求证:不论m为何实数,方程x²+2(m+1)x+2m²+4=0没有实数根

问题描述:

求证:不论m为何实数,方程x²+2(m+1)x+2m²+4=0没有实数根

证明:Δ=4(m+1)^2-8(m^2+2)
=4m^2+8m+4-8m^2-16
=-4m^2+8m-12
=-(4m^2-8m+4)-8
=-4(m-1)^2-8所以说不论m为何实数,方程x²+2(m+1)x+2m²+4=0没有实数根.

△=[2(m+1)]^2-4(2m²+4)
=4[(m^2+2m+1)-(2m^2+4)]
=4(-m^2+2m-3)
=-4(m^2-2m+3)
=-4[(m-1)^2+2]
≤-8
故不论m为何实数,方程x²+2(m+1)x+2m²+4=0没有实数根