已知x1,x2是方程x方+3x+1=0的两实数根,则x1立方+8x+20=
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![已知x1,x2是方程x方+3x+1=0的两实数根,则x1立方+8x+20=](/uploads/image/z/318948-60-8.jpg?t=%E5%B7%B2%E7%9F%A5x1%2Cx2%E6%98%AF%E6%96%B9%E7%A8%8Bx%E6%96%B9%2B3x%2B1%3D0%E7%9A%84%E4%B8%A4%E5%AE%9E%E6%95%B0%E6%A0%B9%2C%E5%88%99x1%E7%AB%8B%E6%96%B9%2B8x%2B20%3D)
已知x1,x2是方程x方+3x+1=0的两实数根,则x1立方+8x+20=
已知x1,x2是方程x方+3x+1=0的两实数根,则x1立方+8x+20=
已知x1,x2是方程x方+3x+1=0的两实数根,则x1立方+8x+20=
x1,x2是方程x方+3x+1=0的两实数根,
即x1^2+3x1+1=0
x1^2+3x1=-1
x1^2=-3x1-1
根据韦达定理得x1+x2=-3
x1^3+8x2+20
=x1*x1^2+8x2+20
=x1*(-3x1-1)+8x2+20
=-3x1^2-x1+8x2+20
=-3x1^2-9x1+8x1+8x2+20
=-3(x1^2+3x1)+8(x1+x2)+20
=-3*(-1)+8*(-3)+20
=3-24+20
=-1
用求根公式解出x1,x2 再代入 有两个结果
因为 x1^2+3x1=-1 ,又由韦达定理有:x1+x2=-3
x1^3+8x2+20 = x1^3+3x1^2-3x1^2+8x2+20 = x1(x1^2+3x1)-3x1^2+8x2+20
= -x1-3x1^2+8x2+20 =-3x1^2-x1-8x1+8x1+8x2+20
=-3(x1^2+3x1)+8x1+8x2+20 =3+8(x1+x2)+20 = 3+8×(-3)+20
=-1