24. If $x, y$ and $z$ are real numbers such that $x+y+z=9$ and $x y+y z+z x=24$, find the largest possible value of $z$.
24. Answer: 5
Since $x+y=9-z, x y=24-z(x+y)=24-z(9-z)=z^{2}-9 z+24$. Now note that $x$ and $y$ are roots of the quadratic equation $t^{2}+(z-9) t+\left(z^{2}-9 z+24\right)=0$. As $x$ and $y$ are real, we have $(z-9)^{2}-4\left(z^{2}-9 z+24\right) \geq 0$, which simplified to $z^{2}-6 z+5 \leq 0$. Solving the inequality yields $1 \leq z \leq 5$. When $x=y=2, z=5$. Hence the largest possible value of $z$ is 5 .