# To find a fifth degree equation by using circles and lines that cannot be solved by radicals

An example quintic whose roots cannot be expressed by radicals is $x^5 - x + 1 = 0$.

I asked a geometry question about a fifth degree equation long time ago . I had an equation in the question. It is $\sin(5\beta)+\sin(\beta)=1$ and It can be expressed as a fifth degree equation $16P^5-20P^3+6P=1$. It was solved by radicals in my previous question.

My Question:

Is it possible to create a fifth degree equation by using circles and lines that cannot be solved by radicals such as $x^5-x+1=0$ ?

In other words;

If an fifth degree equation cannot be solved by radicals, Can we express the roots of the equation (at least one real root) by drawing circles , lines in any way? ( In my example, I used angles to create a fifth degree equation)?

Thanks for answers

• anything you build with circles and lines is solvable by square roots. – mercio Jun 10 '14 at 12:27
• I believe in Schanuel's conjecture. Explanation : Solving a quintic by drawing geometrical objects, one would essentially come up with typical trigonometric function values (which can be replaced by exponential and logarithmic function values), but Schanuel's conjecture implies that the only algebraics expressible in terms of $\exp$ and $\log$ are the solvable ones, which is precisely a "no" to your question. Good question though. – Balarka Sen Jun 10 '14 at 12:27
• @mercio I think he asked for trigonometric solution, rather than just geometric drawings, i.e., angles are involved. For example, trisection of angles are impossible by drawing lines and stuffs but is solvably-possible. – Balarka Sen Jun 10 '14 at 12:27
• @mercio : I am aware that circles and lines is solvable by square roots, I wonder if we try to find angles (trigonometric equations. Please see my previous question). Thanks – Mathlover Jun 10 '14 at 12:33