有这样一道题:已知方程 2x-1=-2x2-α有两个不等实数根,求α的范围.在讲解这一道题时,笔者采用了如下方法: 解原方程等价于2x=-4x2-2α, 设y1=f1(x)=2x,y2=f2(x)=-4x2-2α. 要使原方程有两个不等实根,则需f1(x)与f2(x)的图象有两个不同交点,如图1所示. 由图可知:-2a>1,即α<-1/2. There is such a question: Knowing that the equation 2x-1=-2x2-α has two unequal real numbers, find the range of α. In explaining this question, the author adopted the following method: Solving the original equation is equivalent to 2x = -4x2-2α, Let y1=f1(x)=2x,y2=f2(x)=-4x2-2α. To make the original equation have two unequal real roots, we need f1(x) and f2(x) There are two different intersections in the image, as shown in Figure 1. From the figure we can see: -2a> 1, that is α <- 1/2.