````````````````````````````````````````已知x1,x2是关于x的方程x^2+(m^2)x+n=0的两个实根,y1,y2是关于y的方程y^2+5my+7=0的两个实根,x1-y1=2,x2-y2=2.求m,n

Question

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已知x1,x2是关于x的方程x^2+(m^2)x+n=0的两个实根,
y1,y2是关于y的方程y^2+5my+7=0的两个实根,x1-y1=2,x
2-y2=2.求m,n

Answer 1

由 y^2+5my+7=0有两个实根,(5m)²-4*7≥0
m²≥28/25
由 x1-y1=2,x2-y2=2 ,2个式子相加
(x1+x2)-(y1+y2)=4
由 韦达定理 两根之和等于-b/a
所以式子(x1+x2)-(y1+y2)=4变化
-m²+5m=4
m²-5m+4=0
(m-1)(m-4)=0
因为 m²≥28/25,所以m只能等于4
m=4,m²=16,5m=20
y²+20y+7=0,y=[-20±√(400-28)]/2=-10±√93
由 x1-y1=2,x2-y2=2,所以x1=2+y1,x2=2+y2
x1*x2=(-8+√93)(-8-√93)=64-93=-29
由 韦达定理 x1*x2=c/a=n=-29
结论 m=4,n=-29