**Solution:**

Given quadratic equation: $ax^2 + bx + c = 0$

Let $r_1$ and $r_2$ be the roots of the equation.

Sum of the roots: $r_1 + r_2 = -\frac{b}{a}$

Sum of the squares of their reciprocals: $\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2$

Using the formula $(a+b)^2 = a^2 + b^2 + 2ab$, we can rewrite the above equation as:

$\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2 = \left(\frac{1}{r_1} + \frac{1}{r_2}\right)^2 - 2\left(\frac{1}{r_1}\right)\left(\frac{1}{r_2}\right)$

Simplifying the above equation further:

$\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2 = \left(\frac{r_1 + r_2}{r_1r_2}\right)^2 - \frac{2}{r_1r_2}$

We know that $r_1 + r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$, so substituting these values into the equation:

$\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2 = \left(\frac{-\frac{b}{a}}{\frac{c}{a}}\right)^2 - \frac{2}{\frac{c}{a}}$

$\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2 = \left(\frac{-b}{c}\right)^2 - \frac{2a}{c}$

$\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2 = \frac{b^2}{c^2} - \frac{2a}{c}$

According to the given condition, the sum of the roots of the quadratic equation is equal to the sum of the squares of their reciprocals.

So, we have:

$r_1 + r_2 = \left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2$

Substituting the values of $r_1 + r_2$ and $\left(\frac{1}{r_1}\right)^2 + \left(\frac{1}{r_2}\right)^2$:

$-\frac{b}{a} = \frac{b^2}{c^2} - \frac{2a}{c}$

Multiplying through by $ac$ to eliminate fractions:

$-bc = \frac{b^2a}{c} - 2a^2$

Rearranging the terms:

$0 = \frac{b