Question

Given , $\sqrt{2}=1.414,\sqrt{3}=1.732$ and $\sqrt{6}=2.449$ , find the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 place of decimal

Answer 1

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

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

lets do rationalize the denominator,

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

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

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

$=\frac{\sqrt{3}-\sqrt{2}+1}{4-2\sqrt{3}}$

Lets do rationalize again,

$=\frac{\sqrt{3}-\sqrt{2}+1}{4-2\sqrt{3}}\times \frac{4+2\sqrt{3}}{4+2\sqrt{3}}$

$=\frac{4\sqrt{3}+6-4\sqrt{2}-2\sqrt{2}\sqrt{3}+4+2\sqrt{3}}{16-12}$

$=\frac{6\sqrt{3}-4\sqrt{2}-2\sqrt{6}+10}{4}$

Now, lets take $\sqrt{2}=1.414,\sqrt{6}=2.449$ and $\sqrt{3}=1.732$

$=\frac{6\left(1.732\right)-4\left(1.414\right)-2\left(2.449\right)+10}{4}$

$=\frac{10.392-5.656-4.898+10}{4}$

$=\frac{20.392-10.594}{4}$

$=\frac{9.798}{4}$

$=2.4495$

$=2.450$ corrected to three decimals.

$\therefore \frac{1}{\sqrt{3}-\sqrt{2}-1}=2.45$