已知y=x的平方+bx-c的图像与x轴两交点的坐标分别为(m,0),(-3m,0）（m不等于0) 求证：4c=3b方用9年级的知识回答

已知y=x的平方+bx-c的图像与x轴两交点的坐标分别为(m,0),(-3m,0）（m不等于0) 求证：4c=3b方
用9年级的知识回答

用韦达定理.
此抛物线与X轴相交时,Y=0
所以：x的平方+bx-c=0 此方程两个解就是抛物线与X轴交点坐标.即X1=m,X2=-3m
用韦达定理：XI*X2=C/A=-c/1=-c 将X1=m,X2=-3m代入.得-3m平方=-c（1）
X1+X2=-B/A=-B/1=-b 将X1=m,X2=-3m代入.得-2m=-b（2）
（1）/ （2）平方：[-3m平方/4m平方]=[-c/b平方]
整理得4c=3b平方
额、、韦达定理是九年级知识啊……反正总归会教的、写上去不算错……

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