1. **Problem statement:** Find positive integers $a$ and $b$ with $a \leq b$ satisfying $$a^2 + b^2 + 3ab = 719$$ and then find $a+b$. 2. **Rewrite the equation:** $$a^2 + b^2 + 3ab = 719$$ We consider it as a quadratic in $b$: $$b^2 + 3ab + a^2 = 719$$ 3. **Express as quadratic in $b$:} $$b^2 + 3ab + (a^2 - 719) = 0$$ 4. **Discriminant for $b$ to be integer:** $$\Delta = (3a)^2 - 4 \times 1 \times (a^2 - 719) = 9a^2 - 4a^2 + 2876 = 5a^2 + 2876$$ 5. **$\Delta$ must be a perfect square:** Let $k^2 = 5a^2 + 2876$. 6. **Rearranged:** $$k^2 - 5a^2 = 2876$$ We search for integer pairs $(a,k)$ with $a$ positive and $k^2$ perfect square. 7. **Check positive $a$ values with $a \leq b$ and likely $a+b \approx$ options given:** Test $a$ such that $5a^2 + 2876$ is a perfect square. Try $a=11$: $$5 \times 11^2 + 2876 = 5 \times 121 + 2876 = 605 + 2876 = 3481 = 59^2$$ Perfect square found. 8. **Find $b$:} Solve quadratic: $$b^2 + 3 \times 11 b + 11^2 - 719=0$$ $$b^2 + 33b + 121 - 719=0$$ $$b^2 + 33b - 598=0$$ Discriminant: $$33^2 - 4 \times 1 \times (-598) = 1089 + 2392=3481=59^2$$ Roots: $$b = \frac{-33 \pm 59}{2}$$ Positive root: $$b= \frac{-33 + 59}{2} = \frac{26}{2} = 13$$ 9. Check $a \le b$: $11 \le 13$ true. 10. Compute $a+b$: $$11 + 13 = 24$$ **Answer:** 24, which corresponds to option (B).