Secondary solving method of polynomial

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

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)

• Just by the way, you realise that the example you've given has no real solutions? You want complex ones? – Zain Patel Jul 7 '15 at 18:23
• hint : $$|x+\frac{1}{x}| \geq 2\\x+1+\frac{1}{x}=0\\x+\frac{1}{x}=-1$$and it is not possible – Khosrotash Jul 7 '15 at 18:23
• Choosing a problem with no real roots does not really open the way to methods from calculus. – Yves Daoust Jul 7 '15 at 20:26

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 $$x=e^{\pm i\frac{2\pi}3}.$$

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

You can also complete the square:

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

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

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

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

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

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

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 $1$.