Question

Given, $\sqrt{2}=1.414\mathrm{and}\sqrt{6}=2.449$, find the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal.

Answer 1

$$\begin{gathered} \frac{1}{\sqrt{3}-\sqrt{2}-1} \\ =\frac{1}{\sqrt{3}-\left(\sqrt{2}+1\right)}\times \frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{\sqrt{3}+\left(\sqrt{2}+1\right)} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}+1\right)}^2} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2-2\left(\sqrt{2}\right)\left(1\right)-{\left(1\right)}^2} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{3-2-2\sqrt{2}-1} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{-2\sqrt{2}} \\ =\frac{\sqrt{3}+\left(\sqrt{2}+1\right)}{-2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\sqrt{6}+2+\sqrt{2}}{-4} \\ =\frac{2.449+2+1.414}{-4}\left(\because \sqrt{2}=1.414\mathrm{and}\sqrt{6}=2.449\right) \\ =\frac{5.863}{-4} \\ =-1.465 \end{gathered}$$

Hence, the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal is −1.465.