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What is real root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$?

Some remarks:

  1. I saw this quintic in wikipedia

  2. Real root is given $x=1+{\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{2}}\right)^{4}$ in wikipedia.

  3. I used the transformation $x=y+1$ (Tschirnhaus transformation) and $y^5 + 20 y^3 + 20 y^2 + 30 y + 10=0$. (We can remove the term of degree four.)

  4. Therefore, we have to solve $x^5 + 20 x^3 + 20 x^2 + 30 x + 10=0$ and we have to find $x={\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{2}}\right)^{4}$.

  5. But, I want to know how to solve this without plugging it in and verifying an already known root. Can the depressed quintic be solved? Or does one need to use another method to solve this polynomial?

Stella Biderman
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scarface
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  • You want to know how to solve the quintic for its real root without knowing in advance what the real root is? – Brian Tung Apr 19 '17 at 21:26
  • Yes, exactly @Brian Tung – scarface Apr 19 '17 at 21:30
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    I've editted the question to be clearer – Stella Biderman Apr 19 '17 at 21:32
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    There is no "quadratic formula" for 5th degree polynomials. Individual ones may have special properties that allow them to be factored though. I don't know what this one is, but it probably has something to do with $r-2$ being a partial geometric sequence. – DanielV Apr 19 '17 at 21:42
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    https://math.stackexchange.com/questions/1845342/x-pk-5pl-5pm-5pn-5-is-a-root-of-a-quintic-equation-for-any-p/1847602#1847602 – AlexSam Apr 19 '17 at 23:59

2 Answers2

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First off, I don't know of a systematic way to solve it. But suppose one has the magic inspiration to effect the substitution $x=y+2\,$, then the equation in $y$ turns out to be: $$y^5 + 5 y^4 + 30 y^3 + 90 y^2 + 135 y + 81 = 0$$

At this point, the equation became easy to solve. The ratios between symmetric coefficients are $90/30=3^1\,$, $135/5=3^3\,$, $81/1=3^4\,$, suggesting the substitution $y=3z\,$, which gives: $$3 z^5 + 5 z^4 + 10 z^3 + 10 z^2 + 5 z + 1=0 \;\;\iff\;\; 2z^5 + (z+1)^5=0$$

The latter has the obvious real solution $1+ \cfrac{1}{z}=-\sqrt[5]{2}\,$ which can be rationalized as: $$z = \frac{-1}{1+\sqrt[5]{2}} \cdot \frac{1-\sqrt[5]{2}+\sqrt[5]{2}^2-\sqrt[5]{2}^3+\sqrt[5]{2}^4}{1-\sqrt[5]{2}+\sqrt[5]{2}^2-\sqrt[5]{2}^3+\sqrt[5]{2}^4} = \frac{-1+\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4}{1+2} $$

Reverting back to $x=3z+2$ gives $x=1+\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4\,$.

dxiv
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  • Unexpected solution! +1 – scarface Apr 19 '17 at 23:00
  • @scarface Thanks. As I wrote, however, it's not obvious to me what heuristic could hint at the initial substitution $x=y+2,$, unless one had some clue in advance of what the real root looks like. – dxiv Apr 19 '17 at 23:04
  • @dixiv , I get your heuristic approach. If we know a solution of a quintic in the form $x= 1+\sqrt[5]{r} + \sqrt[5]{r^2}+\sqrt[5]{r^3}+\sqrt[5]{r^r}$, by geometric sequence $x=\dfrac{r-1}{\sqrt[5]{r} -1}$. After, substitute $x=y(r-1)$ and hence $y=\dfrac{1}{\sqrt[5]{r} -1}$. After substitude $y=\dfrac{1}{z}$ and $z=\sqrt[5]{r} -1$. The quintic in terms of $z$ will be $(z+1)^5 = r$. We can solve this quintic. By reverse of the operations $x=(r-1)y=\dfrac{r-1}{z}$. – scarface Apr 19 '17 at 23:52
  • @scarface More or less so, in this case. Same idea can work in more general cases, see for example the link posted by AlexSam. Fact remains, however, that it's hard to tell by just looking at the original equation, that one would expect a root that can be written as a linear combination of powers of $5^{th}$ roots. – dxiv Apr 20 '17 at 00:19
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We try the general method for radical-solvable quintics.

First reduce the quintic by translating the variable to eliminate the fourth-degree term; here that gives

$y^5+py^3+qy^2+ry+s=0$

with $p=20,q=20,r=30,s=10.$

Next, following Wikipedia, we apply the Cayley resolvent. We form the sextic equation

$[P_3(z)]^2-1024z\Delta=0,$

where the coefficients of the cubic polynomial $P_3$ and the discriminant $\Delta$ are determined from $p,q,r,s$ using formulas in the referenced Wikipedia page. If the resolvent sextic has a rational root, this resolvent root together with the coefficients $p,q,r,s$ will provide inputs to a formula for the roots of $y$.

Here we find we can take a shortcut. The constant term in the cubic polynomial is given by

$-p^6+28p^4r-16p^3q^2-176p^2r^2-80p^2qs+224pq^2r-64q^4+400ps^2+320r^3-1600qrs$

and with the coefficient values given above this is zero. So the resolvent sextic must also have a zero constant term and the rational root zero.

When this happens, not only is the quintic solvable but the formula for the roots is simplified in accordance with a formula given in this answer. (That formula is applied with a different resolvent, Watson's resolvent, but Cayley's and Watson's resolvents have zero roots at the same time.) The net result is that one root for $y$ is given by

$y=\sqrt[5]{a_1}+\sqrt[5]{a_2}+\sqrt[5]{a_3}+\sqrt[5]{a_4}$

where the radicands solve a pair of quadratic equations. For the coefficients $p=20,q=20,r=30,s=10$ as given here, the quadratic equations turn out to be

$a^2-4a-32=0\implies-4,8$

$a^2+14a-32=0\implies-16,2$

These radicands agree with the magnitudes and signs of the claimed fifth roots.

Oscar Lanzi
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