Let the roots of the quadratic equation be $\alpha$ and $\beta$. Using Vieta's formulas, their sum is $\alpha + \beta = -b/a$, and their product is $\alpha\beta = c/a$. The given condition requires the sum of the roots to equal the sum of the squares of their reciprocals, so we substitute the values to get $-b/a = ((\alpha + \beta)^2 - 2\alpha\beta) / (\alpha\beta)^2$. Simplifying the right side yields $(b^2/a^2 - 2c/a) / (c^2/a^2)$, which reduces to $(b^2 - 2ac) / c^2$. Equating both sides gives $-bc^2 = a(b^2 - 2ac)$, which rearranges to $2a^2c = b^2a + bc^2$. Dividing the entire equation by $a^2c$ results in the value 2.