Question

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

Answer 1

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

Answer 2

由韦达定理可得
x1+x2=-b/a=-3/2
x1x2=c/a=-2
而x1^5·x2^2+x1^2·x2^5=x1^2x2^2(x1^3+x2^3)
又因为(x1^3+x2^3)=(x1+x2)(x1^2-x1x2+x2^2)=(x1+x2)(x1^2+2x1x2+x2^2-3x1x2)
=(x1+x2)[(x1+x2)^2-3x1x2]=（-3/2）*(9/4+6)=-99/8
所以x1^5·x2^2+x1^2·x2^5=x1^2x2^2(x1^3+x2^3)=4*（-99/8）=-99/2