Example 3 If $a^{2}+11 a=-16, b^{2}+11 b=$ $-16(a \neq b)$, then the value of $\sqrt{\frac{b}{a}}+\sqrt{\frac{a}{b}}$ is $\qquad$
(1990, Shaoxing City Mathematics Competition)
Given the equation, we know that $a$ and $b$ are the two distinct real roots of the quadratic equation $x^{2} + 11x + 16 = 0$. According to the relationship between the roots and coefficients of a quadratic equation, we have
$$
\begin{array}{l} 
a + b = -11, \quad ab = 16. \\
\text { Therefore, } \sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}} = \sqrt{\left(\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}\right)^{2}} \\
= \sqrt{\frac{a^{2} + b^{2}}{ab} + 2} = \sqrt{\frac{(a + b)^{2}}{ab}} \\
= \frac{|a + b|}{\sqrt{ab}} = \frac{11}{\sqrt{16}} = \frac{11}{4}.
\end{array}
$$