Question

Find the value of $\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}$.
$\left[4\right]$

Answer 1

Solution:
$\frac{1}{3-\sqrt{8}}\times \frac{3+\sqrt{8}}{3+\sqrt{8}}=\frac{3+\sqrt{8}}{9-8}$

$=3+\sqrt{8}$

$\frac{1}{\sqrt{8}-\sqrt{7}}\times \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}}=\frac{\sqrt{8}+\sqrt{7}}{8-7}$

$=\sqrt{8}+\sqrt{7}$ [1.5 Marks]

Similarly, by rationalising other terms, we get

$\frac{1}{\sqrt{7}-\sqrt{6}}=\sqrt{7}+\sqrt{6}$

$\frac{1}{\sqrt{6}-\sqrt{5}}=\sqrt{6}+\sqrt{5}$

and $\frac{1}{\sqrt{5}-2}=\sqrt{5}+2$ [1.5 Marks]

So, $\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}$

$=3+\sqrt{8}-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+2)$

$=3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2$

= 5 [1 Mark]