Solve the inequality x^2 - 2x - 1 > 0.

Step-by-step solution

1. 1

   Try factoring first, and see why it fails. Integer factors of $-1$ are only $\pm1$, and no pair sums to $-2$. So this trinomial does not factor over the integers, which means the quadratic formula — not factoring — must supply the boundary points.

2. 2

   Solve the associated equation with the quadratic formula. With $a=1$, $b=-2$, $c=-1$:

   $x=\frac{2\pm\sqrt{(-2)^2-4(1)(-1)}}{2}=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm2\sqrt2}{2}=1\pm\sqrt2$

   The discriminant $8$ is positive but not a perfect square, hence the irrational boundaries.

3. 3

   Note the numerical values of the boundaries. $1-\sqrt2\approx-0.4142$ and $1+\sqrt2\approx2.4142$. Having decimals on hand makes the sign reasoning and the final check concrete.

4. 4

   Use the direction the parabola opens instead of a sign table. The leading coefficient $a=1$ is positive, so the graph is an upward parabola crossing the axis at those two points. An upward parabola is negative between its roots and positive outside them. Since the inequality asks for $>0$, take the outside:

   $x<1-\sqrt2\qquad\text{or}\qquad x>1+\sqrt2$

5. 5

   Confirm with test points. At $x=0$ (inside): $0-0-1=-1<0$, correctly excluded. At $x=-1$ (left): $1+2-1=2>0$ ✓. At $x=3$ (right): $9-6-1=2>0$ ✓.

6. 6

   Note the strictness and write interval notation. The inequality is strict, so the roots themselves — where the expression equals exactly $0$ — are excluded:

   $(-\infty,\,1-\sqrt2)\cup(1+\sqrt2,\,\infty)$

   A scan of $1201$ points across $[-6,6]$ agrees with this set at every point ✓.