4・10 Given the equation $x^{2}-2 x+\lg \left(2 a^{2}-a\right)=0$ has one positive root and one negative root, try to find the range of real number $a$.

 untranslated portion: 

已知方程 $x^{2}-2 x+\lg \left(2 a^{2}-a\right)=0$ 有一个正根和一个负根, 试求实数 $a$ 的值的范围.

The translated portion:

Given the equation $x^{2}-2 x+\lg \left(2 a^{2}-a\right)=0$ has one positive root and one negative root, try to find the range of real number $a$.
[Solution] From the given, we know that $\lg \left(2 a^{2}-a\right)<0$. Therefore, $0<2 a^{2}-a<1$,
Solving this, we get that $a$ should be in the range
$$
-\frac{1}{2}<a<0 \text { or } \frac{1}{2}<a<1 \text {. }
$$