We have#f=X^3+mX-3,m inRR#.How to prove that #m>0# #f# have two roots with equal modules?

Given:

#f(x) = x^3+mx-3" "# with #m in RR#

If #m > 0# then the signs of the coefficients of #f(x)# are in the pattern #+ + -#. With one change of signs, Descartes' Rule of Signs tells us that #f(x)# has exactly one positive real zero.

Also, the pattern of signs of the coefficients of #f(-x)# is #- - -#. With no change of signs, Descartes' Rule of Signs tells us that #f(x)# has no negative real zeros.

Since #f(x)# is of degree #3#, it must have two other zeros which are non-real complex.

Also since the coefficients of #f(x)# are real, such zeros occur in complex conjugate pairs.

So the two roots must be complex conjugates and have equal modulus.

One method involves solving the given cubic directly and observing the resulting roots.

Given:

#f(x) = x^3+mx-3#

Let #x = u+v#

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 #v = -m/(3u)# to make the multiplier of #(u+v)# zero and get:

#u^3-m^3/(27u^3)-3 = 0#

Multiply through by #27u^3# to get:

#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 #m > 0# then the radicand is positive, so this gives real values for #u^3#.

Since #u^3 = (27+sqrt(729+12m^3))/18# is real, it yields one real cube root:

#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 #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

Given that we know that #v = -m/(3u)# and using the symmetry of the derivation, we can tell that a real zero of the original cubic is:

#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 #omega^2 = bar(omega)#, so either they simply have the same modulus or they are the same real number (this latter case is actually not possible as you could observe by considering the coefficients of the imaginary parts).

In any case, #x_2# and #x_3# have the same modulus.