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 values for $a$.


[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
$$-\frac{1}{2}<a<0 \text { or } \frac{1}{2}<a<1 \text {. }$$