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

The correct answer is 9.

Let the roots of the quadratic equation
$x^2+bx+c=0$
be
$\alpha$
and
$\beta$.

From the relations between the roots and coefficients of a quadratic equation, we have:
$\alpha +\beta =-b$
and
$\alpha \beta =c$

Let us define the reciprocals of the roots as:
$u=\frac{1}{\alpha }$
and
$v=\frac{1}{\beta }$

We are given two conditions in the problem:

1. The difference between the reciprocals of the roots is
$\frac{1}{3}$:
$|u-v|=\frac{1}{3}\Rightarrow {(u-v)}^2=\frac{1}{9}$

2. The sum of the reciprocals of the squares of the roots is
$\frac{5}{9}$:
$u^2+v^2=\frac{5}{9}$

Using the algebraic identity
${(u-v)}^2=u^2+v^2-2uv$,
we can substitute the known values:
$\frac{1}{9}=\frac{5}{9}-2uv$
Rearranging the terms to find
$2uv$:
$2uv=\frac{5}{9}-\frac{1}{9}=\frac{4}{9}\Rightarrow uv=\frac{2}{9}$

Since
$uv=\frac{1}{\alpha \beta }=\frac{1}{c}$,
we get:
$\frac{1}{c}=\frac{2}{9}\Rightarrow c=\frac{9}{2}=4.5$

Now, let us find the value of
$u+v$
using the identity
${(u+v)}^2=u^2+v^2+2uv$:
${(u+v)}^2=\frac{5}{9}+\frac{4}{9}=\frac{9}{9}=1$
Taking the square root, we get:
$u+v=\pm 1$

Expressing
$u+v$
in terms of
$b$
and
$c$:
$u+v=\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta \}}{\alpha \beta }=\frac{-b}{c}$
Therefore:
$-\frac{b}{c}=\pm 1\Rightarrow b=\mp c$

Thus,
$b$
can be either
$-4.5$
or
$4.5$.

Let us check the condition for the existence of real roots (i.e., discriminant
$D=b^2-4c\ge 0$):
For both
$b=\pm 4.5$
and
$c=4.5$:
$D={(\pm 4.5)}^2-4\left(4.5\right)=20.25-18=2.25>0$
Since the discriminant is positive, the roots are indeed real.

Finally, we want to find the largest possible value of
$(b+c)$:
- Case 1: If
$b=-4.5$
and
$c=4.5$:
$b+c=-4.5+4.5=0$
- Case 2: If
$b=4.5$
and
$c=4.5$:
$b+c=4.5+4.5=9$

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