- Thread starter
- #1

- Mar 22, 2013

- 573

Find all the closed from of the roots of this solvable quintic

\(\displaystyle 32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0\)

- Thread starter mathbalarka
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- Thread starter
- #1

- Mar 22, 2013

- 573

Find all the closed from of the roots of this solvable quintic

\(\displaystyle 32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0\)

- Thread starter
- #2

- Mar 22, 2013

- 573

\(\displaystyle 32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0\)

\(\displaystyle \Rightarrow (x + 1)(32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1)^2 = 0\)

Expanding this gives

\(\displaystyle 1024 x^{11} - 2816 x^9 + 2816 x^7 - 1232 x^5 + 220 x^3 -11 x + 1 = 0\)

\(\displaystyle \Rightarrow [1024 x^{11} - 2816 x^9 + 2816 x^7 - 1232 x^5 + 220 x^3 -11 x] + 1 = 0\)

Looking at the expression closed with 3-rd bracket carefully, we see that it's one of the Chebyshev polynomials. Hence, letting \(\displaystyle x = \cos(\theta)\), we get

\(\displaystyle \cos(11 \theta) + 1 = 0\)

So, the roots of the equation above are \(\displaystyle \theta = \frac{(2n + 1) \pi}{11}\).

Hence, the (uncertified) roots of the quintic of interest are \(\displaystyle x = \cos \left ( \frac{(2n + 1) \pi}{11} \right )\).

By a quick check, we see that only n = 0, 1, 2, 3, 4 works.

Hence, the roots are \(\displaystyle \cos \left ( \frac{\pi}{11} \right ), \cos \left ( \frac{3 \pi}{11} \right ) , \cos \left ( \frac{5 \pi}{11} \right ) , \cos \left ( \frac{7 \pi}{11} \right ) , \cos \left ( \frac{9 \pi}{11} \right )\)

Balarka

.

- Thread starter
- #3

- Mar 22, 2013

- 573

So, my question is how can we express at least, say, \(\displaystyle \cos (\pi / 11)\) in terms of radicals?

Maybe this one is a challenge too and probably a harder one!