# Working out the discriminant to a polynomial and using for working out “a”

For an equation:

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

i.e.

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

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 $$a$$

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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. math.stackexchange.com/questions/724570/… – 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

The discriminant of

$$x^4+ax^3-b^2$$

is

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

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.

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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