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下题是我们在学习一元二次方程的根的判别式时所常见的: 如果m为有理数,试确定k值,使方程x~2-2mx+10x+4k=0的根是有理数。拿到题目后,有的同学可能会这样解吧! 解原方程即x~2+(10-2m)x+4k=0,要使它的根是有理数,只需其根的判别式△=(10-2m)~2-16k=100-40m+4m~2-16k=4(m~2-10m+25-4k) ①是完全平方式,即m~2-10m+25-4k=0有相等的根,即以m为元的此二次方程的判别式△′=100-4(25-4k)=0,
The following questions are common when we are learning the discriminant of the root of a quadratic equation: If m is a rational number, try to determine the value of k so that the root of the equation x~2-2mx+10x+4k=0 is a rational number. After getting the problem, some students may solve it! The solution to the original equation is x~2+(10-2m)x+4k=0. To make its root a rational number, only its root discriminant △= (10-2m)~2-16k=100-40m+4m~2-16k=4(m~2-10m+25-4k) 1 is completely flat, ie, m~2-10m+25-4k=0. The discriminant △’=100-4(25-4k)=0, which has an equal root, that is, the quadratic equation in m,