(1)x³+ax²+bx+c=0有二重根,試證 4(b²-3ac)(a²-3b)=(9c-ab)² (2)x³+ax²+bx+c=0有三重根,試證 a³=27c及a²=3b?

(1)x^3+ax^2+bx+c=0有二重根，試證
4(b^2－3ac)(a^2－3b)=(9c－ab)^2
Sol
設三根m，m，n，m<>n
2m+n=－a
m^2+2mn=b
m^2n=－c
b^2－3ac
=(m^2+2mn)^2－3(2m+n)(m^2n)
=(m^4+4m^3n+4m^2n^2)－(6m^3n+3m^2n^2)
=m^4-2m^3n+m^2n^2
=m^2(m^2-2mn+n^2)
=m^2(m－n)^2
a^2－3b
=(2m+n)^2－3(m^2+2mn)
=(4m^2+4mn+n^2)－(3m^2+6mn)
=m^2－2mn+n^2
=(m－n)^2
9c－ab
=9(－m^2n)+(2m+n)(m^2+2mn)
=－9m^2n+(2m^3+4m^2n+m^2n+2mn^2)
=2m^3－4m^2n+2mn^2
=2m(m^2－2mn+n^2)
=2m(m－n)^2
So
4(b^2－3ac)(a^2－3b)-(9c－ab)^2
=4m^2(m－n)^2(m－n)^2－[2m(m－n)^2]^2
=0

(2)x^3+ax^2²+bx+c=0有三重根,試證 a^3=27c及a^3=3b?
Sol
設三根m，m，m
3m=－a
3m^2=b
m^3=－c
a^3－27c=(－3m)^3－27(－m^3)=0
a^3=27c
a^2－3b=(－3m)^2－3(3m^2)=0
a^2=3b

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