So I was yet again scrolling through Youtube to see if there were any math equations that I might be able to solve when I came across this equation by Sybermath that I thought that I might be able to solve. The equation was
$$\text{Solve: }x^6+x^4+x^2=3$$
which I thought that I might be able to solve. Here is my attempt at solving the equation:
$$\text{Right away we can see that both}$$
$$x=-1$$
$$\text{and}$$
$$x=1$$
$$\text{are solutions to the equation }x^6+x^4+x^2=3$$
$$\text{However, are there any imaginary solutions to the equation?}$$
$$x^6+x^4+x^2=3$$
$$x^2(x^4+x^2+1)=3$$
$$x^2(x^2(x^2+1)+1)=3$$
$$x^2(x^2(x-i)(x+i)+1)=3$$
$$x^2(x-i)(x+i)+1=\frac{3}{x^2}$$
$$x^2(x-i)(x+i)+\frac{x^2}{x^2}=\frac{3}{x^2}$$
$$x^2(x-i)(x+i)+\frac{x^2-3}{x^2}=0$$
$$x^2(x-i)(x+i)+\frac{(x+\sqrt3)(x-\sqrt3)}{x^2}=0$$
$$(x-i)(x+i)=-\frac{x^2-3}{x^4}$$
$$(x-i)(x+i)=-\left(\frac{1}{x^2}-\frac{3}{x^4}\right)$$
$$(x-i)(x+i)=-1\left(\frac{1}{x^2}-\frac{3}{x^4}\right)$$
$$(x-i)(x+i)=-\frac{1}{x^2}+\frac{3}{x^4}$$
Plugging what we have into the quadratic formula gets us
$$\frac{\pm\sqrt{-4(1)(1)}}{2}=\frac{x^2-3}{x^4}$$
$$\pm i=\frac{x^2-3}{x^4}$$
$$\pm ix^4=x^2-3$$
And plugging what we know here into the quadratic formula gets us
$$\frac{\pm\sqrt{-4(1)(-3)}}{2}=\pm ix^4$$
$$\frac{\pm\sqrt{12}}{2}=\pm ix^4$$
$$\pm\sqrt3=ix^4$$
$$\frac{\sqrt3}{i}=x^4$$
$$-i\sqrt3=x^4$$
Square rooting both sides gets us
$$i\sqrt[4]3=x^2$$
And square rooting again gets us
$$\sqrt[8]3\left(\frac{\sqrt2+i\sqrt2}{2}\right)=x$$
$$\text{Or }\pm\frac{\sqrt[8]3\sqrt2+i\sqrt2}{2}=x$$
My question
Is my solution correct, or what could I do to attain the correct solution(s) or attain it more easily?
Best Answer
Note if you can find $x=1, x=-1$ are roots to the equation, it means
$$x^6+x^4+x^2-3=(x+1)(x-1)P_4(x)$$
You can apply the long division technique, and get
$$P_4(x)=x^4+2x^2+3$$
To find the rest four roots, just solve $P_4(x)=0$, which gives
$$x^4+2x^2+3=(x^2+1)^2+2=0$$ then we get $$\Rightarrow x^2+1=\pm \sqrt2 i$$
and you proceed from here.