本刊1989年第6期刊登《特殊方程的反常规解法》一文,读后深受启发。作为该文的补充,本文再谈几例。例1 解方程4x~2-4x+a~x=0,(a>1)。解:∵(-4)~2-4×4×a~x=16(1-a~x)≥0时,x≤0。而当x≤0时,4x~2-4x+a~x>0,∴方程无解。注:对形如f(x)x~2+g(x)x+h(x)=0的方程,⊿=g~2(x)-4f(x)h(x)≥0仅是它有解的必要条件。有时可借助解决一些非一元二次方程问题。例2 解方程3arcsinx+2arccosx=7π/6。 In the 6th issue of 1989, this article published an article entitled “Anti-conventional Equations for Special Equations”. As a supplement to this article, this article will discuss a few more examples. Example 1 Solution Equation 4x~2-4x+a~x=0, (a>1). Solution: When ∵(-4)~2-4×4×a~x=16(1-a~x)≥0, x≤0. When x ≤ 0, 4x~2-4x+a~x>0, the ∴ equation has no solution. Note: For equations of the form f(x)x~2+g(x)x+h(x)=0, ⊿=g~2(x)-4f(x)h(x)≥0 is only There are necessary conditions for solution. Sometimes it is possible to solve some non-unique quadratic equation problems. Example 2 Solution equation 3arcsinx+2arccosx=7π/6.