Solving $x^2 - x - 1 > 0$ I am having problems understanding how to solve:
$ x^2 - x - 1 > 0 $.
Any help would be much appreciated.
 A: Hint:
$$x^2-x-1 = \left(x-\dfrac{1-\sqrt{5}}{2}\right)\left(x-\dfrac{1+\sqrt{5}}{2}\right)$$
A: Hint
Completing the square, we obtain
$$
x^2-x-1
= \left(x-\frac{1}{2}\right)^2 - \frac{1}{4} - 1
= \left(x-\frac{1}{2}\right)^2 - \frac{5}{4}
=0.
$$
Then, rearrange terms and apply the square root to both side to get
$$
x - \frac{1}{2}
=\pm \frac{\sqrt{5}}{2}
\implies x=\frac {1\pm \sqrt{5}}{2}.
$$
Since the coefficient of $x^2$ is positive, then the expression is positive outside the roots.
A: Graphical method (although this is a bit trivial, it helps understanding quadratic):


*

*Draw the curve of $y = x^2 - x - 2 = (x-2)(x+1)$.

*Move the curve by 1-unit positively in $y$-axis.

*Find the curve sections that are above $x$-axis (which is $y>0$).

*Find the starting point for each section on $x$-axis.
A: Since $x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}$ we must have $|x-\frac{1}{2}|>\frac{\sqrt{5}}{2}$ so $x<\frac{1-\sqrt{5}}{2}$ or $x>\frac{1+\sqrt{5}}{2}$.
A: Hint: you could try to plot your $f(x) = x^2 - x -1$ (can you graph a parabola?) and see the $x$-region where it is positive. Here's a plot made with the help of WA:

You can see that $f(x) > 0$ for $x \in (-\infty , x_-) \cup (x_+, \infty)$, where $x_\pm$ are the positive and negative roots of $f(x)$, respectively.
Cheers!
