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I honestly don't have any idea at all on how to solve this. I am asked to find solutions under $\mathbb{R},\mathbb{Q},\mathbb{C}$ respectively but this seems impossible to solve without a computer.

Does anyone know how to go about this? Thank you

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  • $\begingroup$ It shouldn't be hard to graph it at least and figure out how many roots it has. Because it has a real simple derivative. By the IVT there's at least one root between $-1$ and $0$. I suspect it only has one, but I didn't do the details so not sure. $\endgroup$ – Gregory Grant Jan 17 '16 at 20:16
  • $\begingroup$ Maybe works $(t^{2}+t+1)(t^{3}-t^{2}+1)=0$ $\endgroup$ – pablocn_ Jan 17 '16 at 20:17
  • $\begingroup$ Yeah isn't the derivative always positive? So it can have only one root. $\endgroup$ – Gregory Grant Jan 17 '16 at 20:17
  • $\begingroup$ It has only one real root. Do you think you can factor it? $\endgroup$ – Akiva Weinberger Jan 17 '16 at 20:21
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HINT:

$$t^5+t+1=0\Longleftrightarrow$$ $$\left(t^2+t+1\right)\left(t^3-t^2+1\right)=0$$

Split into two equations:

  • Use the 'abc-formula' to find the solution to: $$t^2+t+1=0$$

  • $$t^3-t^2+1=0\Longleftrightarrow$$


Eliminate the quadratic term by substituting $x=t-\frac{1}{3}$:


$$1-\left(x+\frac{1}{3}\right)^2+\left(x+\frac{1}{3}\right)^3=0\Longleftrightarrow$$ $$x^3-\frac{x}{3}+\frac{25}{27}=0\Longleftrightarrow$$


If $x=y+\frac{\lambda}{y}$ then $y=\frac{1}{2}\left(x+\sqrt{x^2-4\lambda}\right)$:


$$25+\frac{1}{3}\left(-y-\frac{\lambda}{y}\right)+\left(y+\frac{\lambda}{y}\right)^3=0\Longleftrightarrow$$


Multiply both sides by $y^3$ and collect in terms of $y$;

Substitute $\lambda=\frac{1}{9}$ and then $z=y^3$, yielding a quadratic equation in the variable $z$:


$$z^2+\frac{25z}{27}+\frac{1}{729}=0$$

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  • $\begingroup$ Omg. Where did you learn all that? $\endgroup$ – johni Jan 17 '16 at 20:42
  • $\begingroup$ Is "abc-formula" another name for the quadratic formula? I've never heard it called that before $\endgroup$ – ziggurism Jan 18 '16 at 12:51
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    $\begingroup$ @ziggurism yes it is! $\endgroup$ – Jan Eerland Jan 18 '16 at 13:05
  • $\begingroup$ Seems like you could also justify the name for the Pythagorean theorem if you wanted. $\endgroup$ – ziggurism Jan 20 '16 at 21:03
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$$t^5+t+1=(t^2+t+1)(t^3-t^2+1)=0$$

The equation $t^2+t+1=0$ can easily be solved. The equation $t^3-t^2+1=0$ can be solved either numerically or with the (complicated) formula for cubic equations.

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For $\mathbb Q$ one can use the rational root theorem.

This gives that the only $1$ and $-1$ can be rational roots, but they clearly aren't.

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