# 11.1. Solve the equation

$$
x^{4}+2 x \sqrt{x-1}+3 x^{2}-8 x+4=0
$$
Solution. Let's check that $x=1$ is a root. Indeed,

$$
1^{4}+2 \sqrt{1-1}+3 \cdot 1^{2}-8+4=1+3-8+4=0
$$

We will show that there are no other roots. Since the left side of the equation is defined only for $x \geqslant 1$, it is sufficient to consider the case $x>1$. Using the valid inequalities in this case $x^{4}>x^{2}, x \sqrt{x-1}>0$ and $(x-1)^{2}>0$, we obtain for the left side the inequality

$$
x^{4}+2 x \sqrt{x-1}+3 x^{2}-8 x+4>4 x^{2}-8 x+4=4(x-1)^{2}>0
$$

which holds for all $x>1$. Thus, there are no roots for $x>1$.

Answer. $x=1$.