3. In $\triangle A B C$, $\angle C=90^{\circ}$, and the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively. If the sum of the squares of the roots of the equation $c\left(x^{2}+1\right)-2 \sqrt{2} b x-a\left(x^{2}-1\right)=0$ is 10, then the value of $\frac{b}{a}$ is . $\qquad$
3. $\sqrt{15}$.

Given the equation transformed as
$$
(c-a) x^{2}-2 \sqrt{2} b x+(c+a)=0 \text {. }
$$

From $\Delta=8 b^{2}-4\left(c^{2}-a^{2}\right)=4 b^{2}>0$, we know the equation has two real roots $x_{1} 、 x_{2}$, and
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}=\left(\frac{2 \sqrt{2} b}{c-a}\right)^{2}-2 \cdot \frac{c+a}{c-a} \\
=8 \cdot \frac{c^{2}-a^{2}}{(c-a)^{2}}-2 \cdot \frac{c+a}{c-a} \\
=6 \cdot \frac{c+a}{c-a}=10 .
\end{array}
$$

Therefore, $c=4 a, b=\sqrt{15} a \Rightarrow \frac{b}{a}=\sqrt{15}$.