24. Let $a$ and $b$ be two integers. Suppose that $\sqrt{7-4 \sqrt{3}}$ is a root of the equation $x^{2}+a x+b=0$. Find the value of $b-a$.
24. Ans: 5
First we simplify $\sqrt{7-4 \sqrt{3}}$, and write
$$
\sqrt{7-4 \sqrt{3}}=x+y \sqrt{3},
$$
where $x, y$ are rational numbers. Then
$$
7-4 \sqrt{3}=(x+y \sqrt{3})^{2}=x^{2}+3 y^{2}+2 x y \sqrt{3} .
$$

Thus, we have
$$
\left\{\begin{array}{ll}
x^{2}+3 y^{2} & =7 \\
2 x y & =-4
\end{array}\right.
$$

Substituting the second equation into the first one, we get
$$
\begin{aligned}
x^{2}+3\left(-\frac{2}{x}\right)^{2}= & \Longrightarrow x^{4}-7 x^{2}+12=0 \\
& \Longrightarrow\left(x^{2}-4\right)\left(x^{2}-3\right)=0 \\
& \Longrightarrow x=2,-2, \sqrt{3},-\sqrt{3}
\end{aligned}
$$

Since $x$ is a rational number, $x \neq \sqrt{3}$ and $x \neq-\sqrt{3}$. When $x=2, y=\frac{-4}{2 \times 2}=$ -1 . When $x=-2, y=1$. Thus, $\sqrt{7-4 \sqrt{3}}=2-\sqrt{3}$ or $-2+\sqrt{3}$. Since $\sqrt{7-4 \sqrt{3}}>0$, it follows that $\sqrt{7-4 \sqrt{3}}=2-\sqrt{3}$. Thus the 2 nd root of the equation is $2+\sqrt{3}$, since the coefficients of the equation are integers. Therefore, $a=-(2-\sqrt{3})-(2+\sqrt{3})=-4, b=(2-\sqrt{3})(2+\sqrt{3})=1$. Thus, $b-a=5$.