Question

Find the values of each of the following correct to three places of decimals, it being given that $\sqrt{2}=1.4142,\sqrt{3}=1.732,\sqrt{5}=2.2360,\sqrt{6}=2.4492and\sqrt{10}=3.162$.

(i) $\frac{3-\sqrt{5}}{3+2\sqrt{5}}$

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

Answer 1

$\sqrt{2}=1.4142,\sqrt{3}=1.732,\sqrt{5}=2.2360,\sqrt{6}=2.4495and\sqrt{10}=3.162$

(i) $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=\frac{(3-\sqrt{5})(3-2\sqrt{5})}{(3+2\sqrt{5})(3-2\sqrt{5})}$

(Rationalising the denominator)

$=\frac{9-6\sqrt{5}-3\sqrt{5}+2\times 5}{(3{)}^2-(2\sqrt{5}{)}^2}=\frac{9+10-9\sqrt{5}}{9-20}$

$=\frac{19-9\sqrt{5}}{-11}=\frac{19-9\times 2.2360}{-11}$

$=\frac{19-20.124}{-11}=\frac{-1.124}{-11}=0.102$

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

(Rationalising the denominator)

$=\frac{3+2\sqrt{2}+3\sqrt{2}+2\times 2}{(3{)}^2-(2\sqrt{2}{)}^2}=\frac{3+4+5\sqrt{2}}{9-8}$

$=\frac{7+5\sqrt{2}}{1}=7+5\left(1.4142\right)$

$=7+7.0710=14.071$