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1.假設已知化簡后的二次方程 x~2+px+q=0配成完全平方: (x+p/2)~2-(p~2/4-q)=0, -(x+p/2)~2+(p~2/4-4)=0。用y~2表示方程的左端 y~2=(p~2/4-q)-(x+p/2)~2,由此, (x+p/2)~2+y~2=p~2/4-q,所得到的是一个圓的方程,其圓心为点(-p/2,0),半径为r=(p~2/4-q)~(1/p~2/4-q),此圓与Ox軸的交点的横坐标就是二次方程的根。例.图解方程 3x~2-8x-51=0,化簡后的方程是 x~2-2(2/3x)-17=0,或者 -(x-1(1/3))~2+18(7/9)=0。用y~2表示它的左端,得到一个圓的方程 (x-1(1/3)))~2+y~2=18(7/9)。圓心在点(1(1/3),0)半径等于当x=1(1/3)时的y的
1. Assume that the simplified quadratic equation x~2+px+q=0 is assumed to be fully squared: (x+p/2)~2-(p~2/4-q)=0, -( x+p/2)~2+(p~2/4-4)=0. Let y~2 represent the left end of the equation y~2=(p~2/4-q)-(x+p/2)~2, from which (x+p/2)~2+y~2=p ~2/4-q, the result is a circular equation with a center point of (-p/2,0) and a radius of r=(p~2/4-q)~(1/p~2/ 4-q), the abscissa of the intersection of this circle with the Ox axis is the root of the quadratic equation. For example, the equation 3x~2-8x-51=0, the reduced equation is x~2-2(2/3x)-17=0, or -(x-1(1/3))~2+ 18(7/9)=0. The left end is represented by y~2, and a circular equation (x-1(1/3)))~2+y~2=18(7/9) is obtained. The center of the circle at the point (1(1/3),0) is equal to the radius of y when x=1(1/3)