How to Solve a Quartic Polynomial with x^4+1=2x(x^2+1)?

[anemone]

Solve \(\displaystyle x^4+1=2x(x^2+1)\).

 

[MarkFL]

I would first write the quartic in standard form and assume it has two quadratic factors:

\(\displaystyle x^4-2x^3-2x+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1\)

Equating coefficients, we find:

\(\displaystyle a+b=-2\)

\(\displaystyle 2+ab=0\)

We have two solutions, but they are symmetric, hence we may take:

\(\displaystyle a=\sqrt{3}-1,\,b=-(\sqrt{3}+1)\)

Hence:

\(\displaystyle x^4-2x^3-2x+1=(x^2+(\sqrt{3}-1)x+1)(x^2-(\sqrt{3}+1)x+1)=0\)

Application of the quadratic formula on the two factors gives us:

\(\displaystyle x=\frac{1-\sqrt{3}\pm i\sqrt{2\sqrt{3}}}{2},\,\frac{1+\sqrt{3}\pm\sqrt{2\sqrt{3}}}{2}\)

 

[topsquark]

You could try this. But I'll admit that I've never gotten through one of those without making any mistakes. It is, in a word, "tedious."

-Dan

 

[anemone]

You could try this. But I'll admit that I've never gotten through one of those without making any mistakes. It is, in a word, "tedious."

-Dan

Hi Dan,

The link doesn't work...

And my solution is less clever than the approach of MarkFL. (Nerd)

First, I let \(\displaystyle x=\tan \theta\) to transform the original equation and making it as \(\displaystyle \sin^2 2\theta+2\sin 2\theta-2=0\).

I then use the quadratic formula to solve for \(\displaystyle \sin 2\theta\) then \(\displaystyle \tan \theta\) to obtain the numerical approximation of the answers where

\(\displaystyle x=\tan 23.5293^{\circ}=0.4354\) and \(\displaystyle x=\tan 66.4707^{\circ}=2.2966\).

Also, I was not able to find the complex roots as well!

 

[topsquark]

Hi Dan,

The link doesn't work...

It should work now. Thanks for the catch.

-Dan