Solve the equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$
Given equation: $\;$ $\left(x - \sqrt{3}\right)^4 - 5 \left(x - \sqrt{3}\right)^2 + 4 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\left(x - \sqrt{3}\right)^2 = p$ $\;\;\; \cdots \; (2)$
Then $(1)$ becomes
$p^2 - 5p + 4 = 0$
i.e. $\;$ $\left(p - 4\right) \left(p - 1\right) = 0$
i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = 1$
When $\;$ $p = 4$, $\;$ $(2)$ becomes
$\left(x - \sqrt{3}\right)^2 = 4$
i.e. $\;$ $x - \sqrt{3} = \pm 2$
i.e. $\;$ $x = \sqrt{3} + 2$ $\;$ or $\;$ $x = \sqrt{3} - 2$
When $\;$ $p = 1$, $\;$ $(2)$ becomes
$\left(x - \sqrt{3}\right)^2 = 1$
i.e. $\;$ $x - \sqrt{3} = \pm 1$
i.e. $\;$ $x = \sqrt{3} + 1$ $\;$ or $\;$ $x = \sqrt{3} - 1$
$\therefore \;$ The solutions of $(1)$ are
$x = \left\{\sqrt{3} + 2, \sqrt{3} - 2, \sqrt{3} + 1, \sqrt{3} - 1 \right\}$