10. Given that the product of a root of the quadratic trinomial $a x^{2}+b x+b$ and a root of the quadratic trinomial $a x^{2}+a x+b$ is equal to 1. Then the sum of the squares of these two roots is $\qquad$
10.3.

Let $s$ and $\frac{1}{s}$ be the roots of $a x^{2}+b x+b$ and $a x^{2}+a x+b$ respectively. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a s^{2}+b s+b=0, \\
\frac{a}{s^{2}}+\frac{a}{s}+b=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a s^{2}+b s+b=0, \\
b s^{2}+a s+a=0
\end{array}\right. \\
\Rightarrow(a-b) s^{2}+(b-a) s+(b-a)=0 \\
\Rightarrow(a-b)\left(s^{2}-s-1\right)=0 .
\end{array}
$$

If $a=b$, then the original quadratic equation becomes $x^{2}+x+1$, which has no real solutions. Therefore, $a \neq b$.
$$
\begin{array}{l}
\text { Hence } s^{2}-s-1=0 \Rightarrow s-\frac{1}{s}=1 \\
\Rightarrow s^{2}+\frac{1}{s^{2}}=\left(s-\frac{1}{s}\right)^{2}+2=3 .
\end{array}
$$