设x1 x2是一元二次方程2x²-3x-5=0的两个根求x1²+3x2²-3x2

设x1 x2是一元二次方程2x²-3x-5=0的两个根求x1²+3x2²-3x2
设x1 x2是一元二次方程2x²-3x-5=0的两个根求x1²+3x2²-3x2

设x1 x2是一元二次方程2x²-3x-5=0的两个根求x1²+3x2²-3x2
2x²-3x-5=0
2x2²-3x2-5=0
2x2²-3x2=5
x1+x2=3/2
x1*x2=-5/2
x1²+3x2²-3x2
=x1²+x2²+2x2²-3x2
=x1²+x2²+5
=x1²+2x1x2+x2²-2x1x2+5
=(x1+x2)²-2x1x2+5
=(3/2)²-2*(-5/2)+5
=9/4+5+5
=49/4

2x²-3x-5=0
(2x-5)(x+1)=0
2x-5=0或x+1=0
x=2.5或x=-1
所以两个根分别是2.5和-1
x1²+3x2²-3x2
=(2.5)²+3(-1)²-3(-1)
=12.25

2x²-3x-5=0
(2x-5)(x+1)=0
x1=5/2
x2=-1
x1²+3x2²-3x2
把3x2=2x²-5代入
=x1²+3x2²-(2x2²-5)
=x1²+x2²+5
=49/4