Show that x^3+px+q=0 has: (a) one real root if p>0, and (b), three real roots if 4p^3+27Q^2<0

Question

Hello,

Trying to answer the question:

Show that x³+px+q=0 has: (a) one real root if p>0, and (b) three real roots if 4p³+27q²<0

The 'tools' I have at my disposal for the question are the Intermediate Value Theorem, the Mean Value Theorem, Rolle's Theorem, local extrema, and increasing vs. decreasing functions (based on positive or negative derivative).

Answer 1

Let f = x^3 + p x + q

Then f’ = 3 x^2 + p

Setting f’ = 0  There are three cases:  

 p > 0      No solution,  so f is monotone increasing, so just one real root

 p = 0  f = x^3 + q     ,    just one real root for the same reason

 p < 0    , two solutions x = sqrt(z/3)  and – sqrt(z/3)  where z = -p . z is >0

The local minimum of f will occur at   x = sqrt(z/3)  .  plugging in give

 f = (z/3)^3/2 – z (z/3)^1/2 + q this must be > 0 to get three real roots.

Move q to the other side , then square both sides and rearrange terms to get

z^3 - (2/9) z^3 + z^3 < q^2 or

Or  4 z^3   < 27 q^2

which means 27 q^2 + 4 p^2 <0

Answer 2

But the one thing you need which you don't mention is Cardan' solution to the general cubic.

In order to use Cardan's solution you first reduce the general cubic to the form in your problem, i.e you eliminate the term in x² by substitution which reduces the sum of the roots to 0.

In Cardan's formula 81q²+12p³ acts like a discriminant. I will have to look up some more information for you and get back to you...but this should give you a start.

Later:

OK, if p>0 the discriminant will be real and the formula will show that one of the roots is real and the other 2 are the product of the real root with ω and ω² respectively where ω and ω² are the complex roots cube roots of 1.

If the discriminant is less than 0 (as in your problem) then the formula gives 3 complex expressions which are, in fact, real numbers. This is the so-called irreducible cubic. I hope this helps.