Question

设方程甲:x^2-2x-m=0无实根,判断方程乙:x^2+mx+1+2(m^2-1)(x^2+1)=0的根的情况

Answer 1

x^2-2x-m=0无实根说明

（-2）^2+4m＜0

则m＜-1

x^2+mx+1+2(m^2-1)(x^2+1)=0

x^2+mx+1+2m^2x^2-2m^2-2x^2-2=0

(2m^2-1)x^2+mx+-2m^2-1=0

m^2+4(2m^2-1)(2m^2+1)

=16m^4+m^2-4

=16(m^4+1/16m^2+1/1024-1/1024)-4

=16(m^2+1/32)^2-1/64-4>0

所以方程乙有两个不等实根