Common Root of Quadratic Equations

Let common root be $\alpha$. For $3x^2 - 5x + p = 0$, sum of roots $= \alpha + \beta_1 = \frac{5}{3} \implies \beta_1 = \frac{5}{3} - \alpha$. Product of roots $= \alpha \beta_1 = \frac{p}{3} \implies \alpha \left(\frac{5}{3} - \alpha\right) = \frac{p}{3}$.

For $2x^2 - 2x + q = 0$, sum of roots $= \alpha + \beta_2 = 1 \implies \beta_2 = 1 - \alpha$. Product of roots $= \alpha \beta_2 = \frac{q}{2} \implies \alpha(1 - \alpha) = \frac{q}{2}$.

Subtracting products: $\alpha \left(\frac{5}{3} - \alpha\right) - \alpha(1 - \alpha) = \frac{p}{3} - \frac{q}{2}$ $\alpha \left(\frac{5}{3} - 1\right) = \frac{2}{3}\alpha = \frac{p}{3} - \frac{q}{2} \implies \alpha = \frac{p}{2} - \frac{3q}{4}$

Sum of other roots $\beta_1 + \beta_2 = \left(\frac{5}{3} - \alpha\right) + (1 - \alpha) = \frac{8}{3} - 2\alpha$. Substituting $\alpha$: $\beta_1 + \beta_2 = \frac{8}{3} - 2\left(\frac{p}{2} - \frac{3q}{4}\right) = \frac{8}{3} - p + \frac{3q}{2}.$