Question

已知x1、x2为方程x²+x-3=0的两根，求x1³-4x2²+19

Answer 1

x=x1
x1²+x1-3=0
x1²=-x1+3
所以x³=x1*x1²
=x1(-x1+3)
=-x1²+3x1
=-(-x1+3)+3x1
=4x1-3

x=x2
则x2²+x2-3=0
所以-4x2²=4x2-12
且x1+x2=-1
所以原式=4x1-3+4x2-12+19
=4(x1+x2)+4
=-4+4
=0

Answer 2

x1+x2=-1

x1x2=-3
x1^2+2x1x2+x^2^2=1

x1^2+x2^2=7
x1³-4x2²+19
=x1^3+x2^3-x2^3-4x2^2+19
=(x1+x2)(x1^2-x1x2+x2^2)-x2(x2^2+4x)+19
=-10-x2*(x2^2+x2-3+3+3x2)+19
=-10-x2(3+3x2)+19
=-10-3(x2^2+x2-3+3)+19
=-10-9+19
=0
【简化：令x1=x, x2=y】
x^3-4y^2+19
。。。