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

[anemone]

Solve $$x^4+1=2x(x^2+1)$$.

 

[MarkFL]

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

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

Equating coefficients, we find:

$$a+b=-2$$

$$2+ab=0$$

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

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

Hence:

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

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

$$x=\frac{1-\sqrt{3}\pm i\sqrt{2\sqrt{3}}}{2},\,\frac{1+\sqrt{3}\pm\sqrt{2\sqrt{3}}}{2}$$

 

[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 $$x=\tan \theta$$ to transform the original equation and making it as $$\sin^2 2\theta+2\sin 2\theta-2=0$$.

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

$$x=\tan 23.5293^{\circ}=0.4354$$ and $$x=\tan 66.4707^{\circ}=2.2966$$.

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