Secondary solving method of polynomial

Question

$$x+1+\frac{1}{x}=0$$

This is a fairly trivial and possibly bland equation to solve. But for the sake of the question I will display them here:

$$x\left(x+1+\frac{1}{x}\right)=x(0)$$ $$x^2+x+1=0$$ Solve for $x$ using the Quadratic formula. But is there a way other than multiplying by $x$ as the first step, guessing and checking, graphing, or plugging the equation into the Quadratic formula first to solve this equation? No matter how complicated, I would like all possible ways to solve. What I am talking about is the use of derivatives, integrals, etc. If someone has already done your solution, please refrain from putting the same solution in your answer. Thank you. (Note this equation only has complex solutions)

Answer 1

You can also complete the square:

$$x+1+\frac1x=0$$

$$x^2+x+1=0$$

$$(x+0.5)^2+0.75=0$$

$$(x+0.5)^2=-0.75$$

$$x+0.5=\pm \frac{1}{2}\sqrt{3} i$$

$$x=-0.5\pm \frac{1}{2}\sqrt{3} i$$

Answer 2

The terms $x+\dfrac1x$ call for a change of variable with an exponential $x=e^t$ to yield $2\cosh(t)$.

Hence, $$\cosh(t)=-\frac12.$$

As the RHS is negative, the substitution must involve complex numbers and $x=e^{it}$. Then

$$\cos(t)=-\frac12,$$ $$t=\pm\frac{2\pi}3$$ and $$x=e^{\pm i\frac{2\pi}3}.$$

Answer 3

By the form of the equation, $x$ and $\dfrac1x$ are both solutions, and their product is $p:=1$. Their sum is drawn from the equation, $s:=x+\dfrac1x=-1$.

From the sum and the product, we derive the difference, by

$$d^2=s^2-4p=-3.$$

Then $$x=\frac{s\pm d}2=\frac{-1\pm i\sqrt3}2.$$

Answer 4

Multiply by $x-1$ to cause cancellations:

$$(x-1)(x+1+\frac1x)=x^2+x+1-x-1-\frac1x=x^2-\frac1x=0.$$

Then $$x^3=1$$ and the solutions are the complex cubic roots of $1$.