Question 3

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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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