Question

x²-mx+2m-1=0的两个实数根的平方和为7,m为多少?

Answer 1

设x²-mx+2m-1=0的两个实数根为x1,x2,则

｛△=b²-4ac=m²-4(2m-1)≥0；x1+x2=m；x1x2=2m-1｝；m∈(－∞,4-2√3]∪[4+2√3,＋∞),x²1+x²2=(x1+x2)²-2x1x2=m²-2(2m-1)=m²-4m+2=7,m=5或-1,-1(－∞,4-2√3],5不属于(－∞,4-2√3]∪[4+2√3,＋∞),∴m=-1.