已知x₁,x₂是一元二次方程x²+3x+1=0的两实根,则x₁³+8x₂+20=
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 13:06:55
![已知x₁,x₂是一元二次方程x²+3x+1=0的两实根,则x₁³+8x₂+20=](/uploads/image/z/12619862-62-2.jpg?t=%E5%B7%B2%E7%9F%A5x%26%238321%3B%2Cx%26%238322%3B%E6%98%AF%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8Bx%26sup2%3B%2B3x%2B1%3D0%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%A0%B9%2C%E5%88%99x%26%238321%3B%26sup3%3B%2B8x%26%238322%3B%2B20%3D)
已知x₁,x₂是一元二次方程x²+3x+1=0的两实根,则x₁³+8x₂+20=
已知x₁,x₂是一元二次方程x²+3x+1=0的两实根,则x₁³+8x₂+20=
已知x₁,x₂是一元二次方程x²+3x+1=0的两实根,则x₁³+8x₂+20=
答:
x₁,x₂是一元二次方程x²+3x+1=0的两实根
根据韦达定理有:
x₁+x₂=-3
x₁*x₂=1
则:
x₁³+8x₂+20
=(-3x₁-1)x₁+8x₂+20
=-3x1²-x1+8x2+20
=-3(-3x1-1)-x1+8x2+20
=9x1+3-x1+8x2+20
=8x1+8x2+23
=8*(-3)+23
=-1