1. ## Need help with 5th degree equation please!

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!

2. Are you sure that $\displaystyle x_3 ~~$ can be written in the form $\displaystyle a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$ ??

After my calculation , i got
$\displaystyle x_3 = \frac{ 2\sqrt{7} \sin [{ \frac{ \sin^{-1}{ \frac{1}{2\sqrt{7}}}}{3} - \frac{2 \pi}{3}}] + 7 }{3}$

I don't know how to answer your question I'm afraid - this problem was given to me as I posted it by someone else. They deducted all that but have no idea how or if they are correct.

Any chance you could explain to me how you arrived at your answer? (If it's not too much trouble plz!)

Many thanks again!!!

4. Originally Posted by GreekfoX

I don't know how to answer your question I'm afraid - this problem was given to me as I posted it by someone else. They deducted all that but have no idea how or if they are correct.

Any chance you could explain to me how you arrived at your answer? (If it's not too much trouble plz!)

Many thanks again!!!

Hi !

Let's begin in this equation :

$\displaystyle u^3 - (7/3)u + (7/27) = 0$

and recall this identity $\displaystyle \sin{3\theta} = 3\sin{\theta} - 4 \sin^3{\theta}$

Change $\displaystyle \sin{\theta} = ku$

$\displaystyle u^3 - [3/(4 k^2)]u + \sin{3\theta} /(4k^3) = 0$

then compare with the equation $\displaystyle u^3 - (7/3)u + (7/27) = 0$ so we have

$\displaystyle 3/(4k^2) = 7/3~~\implies k = 3/(2\sqrt{7})$

And $\displaystyle \sin{3\theta} = (7/27)(4)(9/28)(3/(2\sqrt{7})$

$\displaystyle sin{\theta} = \sin[\frac{\sin^{-1}{ \frac{1}{2\sqrt{7}}}}{3}- \frac{2 \pi}{3}]$

$\displaystyle \sin{\theta} = ku \implies u = (1/k) \sin(...)$

Finally , sub. the value of u into $\displaystyle x = u + 7/3$ then the three roots are all obtained .

5. Many thanks again simplependulum! I must admit I did not understand half of the steps, but it is more than good enough!!

However, I have a question: The answer you gave me in your first post, I used the calculator to find the answer as a number and I got 2.380635654, which is none of the roots! I am looking for 0.753020375967

6. Originally Posted by GreekfoX
Many thanks again simplependulum! I must admit I did not understand half of the steps, but it is more than good enough!!

However, I have a question: The answer you gave me in your first post, I used the calculator to find the answer as a number and I got 2.380635654, which is none of the roots! I am looking for 0.753020375967
If you input $\displaystyle \sin{2\pi}$ , then the value you get is not zero , right ?

7. Originally Posted by simplependulum
Are you sure that $\displaystyle x_3 ~~$ can be written in the form $\displaystyle a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$ ??

After my calculation , i got
$\displaystyle x_3 = \frac{ 2\sqrt{7} \sin [{ \frac{ \sin^{-1}{ \frac{1}{2\sqrt{7}}}}{3} - \frac{2 \pi}{3}}] + 7 }{3}$

I am referring to this one. When I hit it on the calculator I get 2.389635654...