11.9 Form a biquadratic equation if the numbers $\sqrt{3}-1$ and $\sqrt{3}+1$ are its roots.
## 11.9 The general form of a biquadratic equation is: $a x^{4}+b x^{2}+c=0$. According to the condition, it has roots $x_{1}=\sqrt{3}-1$ and $x_{2}=\sqrt{3}+1$. The roots of a biquadratic equation are pairwise opposite; therefore, there are also roots $x_{3}=-(\sqrt{3}-1)$ and $x_{4}=-(\sqrt{3}+1)$. Assuming $a=1$, we can represent the left side of the equation as a product:

$$
\begin{aligned}
& (x-(\sqrt{3}-1))(x+(\sqrt{3}-1))(x-(\sqrt{3}+1))(x+(\sqrt{3}+1))=0 \\
& \left(x^{2}-(\sqrt{3}-1)^{2}\right)\left(x^{2}-(\sqrt{3}+1)^{2}\right)=0 \\
& \left(x^{2}-4+2 \sqrt{3}\right)\left(x^{2}-4-2 \sqrt{3}\right)=0 ;\left(x^{2}-4\right)^{2}-(2 \sqrt{3})^{2}=0 \\
& x^{4}-8 x^{2}+16-12=0 ; x^{4}-8 x^{2}+4=0
\end{aligned}
$$