G2.4 In Figure 3, $A B C$ is an isosceles triangle with $\angle A B C=u^{\circ}, A B=B C=a$ and $A C=b$. If the quadratic equation $a x^{2}-\sqrt{2} \cdot b x+a=0$ has two real roots, whose absolute difference is $\sqrt{2}$, find the value of $u$.
Let the roots be $\alpha, \beta$.
$$
\begin{array}{l}
|\alpha-\beta|=\sqrt{2} \\
(\alpha-\beta)^{2}=2 \\
(\alpha+\beta)^{2}-4 \alpha \beta=2 \\
\left(\frac{\sqrt{2} b}{a}\right)^{2}-4=2 \\
b^{2}=3 a^{2} \\
\cos u^{\circ}=\frac{a^{2}+a^{2}-b^{2}}{2 a^{2}} \\
=\frac{2 a^{2}-3 a^{2}}{2 a^{2}} \\
\quad=-\frac{1}{2} \\
u=120
\end{array}
$$