Question

已知方程2x²+4x-3=0,不解方程,求出下列各式值

(1)x1+x2
(2)X1·x2
（3）X1²+x2²

Answer 1

x1,x2是方程2x²+4x-3=0两根，
1）x1+x2=-4/2=-2
2)x1*x2=-3/2
3)x1²+x2²=(x1+x2)²-2x1x2
=4+3=7

Answer 2

x1+x2=-4/2=-2

x1*x2=-3/2

X1²+x2²=(x1+x2)^2-2x1*x2=4+(3/2)*2=7

望采纳

Answer 3

据韦达定理有
(1)x1+x2=-4/2=-2
(2)X1·x2=-3/2
（3）X1²+x2²=(x1+x2)^2-2*X1·x2=7

Answer 4

x1+x2=-4/2=-2

x1*x2=-3/2

X1²+x2²=(x1+x2)^2-2x1*x2=4+(3/2)*2=7

Answer 5

x1+x2＝-4/2＝-2
X1·x2=-3/2
X1²+x2²=(x1+x2)^2-2x1x2=(-2)^2-2*(-3/2)=7