Find quadratic polynom?

Q(x)=-0.397422994x² - 0.790727208x + 0.264948663.

The roots of the cubic (I found them by Cardano's method)

are u = 2cos 40°, v = 2cos80°, w = -2cos20° (u>v>w).

I got 3 equations in the 3 unknowns a, b, c (where Q(x) =

ax² +bx +c) and the answer

by heavy, numerical algebra (done by hand with a pocket

calculator, as I haven't got a software package, like Mathlab,

etc to assist me). I tried keeping the trig forms but found the

algebra even messier than

with numbers. No doubt keeping the trig forms simplifies

somewhat, and keeping the answers in trig form does give you

the exact solution. If you want to go down this road, I've given

you enough information to carry on. BTW, I've found c=-2a/3

by using the theory of eqns (sum of roots, product of roots, etc)

on the cubic.

My quadratic is clearly wrong. I should have checked it, like you did.

After going through all that number work I didn't have the will to

do any more of the same. Your hint that the quadratic has integer

coefficients is the best news I've had today. I might have a crack at

it later on. [I am partially sighed, and my eye tires quickly].

Hi Oregfiu,

Your belief that Q(x) has integer coefficients gave me courage to

tackle this using the exact trig forms, and it paid off. It was rather

easier than I expected.

Q(u)=v, Q(v)=w, Q(w)=u give (after simplifying):

asin40+bsin20=cos50=sin40

asin80-bcos50=cos10=sin80

By inspection these equations are satisfied by

a=1, b=0. So Q(x)=x² - 2. [c=-2a; to get this I

evaluated ∑u² =6, using the theory of eqns on

the cubic. If you want details post a note and I'll

fill you in].

Regards, John

I guess this is cheating but... call x = 2 cos t.

Then 8 cos^3 t - 6 cos t + 1 =0 <===> cos(3t) = -1/2.

So t = 2pi/9, 4pi/9 or 8 pi/9.

notice that cos 16 pi/ 9 = cos 2 pi/9, so that t ---> 2t gives a circular correspondance.

Starting from P( cos t) = cos 2t for P = 2X^2 - 1, you can find Q.

Set cos t = u/2 so that P(cos(t)) = v / 2.

Then 2(u/2)^2 - 1 = v/2 or u^2 - 2 = v^2. Same for (v,w) and (w,u). So Q(x) = x^2 - 2