24. If $a$ and $b$ are positive real numbers, and the equations $x^{2}+a x+2 b=0$ and $x^{2}+2 b x+a=0$ both have real roots, then the minimum value of $a+b$ is
(A) 2;
(B) 3;
(C) 4;
(D) 5;
(E) 6
$\mathrm{E}$
24. Since the two equations have real roots, their corresponding discriminants are not less than 0, i.e., $\mathrm{a}^{2}-8 \mathrm{~b} \geqslant 0$ and $\mathrm{b}^{2}-\mathrm{a} \geqslant 0$. Using the fact that $\mathrm{a}, \mathrm{b}$ are positive real numbers and the transitivity of inequalities, we get $\mathrm{a}^{4} \geqslant 64 \mathrm{~b}^{2} \geqslant 64 \mathrm{a}$ and $\mathrm{b}^{4} \geqslant \mathrm{a}^{2} \geqslant 8 \mathrm{~b}$. Solving $\mathrm{a}^{4} \geqslant 64 \mathrm{a}$ and $b^{4} \geqslant 8 b$ and noting that $a>0, b>0$, we get $a \geqslant 4$, $b \geqslant 2$. Thus, $a+b \geqslant 6$, meaning the minimum value of $a+b$ is 6, i.e., (E).

Another solution: We can take $\mathrm{a}$ as the horizontal coordinate and $\mathrm{b}$ as the vertical coordinate to draw the regions $b \leqslant \frac{1}{8} a^{2}$ and $a \leqslant b^{2}$, and let $u=a+b$. Draw the family of parallel lines $b=-a+u$, and the minimum value of $u$ is obtained by the line $b=-a+u$ passing through the intersection of $b=\frac{1}{8} a^{2}$ and $a=b^{2}$. It is easy to find that $u=6$, i.e., (E).