Find the values of each of the following correct to three places of decimals, it being given that

Solution:

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

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$3-2 \sqrt{5}$

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

Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(3-\sqrt{5})(3-2 \sqrt{5})}{9-20}$

$=\frac{(9-6 \sqrt{5}-3 \sqrt{5}+10)}{-11}$

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

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

$=\frac{(9(2.236)-19)}{11}$

$=\frac{(20.124-19)}{11}$

$=\frac{1.124}{11}$

$=0.102$

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

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$3+2 \sqrt{2}$

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

As we know, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\frac{(1+\sqrt{2})(3+2 \sqrt{2})}{9-8}$

$=3+2 \sqrt{2}+3 \sqrt{2}+4$

$=7+5 \sqrt{2}$

$=7+7.07=14.07$