已知x1x2是方程2x^2+3x-4=0的两个根,求x1^5·x2^2+x1^2·x2^5的值

题目

已知x1x2是方程2x^2+3x-4=0的两个根,求x1^5·x2^2+x1^2·x2^5的值

答案

x1+x2=-3/2
x1*x2=-4/2=-2
x1^5·x2^2+x1^2·x2^5
=x1²x2²(x1³+x2³)
=(x1x2)²(x1+x2)(x1²-x1x2+x2²)
=(x1x2)²(x1+x2)((x1+x2)²-3x1x2)
=(-2)²(-3/2)((-3/2)²-3*(-2))
=-6(9/4+6)
=-99/2