Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For an equation:

$$ x-b^2/x^3+a=0 \\$$


$$ x^4-b^2+ax^3=0 \\$$

If the discriminant is positive (i.e. $> or =0$) for real roots, what is the discriminant for these equations? Can you use the discriminant to solve the inequality for $$a$$

share|cite|improve this question
Do you mean discriminant? – vadim123 Mar 24 '14 at 15:05
Yes, I was told to ask a separate question regarding this from a previous question.… – John Mar 24 '14 at 15:09
Basically from that post; the helper provided information of getting from $$x^3−bx+a=0$$ to : $$a^2/4−b^3/27≥0$$; I would like to do the same for the equation above and then solve for "a" – John Mar 24 '14 at 15:11
You are looking for a discriminant for a quartic, which is substantially more complicated than a discriminant for a cubic. – vadim123 Mar 24 '14 at 15:13
up vote 1 down vote accepted

The discriminant of



$$-b^4(256b^2 + 27a^4)$$

So the discriminant is never positive.

The discriminant is only 0 for b = 0. In this case the equation has the solutions 0 (triple root) and -a (simple root). For $b\ne0$ , the equation has 2 real solutions because the discriminant is negative. Note, that the original equation has only the solution -a in the case b=0. The multiplication with $x^3$ changes the set of solutions in this case.

With the Descartes-sign-rule we can conclude that the equation has a positive and a negative solution, if $b\ne0$.

I forgot the case a = b = 0. In this case, the only solution is 0.

share|cite|improve this answer
Hi Peter, Thank you for the response. What would happen if I hadn't multiplied by $$x^3$$ i.e. used: $$ x-b^2/x^3+a=0 \\$$, how would that look? – John Mar 24 '14 at 15:35
The multiplication is valid, if it is clear that $x\ne0$. If $b\ne0$, this follows easily. – Peter Mar 24 '14 at 16:20
The discriminant is only defined for polynomials, so the multiplication is not only correct, but necessary. A problem only can occur, when the polynomial has root 0. A multiplication with a non-zero number does not change the set of solutions. – Peter Mar 24 '14 at 16:24
I understand the last part, thank you re: the necessity of multiplying by $$x^3$$ However, I am not sure if I 100% understand the answer with regards to the 2 real solutions for $$b≠0$$ What are the two real solutions and if I use the two real solutions and resolve for $$a$$ what does $$a$$ equate to? – John Mar 24 '14 at 17:35
Hi @Peter, I was wondering for the two real solutions, what $$a$$ would equate to? – John Mar 25 '14 at 10:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.