4. The real roots of the equation $x^{2}-8[x]+7=0$ are
$\qquad$ . (Here $[x]$ denotes the greatest integer less than or equal to $x$)
4. $1, \sqrt{33}, \sqrt{41}, 7$.

Solution: Let $n=[x]$, from $x^{2}+7=8 n$ we know $n>0$. Using $x-1<n \leqslant x$, we have
$$
n^{2}+7 \leqslant x^{2}+7<(n+1)^{2}+7=n^{2}+2 n+8
$$

Solving this, we get $1 \leqslant n<2$ or $4<n \leqslant 7$, so $n=1,5,6,7$. Thus, $x^{2}=1,33,41,49$
Therefore, $x=1, \sqrt{33}, \sqrt{41}, 7$ are the solutions.