一、从几道错例谈起近年来,关于一元二次方程有有理根的问题,许多书刊资料均有所涉及,但常见到将“判别式”错用在“完全平方数”上的解法和证法。下面略举几例加以分析。例1 若α是有理数,旦方程x~2-3(α-2)x+α~2-2α+2k=0有有理根,求k的值。解:△=9(α-2)~2- 4(α~2-2α+2k)=5α~2-28α+36-8k 当5α~2-28α+36-8k为完全平方式时,方程有有理根,要使5α~2-28α+36-8k为完全平方式,必须△′=(-28)~2-4×5×(36-8k)=0,∴ k=-2/5。这个解法是错误的。事实上,当k=-2/5时,方程即为x~2-3(α-2)x+α~2-2α-4/5=0,判别式△=5α~2-28α+196/5=1/(5α-14)~2,方程的两根为x=1/2[3(α-2)± I. Talking from several wrong cases In recent years, there have been questions about the rational root of the one-dimensional quadratic equation. Many books and publications have been involved, but it is common to use the “discriminant” method to solve the problem of “complete squares”. And proofs. Here are a few examples to analyze. Example 1 If α is a rational number, if the equation x~2-3(α-2)x+α~2-2α+2k=0 has a rational root, find the value of k. Solution: △=9(α-2)~2- 4(α~2-2α+2k)=5α~2-28α+36-8k When 5α~2-28α+36-8k is completely flat, the equation There is a rational root, to make 5α~2-28α+36-8k completely flat, it must be Δ′=(-28)~2-4×5×(36-8k)=0, ∴k=-2/5 . This solution is wrong. In fact, when k=-2/5, the equation is x~2-3(α-2)x+α~2-2α-4/5=0, and the discriminant is Δ=5α~2-28α+196. /5=1/(5α-14)~2, two equations are x=1/2[3(α-2)±