We havef=X^3+mX-3,m inRR.How to prove that m>0 f have two roots with equal modules?
2 Answers
See explanation...
Explanation:
Given:
f(x) = x^3+mx-3" " withm in RR
If
Also, the pattern of signs of the coefficients of
Since
Also since the coefficients of
So the two roots must be complex conjugates and have equal modulus.
For interest, here's a method using direct solution of the cubic...
Explanation:
One method involves solving the given cubic directly and observing the resulting roots.
Given:
f(x) = x^3+mx-3
Let
We want to solve:
0 = (u+v)^3+m(u+v)-3
color(white)(0) = u^3+v^3+(3uv+m)(u+v)-3
Add the constraint
u^3-m^3/(27u^3)-3 = 0
Multiply through by
27(u^3)^2-81(u^3)-m^3 = 0
Using the quadratic formula, we find:
u^3 = (81+-sqrt(6561+108m^3))/54
color(white)(u^3) = (81+-3sqrt(729+12m^3))/54
color(white)(u^3) = (27+-sqrt(729+12m^3))/18
Note that if
Since
u_1 = root(3)((27+sqrt(729+12m^3))/18)
and two non-real complex cube roots:
u_2 = omega root(3)((27+sqrt(729+12m^3))/18)
u_3 = omega^2 root(3)((27+sqrt(729+12m^3))/18)
where
Given that we know that
x_1 = root(3)((27+sqrt(729+12m^3))/18)+root(3)((27-sqrt(729+12m^3))/18)
with the other two zeros being:
x_2 = omega root(3)((27+sqrt(729+12m^3))/18)+omega^2 root(3)((27-sqrt(729+12m^3))/18)
x_3 = omega^2 root(3)((27+sqrt(729+12m^3))/18)+omega root(3)((27-sqrt(729+12m^3))/18)
Note that these latter two zeros are complex conjugates of one another since
In any case,