Solve $x^4+4x^2+6=x$ for all Complex Numbers

[anemone]

Solve for all complex numbers $x$ such that $x^4+4x^2+6=x$.

 

[MarkFL]

My solution:

Spoiler

I would arrange the equation as:

\(\displaystyle x^4+4x^2-x+6=0\)

Then I would assume a possible factorization of the form:

\(\displaystyle x^4+4x^2-x+6=\left(x^2+ax+2\right)\left(x^2+bx+3\right)\)

Expand the right side:

\(\displaystyle x^4+4x^2-x+6=x^4+(a+b)x^3+(ab+5)x^2+(3a+2b)x+6\)

Equating the right coefficients yields the linear system:

\(\displaystyle a=-b\)

\(\displaystyle 3a+2b=-1\)

This implies $a=-1,\,b=1$ and so we have:

\(\displaystyle x^4+4x^2-x+6=\left(x^2-x+2\right)\left(x^2+x+3\right)=0\)

Application of the quadratic formula on each quadratic factor in turn yields:

\(\displaystyle x=\frac{1\pm\sqrt{7}i}{2}\)

\(\displaystyle x=\frac{-1\pm\sqrt{11}i}{2}\)

 

[anemone]

Solve for all complex numbers $x$ such that $x^4+4x^2+6=x$.

Very well done, MarkFL!

My solution:

Spoiler

Completing the square on the LHS we get:
$(x^2+2)^2+2=x$, so

$(x^2+2)^2+2-x=0$,

$(x^2+2)^2-x^2+x^2+2-x=0$,

$(x^2+2+x)(x^2+2-x)+x^2+2-x=0$ and this can be easily factorized as

$(x^2+2-x)(x^2+2+x+1)=0$

$(x^2+3+x)(x^2+2-x)=0$ and solving each factor for the corresponding complex solutions we get

$x=\dfrac{1\pm i\sqrt{7}}{2}$ or $x=\dfrac{-1\pm i\sqrt{11}}{2}$.