Andrew J.

asked • 11d# 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

Hello,

Trying to answer the question:

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

## 2 Answers By Expert Tutors

Richard P. answered • 11d

PhD in Physics with 10+ years tutoring experience in STEM subjects

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

Andrew J.

Thank you Richard! I believe this is the solution I finally arrived at and posted as a comment below my question above. Much appreciated!10d

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 x^{2} by substitution which reduces the sum of the roots to 0.

In Cardan's formula 81q^{2}+12p^{3} 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 ω^{2} respectively where ω and ω^{2} 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.

Andrew J.

Thank you Paul. I haven't covered Cardan's solution yet, but it sounds like I'm still good to go given the requirement in the question for 3 real roots, which requires that -p be greater than zero.10d

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William W.

11d