Find x for the equation: x+1/x = -1 (x plus one divided by x equals negative one)

We must put a notch somewhere - My answer was the most succinct.  

I think I am going to faint.  

A.  If you mean x + (1/x) = -1 then:

Multiply all terms by x:

x² + 1 = -x

Add x to both sides:

x² + x + 1 = 0

Solve:

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.866\: \!025\: \!403\: \!785}}{i}\right)\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.866\: \!025\: \!403\: \!785}}{i}\\
\end{array} \right\}$$

B.  If you mean (x + 1)/x = -1

Multiply both sides by x

x + 1 = -x

Add x to both sides

2x + 1 = 0

Subtract 1 from both sides

2x = -1

Divide both sides by 2

x = -1/2

.

Find x for the equation: x+1/x = -1 (x plus one divided by x equals negative one)

$$\small{\text{
set $\frac{1}{x} = -1-x$ and see there is no cut between function $\frac{1}{x}$ and line $-1-x$
}}$$

For

x+(1/x) = -1

Alan and Heureka are both telling you that there are no real solutions.

Alan has said x cannot equal 0 because you cannot divide by 0

Alan has rearranaged the equation to give a quadratic.

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

For any quadratic you can determine the nature of the roots by examining the discriminate

$$\\\triangle =b^2-4ac\\
\triangle =1-4=-3\\$$

Since the discriminant (which is under a square root) is negative, there are no real roots.

Thanks for that reminder about the discriminant, Melody......this should always be kept in mind when searching for "real" roots in a quadratic (it can save us some unnecessary work !!!)

Don't faint.......there's only me to revive you, and I'm afraid it's a long trip to Sydney.....!!!

A mild swoon WILL be permitted, however.....