一元二次方程x~2+px+q=0(p,q不为零,p~2-4q≥0)的实数根可用下述图解方法求得。以点A(0,1)和点B(-p,q)为直径的两端作圆,则该圆与x轴的交点的横坐标就是一元二次方程x~2+px+q=0的实数根。证明 AB的中点坐标为(-(1/2)p,(1/2)(1+q)),AB线段的长为 |AB|=(p~2+(1-q)~2)~(1/2), 故以AB为直径的圆的方程为(x+(1/2)p)~2+[y-(1/2)(1+q)]~2 The root of the real number of the quadratic equation x~2+px+q=0 (p, q is not zero, p~2-4q≥0) can be obtained by the following graphical method. Let A(0,1) and B(-p,q) be circles at the two ends of the diameter. The abscissa at the intersection of the circle and the x-axis is a quadratic equation x~2+px+q=0. The real number of roots. Prove that the midpoint coordinate of AB is (-(1/2)p,(1/2)(1+q)) and the length of AB segment is |AB|=(p~2+(1-q)~2) ~(1/2), so the equation for the circle with AB diameter is (x+(1/2)p)~2+[y-(1/2)(1+q)]~2