JAMB 2001 · UME · Q5

Given that $p = 1 + \sqrt2$ and $q = 1 - \sqrt2$ , evaluate $\dfrac{p^2 - q^2}{2pq}$ .

Worked solution (try it first)

Factorise the top as a difference of two squares: $p^2 - q^2 = (p - q)(p + q)$ .
$p - q = 2\sqrt2$ and $p + q = 2$ , so $p^2 - q^2 = 4\sqrt2$ .
$pq = (1 + \sqrt2)(1 - \sqrt2) = 1 - 2$ , which is $-1$ , so $2pq = -2$ .
So the value is $\dfrac{4\sqrt2}{-2} = -2\sqrt2$ , option C.