已知关于x的一元二次方程x2-mx+n=0（m.n为常数）的两根分别为x1.x2，是利用求根公式说明：x1+x2=m，x2=n
\u5df2\u77e5\u5173\u4e8ex\u7684\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0bx2+mx+n-1=0, \u82e5\u65b9\u7a0b\u7684\u4e24\u4e2a\u8ddf\u5206\u522b\u4e3a1,-2,\u6c42m n\u7684\u503c\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0bx²+mx+n-1=0\u7684\u4e24\u6839\u5206\u522b\u662f1\uff0c-2\uff0c
\u7531\u97e6\u8fbe\u5b9a\u7406\uff1a
1-2=-m\uff0c 1*(-2)=n\uff0c
\u6240\u4ee5 m=1\uff0cn=-2\u3002
mn=-2\u3002
\u6839\u636e\u97e6\u8fbe\u5b9a\u7406\u5f97\uff1a
x1+x2\uff1d\uff0dm\uff0cx1x2\uff1dn
\u56e0\u4e3a\u5173\u4e8ex\u7684\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0bx²+mx+n=0\u7684\u4e24\u6839\u4e3ax1=2,x2=1
\u6240\u4ee52+1\uff1d\uff0dm\uff0c2\u00d71\uff1dn
\u6240\u4ee5m\uff1d\uff0d3\uff0cn\uff1d2
x1+x2=(m+sqrt(m^2-4n))/2+(m-sqrt(m^2-4n))/2=m
x1.x2=(m+sqrt(m^2-4n))/2+(m-sqrt(m^2-4n))/2=4n/4=1 ； （利用平方差公式）
x1=[m+sqrt(m^2-4n)]/2 x2=[m-sqrt(m^2-4n)]/2; x1+x2=m
x1*x2=[m^2-(m^2-4n)]/4=n
利用伟达定理证明
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