Hey all! I am new here and this site has been recommended by a friend of mine. I have been given a Math problem to solve but since I haven't had any Math practice for some years now and since this stuff is way out of my reach, I thought I would turn to you guys! If someone can help I would really appreciate it!
So, here is the problem:
To find the real root of the equation of 5th degree as below
x^5 -13.R^2.x^4 +63.R^4.x^3 -140.R^6.x^2 +140.R^8.x -49.R^10 =0 (1)
The real roots are three:
X1 = R^2. [3- sq.rt.2] = 1.5857865
X2 = R^2. [3+ sq.rt.2] = 4.4142135
X3 = R^2. [a + b.sqrt.2 +c.sqrt.3 + d.sqrt.6] = 0.753020375967
X4 = 2.445
X5 = 3.802
Setting X1, X2 in equation (1) then we have equation (1a) as:
E^3 - 7.E^2 + 14.E -7 = 0
with roots as above (X3, X4, X5)
By setting E=Ψ + 7/3 in equation (1a) then we have equation (1b) as:
ψ^3 - [7/3].ψ + [7/27] = 0
with the three roots:
ψ1 = -1.5857
ψ2 = 0.11151
ψ3 = 1.46849
So, I know this equation is solved (the roots are given there) but what I'm looking for is the trigonometric method. Specifically, I am trying to confirm the answer given for root X3 = 0.753......
Could someone help me reach the answer for X3 using the equation (1b) ? I would be very grateful to anyone who can show me the method to do so. Also, the person who has the problem said is trying to solve it trigonometrically (and not algebraically - if that makes any sense to you guys!) and that the answer he is looking for is not in the form 0.753...... but rather contains roots and fractions and is rather long (compared to a few decimals!).
Many thanks in advance! I hope someone understands what I'm talking about and can help me! Thanks!