Question

已知方程3x²+4x-5=0的两根为X1.X2,求下列各式的值;

Answer 1

据根与系数的关系有：x1+x2=-4/3 x1x2=-5/3

(1) X1²+X2²=(x1+x2)^2-2x1x2=16/9+10/3=46/9

(2) 1/X1+1/X2 =(x2+x1)/x1x2=(-4/3)/(-5/3)=4/5

(3) (x1-x2)^2=(x1+x2)^2-4x1x2=16/9+20/3=76/9

(4) (X1+1)(X2+1)=x1x2+x1+x2+1=-5/3-4/3+1=-2