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Please help, equation $3x-x^3=1$ has three roots. Interesting fact that $|x_3|= x_1+x_2$. Is it possible to reduce this equation to a quadratic?

Simply Beautiful Art
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Nioko
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    Interesting fact that ... Not too surprising since $x_1+x_2+x_3=0$ by Vieta's formulas. – dxiv Jul 18 '17 at 08:03
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    "reduce this equation to a quadratic": what do you mean exactly ? Cubics can be solved in closed-form in all cases. –  Jul 18 '17 at 13:56
  • Basically I attempt to find purely geometric solution to 3x-x^3=h, where 2<h<0. Geometrically simple relations between roots suggest possibility without involvement of cubic curves. – Nioko Jul 19 '17 at 01:37
  • mistake (typo ) 2>h>0 – Nioko Jul 19 '17 at 01:54

2 Answers2

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What we can do is use the root coefficient relationships to express any two roots of the cubic equation in terms of the third root. Let the cubic equation be $ax^3+bx^2+cx+d=0$ and let $r$ be any root. Then the remaining two roots must sum to $-(ra+b)/a$ and their product must be $-d/(ra)$ leading to the quadratic equation:

$(ra)x^2+(r^2a+rb)x-d=0$

and the other two roots are:

$r_{\pm} = \frac{-(r^2a+rb)\pm \sqrt{(r^2a+rb)^2+4rad}}{2ra}$

In some applications, such as the thermodynamics problem of modeling vapor liquid equilibrium with a cubic equation of state, we can set up cubic equations to have a predefined root (corresponding to one phase in the equilibrium problem) and use the above scheme to get the other roots.

BrownianBread
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Oscar Lanzi
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  • Awesome! Empirically I found that 1/r1 = r3-1, 1/r2 = 1-r1 and 1/r3 = r2-1. But solution you provided give much more general relations between roots! – Nioko Jul 19 '17 at 01:15
  • Never would think that this observation can assist to improve prediction of phase change of state of matter. Mathematical modeling seem to be useful. – Nioko Jul 19 '17 at 01:19
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No, because our equation has three roots.

If you mean if is it possible to write

this equation in the form $(x-a)(x^2+bx+c)=0$, where $\{a,b,c\}\subset\mathbb Q$,

then it's impossible again because our equation has no rational roots.

Michael Rozenberg
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