Jambmaths question:
Given that $p=1+\sqrt{2}\text{ and }q=1-\sqrt{2}$ evaluate $\frac{{{p}^{2}}-{{q}^{2}}}{2pq}$
Option A:
$2(2+\sqrt{2})$
Option B:
$-2(2+\sqrt{2})$
Option C:
$2\sqrt{2}$
Option D:
$-2\sqrt{2}$
Jamb Maths Solution:
$\begin{align} & \frac{{{p}^{2}}-{{q}^{2}}}{2pq}=\frac{(p+q)(p-q)}{2pq} \\ & =\frac{(1+\sqrt{2}+1-\sqrt{2})(1+\sqrt{2}-1+\sqrt{2})}{2(1+\sqrt{2})(1-\sqrt{2})} \\ & =\frac{2(2\sqrt{2})}{2(1-2)}=-2\sqrt{2} \end{align}$
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