12. Given that $a$ and $b$ are two real roots of the equation $x^{4} + m = 9 x^{2}$, and they satisfy $a + b = 4$. Then the value of $m$ is $\qquad$.
12.12.25.

The equation can be transformed into $x^{4}-9 x^{2}+m=0$. Then $a^{2}, b^{2}$ are the roots of $y^{2}-$ $9 y+m=0$, where $y=x^{2}$. By Vieta's formulas, we get
$$
a^{2}+b^{2}=9, a^{2} b^{2}=m .
$$

Given $a+b=4$, then $(a+b)^{2}=16$, we can obtain $a b=\frac{7}{2}$.