pure math的問題

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