Question

一元二次方程题

已知x1 x2为方程x平方+3x+1=0的两实根。则x1三次方+8x2+20=？

求简单方法

Answer 1

x1三次方+8x2+20
=x1 * x1平方 +8x2+20
=x1* (-1-3x1) +8x2+20
= -3 * x1平方 - x1 +8x2+20
= -3 * (-1-3x1) - x1 +8x2+20
=8 * (x1+x2) +3 + 20
= 8 * (-3) + 23
= -1

Answer 2

解x1 x2为方程x平方+3x+1=0的两实根
∴x1+x2=-3 x1^2+3x1+1=0 乘以x1得x1^3=-3x1^2-x1
x1三次方+8x2+20=-3x1^2-x1+8x2+20
x1^2+3x1+1=0 3x1^2+9x1+3=0 -3x1^2=9x1+3
x1三次方+8x2+20=-3x1^2-x1+8x2+20= 8x1+8x2+23=-24+23=-1