Question

若方程2x²+6x-1=0两根为x1x2,求:(1)x1x2²+x2x1²

若方程2x²+6x-1=0两根为x1x2,求:(1)x1x2²+x2x1²（2）x2²+x1²（3）|x1-x2|

Answer 1

你好，数学之美为您解答
根据韦达定理可得
x1x2=-2 x1+x2=-3
所以
(1) x1x2²+x2x1²=x1x2(x1+x2)=(-2)*(-3)=6
(2）x2²+x1²=(x1+x2)²-2X1X2=9-2*(-2)=13
(3）|x1-x2|=√(x1-x2)²=√(x1+x2)²-4x1x2=√(9+8)=√17

Answer 2

x1*x2=-1/2且x1+x2=-3
(1)
x1*x2²+x2*x1²
=x1x2(x2+x1)
=(-1/2)*(-3)
=3/2
(2)
x2²+x1²
=x2²+2x1x2+x1²-2x1x2
=(x1+x2)²-2x1x2
=(-3)²-2(-1/2)
=10
(3)
(|x1-x2|)²
=(x1+x2)²-4x1x2
=9+2=11
所以|x1-x2|=√11