关于x的方程x2+(m+1)x+
已知x^2+(m+1)x+(m^2+m-8)=0有两个实数根
所以,△=(m+1)^2-4(m^2+m-8)≥0
===> m^2+2m+1-4m^2-4m+32≥0
===> 3m^2+2m-33≤0
===> (m-3)(3m+11)≤0
===> -11/3≤m≤3……………………………………………………(1)
又,由一元二次方程根与系数的关系有:
x1+x2=-(m+1)
x1x2=m^2+m-8
已知3x1=x2*(x1-3)
===> 3x1=x1x2-3x2
===> 3(x1+x2)=x1x2
===> -3(m+1)=m^2+m-8
===> m^2+4m-5=0
===...全部
已知x^2+(m+1)x+(m^2+m-8)=0有两个实数根
所以,△=(m+1)^2-4(m^2+m-8)≥0
===> m^2+2m+1-4m^2-4m+32≥0
===> 3m^2+2m-33≤0
===> (m-3)(3m+11)≤0
===> -11/3≤m≤3……………………………………………………(1)
又,由一元二次方程根与系数的关系有:
x1+x2=-(m+1)
x1x2=m^2+m-8
已知3x1=x2*(x1-3)
===> 3x1=x1x2-3x2
===> 3(x1+x2)=x1x2
===> -3(m+1)=m^2+m-8
===> m^2+4m-5=0
===> (m-1)(m+5)=0
===> m1=1,m2=-5
联系(1)就有:m=1
所以:x^2+2(m+n)+5m+2n-4=0 ===> x^2+2(1+n)x+(2n+1)=0
===> x^2+(2n+2)x+(2n+1)=0
===> [x+(2n+1)]*(x+1)=0
===> x1=-(2n+1),x2=-1
已知有一根大于-1且小于2,则:-1<-(2n+1)<2
===> -2<2n+1<1
===> -3<2n<0
===> -3/2<n<0
已知n为整数
所以,n=-1。
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