2x²+3x-4=0 有两个根X1 X2 X1²+X2²=?

2x²+3x-4=0 有两个根X1 X2 X1²+X2²=?
2x²+3x-4=0 有两个根X1 X2 X1²+X2²=?

2x²+3x-4=0 有两个根X1 X2 X1²+X2²=?
由韦达定理：
x1+x2=-3/2
x1x2=-4/2=-2
所以,x1²+x2²=(x1+x2)²-2x1x2=(-3/2)²-2(-2)=25/4

由根与系数的关系，有
x1+x2=-3/2 , x1·x2=-4/2=-2
∴x1²+x2²
=(x1+x2)²-2x1·x2
=(-3/2)²-2×(-2)
=(9/4)+4
=25/4
=6.25

4分之25

25/4

由韦达定理得
X1 + X2 = -3/2
X1 * X2 = -2
则
X1²+X2²=(X1 + X2 )² - 2 * X1 * X2
=9/4 + 4
=25/4

x1^2+x2^2=25/4
x1+x2=-b/a=-3/2
x1*x2=c/a=-2
故x1^2+x2^2=(x1+x2)^2-2*x1*x2=9/4+4=25/4