要判别有理系数一元二次方程ax~2+bx+c=0(a≠0)有无有理根,只要看它的判别式△=b~2-4ac是不是有理数的完全平方。如果a、b、c是常数,由△是否是平方数立刻可以求得,如果a、b、c不是常数,它的判别式含有参数t,当△=pt+q(p≠0)时,只要令pt+q=k~2,k是有理数,便得t=(k~2-q)/p,原方程根就是有理根,当△=pt~2+qt+k (p≠0)时,问题就没有那么简单了。本文就这种情况介绍求有理系数一元二次方程有理根的方法。预备知识第一,如果p为有理数的完全平方,即p=m~2,可设pt~2+qt+k=(mt±n)~2,整理化简得t=(n~2-k)/(q±2mn),即当(?)的有 To determine whether rational coefficients have a rational quadratic equation ax~2+bx+c=0 (a ≠ 0) whether or not there is a rational root, just look at its discriminant △ = b ~ 2-4ac is not a rational square. If a, b, and c are constants, it can be obtained immediately from whether or not △ is a square number. If a, b, and c are not constants, their discriminants contain the parameter t when △=pt+q(p≠0). As long as we let pt+q=k~2, k be a rational number, we get t=(k~2-q)/p, and the original equation root is a rational root, when △=pt~2+qt+k (p≠0) The problem is not so simple. In this paper, we introduce the rational roots of rational quadratic equations. Prerequisite knowledge First, if p is the full square of rational numbers, that is, p=m~2, we can set pt~2+qt+k=(mt±n)~2, and simplify t=(n~2-k )/(q±2mn), which means when (?)