find the value of
Question:

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

Solution:

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

$=\frac{1}{\sqrt{3}-(\sqrt{2}+1)} \times \frac{\sqrt{3}+(\sqrt{2}+1)}{\sqrt{3}+(\sqrt{2}+1)}$

$=\frac{\sqrt{3}+(\sqrt{2}+1)}{(\sqrt{3})^{2}-(\sqrt{2}+1)^{2}}$

$=\frac{\sqrt{3}+(\sqrt{2}+1)}{(\sqrt{3})^{2}-(\sqrt{2})^{2}-2(\sqrt{2})(1)-(1)^{2}}$

$=\frac{\sqrt{3}+(\sqrt{2}+1)}{3-2-2 \sqrt{2}-1}$

$=\frac{\sqrt{3}+(\sqrt{2}+1)}{-2 \sqrt{2}}$

$=\frac{\sqrt{3}+(\sqrt{2}+1)}{-2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{\sqrt{6}+2+\sqrt{2}}{-4}$

$=\frac{2.449+2+1.414}{-4} \quad(\because \sqrt{2}=1.414$ and $\sqrt{6}=2.449)$

$=\frac{5.863}{-4}$

$=-1.465$

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