$x^5+x-1=0\;,\;x\in\mathbb{R}$

Question

Given the following equation $$x^5+x-1=0\;,\;x\in\mathbb{R}$$

How to prove that (unevaluated)

$$x=\dfrac13\left(-1+\sqrt[3]{\dfrac{25}2-\dfrac{3\sqrt{69}}2}+\sqrt[3]{\dfrac12\left(25+3\sqrt{69}\right)}\right).$$

Any hint would be appreciated.

Hint: you can factor the polynomial. – Nishant Jun 29 '14 at 18:47 — Nishant, Jun 29 '14 at 18:47

score 9 · Accepted Answer · answered Jun 29 '14 at 18:52

9

Hint: $x^5 + x - 1 = (x^3 + x^2 -1)(x^2-x+1)$. Solving the first of these will give your solution.

answered Jun 29 '14 at 18:52

Mathmo123

4

And it should be established that the other four roots are not real, so that we may conclude $x$ is this number with certainty. – 2'5 9'2 Jun 29 '14 at 19:10

1 Answers1