Question

已知方程x²+ax+b=0的两个根为X1,X2,且4x1+x2=0,又知根的判别式△=25,求a、b的值

Answer 1

判别式=a^2-4b=(x1+x2)^2-4x1x2=(x1-x2)^2=25

得：x1-x2=5或x1-x2=-5

当x1-x2=5时,由4x1+x2=0,解得：x1=1,x2=-4,得a=-(x1+x2)=3,b=x1x2=-4

当x1-x2=-5时,由4x1+x2=0,解得：x1=-1,x2=4,得a=-(x1+x2)=-3,b=x1x2=-4