✔ 最佳答案
a. Show that the polynomial equation f(x)=0 has a repeated root x=p if, and only if f(p)=0 and f'(p)=0.
=>) If f(x)=0 has a repeated root x=p
f(x) can be wriiten as f(x)=(x-p)^2*Q(x) , for some polynomial Q(x)
So,f(x)=(x-p)^2*Q(x)
f'(x)=(x-p)^2*Q'(x)+Q(x)[2(x-p)]
by putting x=p into f(x) and f'(x)
gives f(p)=f'(p)=0
<=) If f(p)=0 and f'(p)=0
By f(p)=0 , x=p is a root of f(x)
Let f(x)=(x-p)H(x) , for some polynomial H(x)
Then, f'(x)=(x-p)H'(x)+H(x)
and since f'(p)=0 , we have H(p)=0
Hence , it shows that x=p is a root of H(x)
i.e. H(x)=(x-p)P(x) , for some polynomial P(x)
f(x) can thus be wriiten as f(x)=(x-p)^2*P(x)
which shows that f(x)=0 has a repeated root x=p
b. Hence, or otherwise, show that, if the equation
x^3 + ax^2 + bx + c = 0 has a repeated root and a^2 is not equal to 3b
Let f(x)=x^3 + ax^2 + bx + c and p be the repeated root of f(x)
By (a) , we have f(p)=0 and f'(p)=0
For f(p)=0 ,
p^3+ap^2+bp+c=0---(1)
For f'(p)=0
Since f'(x)=3x^2+2ax+b
3p^2+2ap+b=0---(2)
(1)*3-(2):
ap^2+2bp+3c=0---(3)
(2)*a-(3)*3:
(6b-2a^2)p+(9c-ab)=0
p=(9c-ab)/2(a^2-3b)
put p=(9c-ab)/2(a^2-3b) into (2)
a[(9c-ab)/2(a^2-3b)]^2+2b[(9c-ab)/2(a^2-3b)]+3c=0
a(9c-ab)^2+2b(9c-ab)[2(a^2-3b)]+3c[2(a^2-3b)]^2=0
a(9c-ab)^2+4b(9c-ab)(a^2-3b)+12c[(a^2-3b)]^2=0
a(9c-ab)^2+(a^2-3b)[4b(9c-ab)+12c(a^2-3b)]=0
a(9c-ab)^2+(a^2-3b)(-4ab^2+12a^2c]=0
(9c-ab)^2+(a^2-3b)(12ac-4b^2)=0
(9c-ab)^2-4(a^2-3b)(b^2-3ac) = 0