Common Root of Three Quadratics

Let the common negative root be $\alpha < 0$. $\alpha^2 + m\alpha + 9 = 0$ $\alpha^2 + n\alpha + 17 = 0$ $\alpha^2 + (m+n)\alpha + 35 = 0$

Adding the first two equations gives $2\alpha^2 + (m+n)\alpha + 26 = 0$. Subtracting the third equation from this sum: $\alpha^2 - 9 = 0 \implies \alpha = -3 \quad (\text{since } \alpha < 0)$

Substitute $\alpha = -3$ into the equations: $(-3)^2 - 3m + 9 = 0 \implies 3m = 18 \implies m = 6$ $(-3)^2 - 3n + 17 = 0 \implies 3n = 26 \implies n = \frac{26}{3}$

Then $2m + 3n = 2(6) + 3\left(\frac{26}{3}\right) = 12 + 26 = 38$.