Question

Find the values the following correct to three places of decimals, it being given that ${\displaystyle \sqrt{2}=1.4142}$, ${\displaystyle \sqrt{3}=1.732}$, ${\displaystyle \sqrt{5}=2.2360}$, ${\displaystyle \sqrt{6}=2.4495}$ and ${\displaystyle \sqrt{10}=3.162}$

${\displaystyle \frac{3-\sqrt{5}}{3+2\sqrt{5}}}$

Answer 1

We know that rationalization factor for ${\displaystyle 3+2\sqrt{5}}$ is ${\displaystyle 3-\sqrt{5}}$. We will multiply numerator and denominator of the given expression ${\displaystyle \frac{3-\sqrt{5}}{3+2\sqrt{5}}}$ by ${\displaystyle 3-2\sqrt{5}}$ to get

${\displaystyle \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}=\frac{{\left(3\right)}^2-3\times 2\times \sqrt{5}-3\times \sqrt{5}+2\times {\left(\sqrt{5}\right)}^2}{{\left(3\right)}^2-{\left(2\sqrt{5}\right)}^2}}$

${\displaystyle =\frac{9-9\sqrt{5}+10}{9-20}}$

${\displaystyle =\frac{19-9\sqrt{5}}{-11}}$

${\displaystyle =\frac{9\sqrt{5}-19}{11}}$

Putting the value of ${\displaystyle \sqrt{5}}$, we get

${\displaystyle \frac{9\sqrt{5}-19}{11}=\frac{9\left(2.236\right)-19}{11}}$

${\displaystyle =\frac{20.124-19}{11}}$

${\displaystyle =\frac{1.124}{11}}$

= 0.102

Hence the given expression is simplified to 0.102