已知 $a$ 和 $b$ 是二次方程 $x^2 - 2x - 1 = 0$ 的两个根，且 $a > b$。 证明 $\frac{a}{b} = -3 - 2\sqrt{2}$。

1. Find the roots of the quadratic equation using the quadratic formula

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 4}}{2}$
$x = \frac{2 \pm \sqrt{8}}{2}$
$x = \frac{2 \pm 2\sqrt{2}}{2}$
$x = 1 \pm \sqrt{2}$

2. Identify $a$ and $b$ using the given condition $a > b$

Since $a > b$, we have:
$a = 1 + \sqrt{2}$
$b = 1 - \sqrt{2}$

3. Set up the fraction $\frac{a}{b}$ and rationalise the denominator

$\frac{a}{b} = \frac{1 + \sqrt{2}}{1 - \sqrt{2}}$
Multiply numerator and denominator by the conjugate of the denominator, $(1 + \sqrt{2})$:
$= \frac{(1 + \sqrt{2})(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}$

4. Expand and simplify to show the final result

$= \frac{1 + 2\sqrt{2} + (\sqrt{2})^2}{1^2 - (\sqrt{2})^2}$
$= \frac{1 + 2\sqrt{2} + 2}{1 - 2}$
$= \frac{3 + 2\sqrt{2}}{-1}$
$= -3 - 2\sqrt{2} \text{ (shown)}$