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For a cubic equation:

$$x^3-xb+a=0 \\$$

EDIT: the above equation has three real solutions for x.

one of the solutions is: $$a=2\cdot \left(\frac b3\right)^{3/2}$$

EDIT: "one of the solutions is:" should read, "by solving for x, the answer to "a" was found.

How does one arrive at this?

Could anyone please show me the working out please.

John
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  • Solution of this cubic is for $x$, not $a$. Even though, that's definitely wrong. Substitute and check by yourself. – user88595 Mar 24 '14 at 11:19
  • Thanks for the response. The information I've been given is that there are three real solutions for "x", then apparently from this, they've solved for "a", as above. I would like to find out how they arrived at the value for "a". – John Mar 24 '14 at 11:22
  • There are not necessarily three real solutions. There are one, two or three depending on $a$ and $b$. Now if you want to solve for $a$, the equation is linear and the solution is simply $xb - x^3$. Where did you see this?? I think you question is lacking context. – user88595 Mar 24 '14 at 11:29
  • This was given as a part of an algorithm which computes energy terms. All I was given is that there are three real solutions, and they then went on to describe "a" after finding the three real solutions, that is as much information as I have. – John Mar 24 '14 at 11:34

2 Answers2

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I think I understood what you meant. What you wrote is certainly not a solution of the cubic. It is a condition. For the equation $$x^3 - bx + a = 0$$ to have only real solutions, you need the determinant to be positive, that is : $$\frac{a^2}{4} - \frac{b^3}{27}\ge 0$$

It seems that you want to solve the equality $\dfrac{a^2}{4} - \dfrac{b^3}{27} = 0$ which is solved when $a = 2\bigg(\dfrac{b}{3}\bigg)^{3/2}$ which is what you have.

When the determinant is zero, or when the above equation is satisfied, then you have a exactly two (distinct) real solutions as your expression can be factorised : $$x^3 - bx + 2\bigg(\dfrac{b}{3}\bigg)^{3/2} = \bigg(x - \sqrt{\frac{b}{3}}\bigg)^2\bigg(x + 2\sqrt{\frac{b}{3}}\bigg)$$

Here's an illustrative plot using WA which shows what happens when $b = 5$ for example.

user88595
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  • That is quite something! Would it be possible to find an answer for "a" in similar fashion using; $$x-(b^2/x^3)+a=0$$ – John Mar 24 '14 at 11:56
  • I really don't think that's related to the initial problem. Unless you have specific conditions which I can't think off. – user88595 Mar 24 '14 at 12:12
  • Is it possible to put a constraint of maximum value of "a"... if that makes sense? – John Mar 24 '14 at 12:20
  • $b$ gives the shape of the function for small $x$ and $a$ gives the "altitude" of the function, that is how high (or low) the plot will be. You could set a maximum value so that it doesn't go too high in a certain range in $x$. – user88595 Mar 24 '14 at 13:03
  • The conditions that I am aware of are that $$a\ge 0$$ , I thought perhaps in the same way it was possible to use the conditions to arrive at "a" for the cubic equation, it may also be possible to get a value for "a" with this equation. – John Mar 24 '14 at 13:39
  • You will need more conditions than this. – user88595 Mar 24 '14 at 13:41
  • Let me think this through and I'll get back to you shortly – John Mar 24 '14 at 13:42
  • If the condition was that the discriminant is >0, would it then be possible to ask what "a" is? – John Mar 24 '14 at 14:06
  • You could solve the inequality, sure. – user88595 Mar 24 '14 at 14:11
  • Would you be able to point me in the right direction as to how I could solve the inequality? I have to apologise as I have no formal maths training since I was 16, so am not aware of some of the terms used above. – John Mar 24 '14 at 14:14
  • It's like solving an equality but beware of special rules like taking square root to solve for $a$. eg $x^2\ge9 \Rightarrow x\ge3\text{ or } x\le -3$. – user88595 Mar 24 '14 at 14:24
  • Alternatively use Wolfram Alpha : http://www.wolframalpha.com/input/?i=solve+a^2%2F4+-+b^3%2F27+\ge+0%2C+a – user88595 Mar 24 '14 at 14:26
  • Great, I'll do that. would you be able to tell me what the discriminant is for $$x−(b^2/x^3)+a=0$$ – John Mar 24 '14 at 14:30
  • Discriminants are proper to the degree of polynomial. This one is a quartic so the discriminant is very different and too ugly to learn or even write it. Here's a link to what it looks like http://en.wikipedia.org/wiki/Discriminant#Formulas_for_low_degrees – user88595 Mar 24 '14 at 14:34
  • If you want to know more about generating discriminants I recommend you create a new question as it isn't my area of expertise. – user88595 Mar 24 '14 at 14:36
  • No problem at all. If I've understood correctly, for a cubic equation; $$ax^3+bx^2+cx+d$$ the discriminant is: $$b^2c^2 - 4ac^3 - 4b^3d - 27a^2d^2 + 18abcd$$ , using that you can get: $$a^2/4 - b^3/27$$ – John Mar 24 '14 at 14:40
  • If you look at last link you will see what the general formula for the discriminant of a cubic is. Discriminant will only be in terms of the coefficients, not $x$. If you want to know more, please ask a new question. – user88595 Mar 24 '14 at 14:43
  • many thanks for all your help... I will ask a new question later on. – John Mar 24 '14 at 14:47
  • Hi Sorry for the late question: how did you get; $$a^2/4−b^3/27≥0$$ – John Mar 24 '14 at 15:55
  • From Cardano's method : http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method – user88595 Mar 24 '14 at 16:05
  • I just realised this minus in the middle is actually a plus, sorry for the typo. – user88595 Mar 24 '14 at 16:06
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As one of the roots is multiple (why? Otherwise we cannot deduce an equality for $\;a\;$ by only looking at the cubic's discriminant), the cubic's derivative also vanishes at the multiple root $\;\alpha\;$ , say:

$$\begin{cases}I&\alpha^3-b\alpha+a=0\\{}\\II&3\alpha^2-b=0\end{cases}\;\stackrel{I I}\implies \alpha=\sqrt\frac b3\stackrel I\implies$$

$$\stackrel I\implies \left(\frac b3\right)^{3/2}-\frac{b^{3/2}}{\sqrt3}+a=0\implies$$

$$a=\frac{-b\sqrt b+3b\sqrt b}{3\sqrt3}=2\left(\frac b3\right)^{3/2}$$

DonAntonio
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  • Hi DonAntonio, Again I can't thank you enough. Could I please address the same point above that I made to user88595: Would it be possible to find an answer for "a" in similar fashion using; $$ x−(b^2/x^3)+a=0 \$$ – John Mar 24 '14 at 12:07
  • I can only guess what you meant by that, @John, since it really isn't clear: now you have a rational function $$x-\frac{b^2}{x^3}+a=0\iff x^4+ax^3-b^2=0$$ If you give some conditions perhaps something can be said about $;a;$ ... – DonAntonio Mar 24 '14 at 12:10
  • I was told that in the same way that I could find "a" in the previous equation, I could use the same conditions to find "a" again for the latter. Is it possible to put a constraint of maximum value of "a"... if that makes sense? – John Mar 24 '14 at 12:15
  • Perhaps @John, I can't say for sure... – DonAntonio Mar 24 '14 at 12:22