Question

If $\sqrt{2}=1.414,\sqrt{3}=1.732,\sqrt{5}=2.236$ and $\sqrt{6}=2.449,$ find the value of $\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}$ . [4 Marks]

Answer 1

The given expression :

$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}$

On rationalizing,

$\Rightarrow \frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$ [1 Mark]

$\Rightarrow \frac{4+3+4\sqrt{3}}{4-3}+\frac{4+3-4\sqrt{3}}{4-3}+\frac{3+1-2\sqrt{3}}{3-1}$ [1 Mark]

$\Rightarrow \frac{7+4\sqrt{3}}{1}+\frac{7-4\sqrt{3}}{1}+\frac{4-2\sqrt{3}}{2}$ [1 Mark]

$\Rightarrow 7+4\sqrt{3}+7-4\sqrt{3}+2-\sqrt{3}$

$\Rightarrow 16-\sqrt{3}\Rightarrow 16-1.732=14.268$

$\therefore \text{Answer = 14.268}$ [1 Mark]