Let the roots of f(x) be r,s,t.
Use capital letters to denote the roots of g(x), and these are the squares of their respective lowercase variable name. So R,S,T are the roots of g. R=r^2, S=s^2, T=t^2
Use Vieta's Formulas to relate the coefficients to the roots.
for f(x)
r+s+t = -a
rs+rt+st = b
rst = -c
for g(x)
R+S+T = -b
RS+RT+ST = c
RST = -a So a <= 0
c^2 = -a
so a is the negative of a perfect square
and since c≠0, a≠0 so a < 0
We have r+s+t = -a and we can square both sides
(r+s+t)^2 = r^2+s^2+t^2 + 2(rs+rt+st) = a^2
(-b) + 2(b) = a^2
So b = a^2
a = -c^2
b = c^4
All coefficients in terms of c
f(1)=0 means that a + b + c = -1
so c^4 - c^2 + c + 1 = 0
which is factorable to
(c+1)(c^3 - c^2 + 1) = 0
The roots of the cubic in c are about c≈-0.75488 and 2 complex roots.
c = -1
So (a,b,c) = (-1, 1, -1)
and the requested sum a^2023+b^2023+c^2023 = -1.
f(x)=x^3 - x^2 + x - 1;
= (x - 1) (x^2 + 1) roots are {1, i, -i}
g(x)=x^3 + x^2 - x - 1
= (x - 1) (x + 1)^2 roots are {1, -1, -1}
Indeed the 3 roots of g are squares of the roots of f
|
|
Posted by Larry
on 2023-10-04 16:21:16 |
Analytic solution
