How can you solve $x^2+2x^{-1}-1=0$? How can you solve $x^2+2x^{-1}-1=0$?
 A: $$x^2+2x^{-1}-1=0\implies x^3-x+2=0$$
Check this cubic's roots are awfully ugly, but it has at least one real root.
A: $$x^3-x+2=0$$
use the Newton_Raphson Method
$$y=x^3-x+2$$
$$x_{n+1}=x_n-\frac{y_n}{y'_n}$$
when you find the first root, use the long division to reduce the cubic equation to second and use the quadratic formula to get the other roots
A: since the real root must be negative, call it $-p$ for positive real $p$ so we have:
$$
p^3-p-2 =0
$$
this gives the iterative solution:
$$
p_{n+1} = \sqrt \frac{p_n+2}{p_n}
$$
which is suitable for quick estimate on a hand calculator
A: A good first step is to multiply both sides by x, in order to make every term a non-negative power of x;
$$ x^3-1x + 2=0 $$
While this is of a low enough degree that there is a cubic formula we can use, it's giant and ugly, so instead we can play with the equation (a lot of fun!) and find solutions. I'll say right here, the essence of the solution strategy is to factor things out.
Since there is no set of factors that jumps off the page for the left side of this equation, we can manipulate it some more, putting the constants on the right side and factoring everything that involves x:
$$ x(x^2-1)=-2 \rightarrow x(x-1)(x+1)=-2$$
So, are there any integer values of x for which this equation holds true? 
$x=1 \rightarrow x(x+1)=2 $, but that is not $-2$ nor does it use all three of the factors. In order to have an odd number of negative factors, we need $ (x-1)x(x+1)$ to equal $ -1\cdot0\cdot1 $, which is zero, $-3\cdot-2\cdot-1$, which equals -6, or a triplet even further to the left of the origin, which will give us a product with an even greater absolute value.
Thus, if you're looking for an integer solution, there is none, so you'll need to either use a TI-89, go to http://www.wolframalpha.com/, or go here for the cubic formula: http://www.math.vanderbilt.edu/~schectex/courses/cubic/
Note that $x=0$ is okay when you have uniformly non-negative powers of x, but if you got there from an equation that had a negative power of x, then you need to exclude any solutions that would produce infinity when plugged into the original equation.
A: The real root to this equation is
$$-\frac{\sqrt[3]{9-\sqrt{78}}+\sqrt[3]{9+\sqrt{78}}}{3^{2/3}}$$
or written in another way
$$ -\sqrt[3]{1-\frac{\sqrt{78}}{9}}-\sqrt[3]{1+\frac{\sqrt{78}}{9}}$$
