfrac{1}{n} - \tfrac{1}{m}\right)^2 = \tfrac{1}{n^2} + \tfrac{1}{m^2} - \tfrac{2}{mn}$

Substituting known values:

$\left(\tfrac{1}{3}\right)^2 = \tfrac{5}{9} - \tfrac{2}{mn}$

$\tfrac{1}{9} = \tfrac{5}{9} - \tfrac{2}{mn}$

$\tfrac{2}{mn} = \tfrac{5}{9} - \tfrac{1}{9} = \tfrac{4}{9}$

$mn = \tfrac{2 \times 9}{4} = \tfrac{9}{2}$

Therefore: $c = \tfrac{9}{2}$

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Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$:

$\left(\tfrac{1}{n} + \tfrac{1}{m}\right)^2 = \tfrac{1}{n^2} + \tfrac{1}{m^2} + \tfrac{2}{mn}$

We know:

$\tfrac{1}{n^2} + \tfrac{1}{m^2} = \tfrac{5}{9}$

$\tfrac{2}{mn} = \tfrac{4}{9}$ (from previous calculation)

Therefore:

$\left(\tfrac{1}{n} + \tfrac{1}{m}\right)^2 = \tfrac{5}{9} + \tfrac{4}{9} = 1$

Taking the square root: $\tfrac{1}{n} + \tfrac{1}{m} = \pm 1$

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Since $\tfrac{1}{n} + \tfrac{1}{m} = \tfrac{m + n}{mn}$, we have:

$\tfrac{m + n}{mn} = \pm 1$

Case 1: $\tfrac{m + n}{\tfrac{9}{2}} = 1$

$m + n = \tfrac{9}{2}$

Since $m + n = -b$, we get $b = -\tfrac{9}{2}$

Case 2: $\tfrac{m + n}{\tfrac{9}{2}} = -1$

$m + n = -\tfrac{9}{2}$

Since $m + n = -b$, we get $b = \tfrac{9}{2}$

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We have $c = \tfrac{9}{2}$ in both cases.

Case 1: $b = -\tfrac{9}{2}$, so $b + c = -\tfrac{9}{2} + \tfrac{9}{2} = 0$

Case 2: $b = \tfrac{9}{2}$, so $b + c = \tfrac{9}{2} + \tfrac{9}{2} = 9$

Therefore, the largest possible value of $(b + c)$ is $9$.