已知x1,x2是方程4x^2-4mx+m+2=0的两个实根,x1^2+x2^2=4,求m值
问题描述:
已知x1,x2是方程4x^2-4mx+m+2=0的两个实根,x1^2+x2^2=4,求m值
答
‘根据题意得x1+x2=4m/4=mx1*x2=(m+2)/4x1^2+x2^2=4x1^2+x2^2+2x1*x2-2x1*x2=4(x1+x2)²-2x1*x2=4m²-(m+2)/2=42m²-m-2=82m²-m-10=0(2m-5)(m+2)=0m=5/2或m=-2