Question 1: State with reason which of the following are surds.

(i) $\sqrt{18}$ $= \sqrt{9 \times 2} = 3\sqrt{2}$. Therefore it is surd.

(ii) $\sqrt[3]{7}$. This is a surd.

(iii) $\sqrt{2+{\sqrt{3}}}$: We observe that $\sqrt{2+{\sqrt{3}}}$ is an irrational number. But, $2+{\sqrt{3}}$ is not a rational number. Hence $\sqrt{2+{\sqrt{3}}}$ is not a surd.

(iv) $\sqrt[3]{\pi}$. $\pi$ cannot be expresses as a rational number under a root sign. Therefore $\sqrt[3]{\pi}$ is not a surd.

(v) $\sqrt{\frac{3}{16}}$ $=$ $\sqrt{\frac{3}{4^2}}$ $=$ $\frac{1}{4}$ $\sqrt{3}$. Therefore it is a surd.

Question 2: Simplify the following:

(i) $(\sqrt[4]{5})^4$ $=$ $\Big[$ $(5)$ $^{\frac{1}{4}} \Big]^4$ $= 5$

(ii) $\sqrt[3]{54}$ $= (54)^{\frac{1}{3}} = (3 \times 3 \times 3 \times 2)^{\frac{1}{3}} = 3(2)^{\frac{1}{3}} = 3 \sqrt[3]{2}$

(iii) $\sqrt[4]{1875}$ $= (25 \times 25 \times 3)^{\frac{1}{4}} = (5^4)^{\frac{1}{4}} (3)^{\frac{1}{4}} = 5(3)^{\frac{1}{4}} = 5 \sqrt[4]{3}$

(iv)  $\sqrt[5]{\sqrt[3]{12}}$ $= \{ (12)^{\frac{1}{3}} \}^{\frac{1}{15}} = (12)^{\frac{1}{15}} = \sqrt[15]{12}$

(v) $4\sqrt[5]{64}$ $= 4 (64)^{\frac{1}{5}} = 4 (2^5 . 2)^{\frac{1}{5}} = 8(2)^{\frac{1}{5}} = 8\sqrt[5]{2}$

Question 3: Express the following as pure surds:

(i) $5\sqrt[3]{4}$ $= \sqrt[3]{5^3 \times 4} = \sqrt[3]{125 \times 4} = \sqrt[3]{500}$

(ii) $\frac{3}{2}$ $\sqrt{\frac{3}{2}} = \sqrt{\frac{3^2 \times 3}{2^2 \times 2}} = \sqrt{\frac{27}{8}}$

(iii) $\frac{1}{4}$ $\sqrt[3]{128} =$ $\sqrt[3]{\frac{128}{4 \times 4 \times 4}}$ $= \sqrt[3]{3}$

(iv) $\frac{1}{7}$ $\sqrt[3]{1029} =$ $\sqrt[3]{\frac{1029}{7 \times 7 \times 7}}$ $= \sqrt[3]{3}$

(v) $\frac{a}{b} \sqrt[3]{\frac{b^4}{a^4}}$ $=$ $\sqrt[3]{\frac{a^3b^4}{b^3a^4}}$ $=$ $\sqrt[3]{\frac{b}{a}}$

Question 4: Express each of the following as a mixed surd in simplest form:

(i) $\frac{3}{4}$ $\sqrt[3]{128} =$ $\sqrt[3]{\frac{3 \times 3 \times 3 \times 128}{4 \times 4 \times 4}}$ $= 3 \sqrt[3]{2}$

(ii) $6 \sqrt[7]{384} = 6 \sqrt[7]{2^7 \times 3} = 12 \sqrt[7]{3}$

(iii) $\sqrt[5]{192} = \sqrt[5]{2^5 \times 2 \times 3} = 2 \sqrt[5]{6}$

(iv) $\sqrt{112} = \sqrt{2^4 \times 7} = 4 \sqrt{7}$

(v) $\sqrt[4]{243} = \sqrt[4]{3 \times 3^4} = 3 \sqrt[4]{3}$

Question 5: Convert:

(i) $\sqrt[3]{9}$ into a surd of order $6$

$\Rightarrow \sqrt[3]{9} = (9)^{\frac{1}{3}} = (9^2)^{\frac{1}{6}} = \sqrt[6]{81}$

(ii) $\sqrt{3}$ into a surd of order $8$

$\Rightarrow \sqrt{3} = (3)^{\frac{1}{2}} = (3^4)^{\frac{1}{8}} = \sqrt[8]{81}$

(iii) $\sqrt{5}$ and $\sqrt[3]{7}$ into surds of the same but smallest order

LCM of $2$ and $3$  is $6$

$\Rightarrow \sqrt{5} = (5)^{\frac{1}{2}} = (5^3)^{\frac{1}{6}} = \sqrt[6]{125}$

$\Rightarrow \sqrt[3]{7} = (7)^{\frac{1}{3}} = (7^2)^{\frac{1}{6}} = \sqrt[6]{49}$

(iv) $\sqrt[4]{6}$ and $\sqrt[8]{12}$ into surds of the same but smallest order

LCM of $4$ and $8$  is $8$

$\Rightarrow \sqrt[4]{6} = (6)^{\frac{1}{4}} = (6^2)^{\frac{1}{8}} = \sqrt[8]{36}$

$\Rightarrow \sqrt[8]{12} = (12)^{\frac{1}{8}} = (12)^{\frac{1}{8}} = \sqrt[8]{12}$

(v) $\sqrt[5]{4}$ into a surd of order $15$

$\Rightarrow \sqrt[5]{4} = (4)^{\frac{1}{5}} = (4^3)^{\frac{1}{15}} = \sqrt[15]{64}$

Question 6: Which is greater?

(i) $\sqrt[4]{5} \ or \ \sqrt[3]{4}$

LCM of $4$ and $3$  is $12$

$\Rightarrow \sqrt[4]{5} = (5)^{\frac{1}{4}} = (5^3)^{\frac{1}{12}} = \sqrt[12]{125}$

$\Rightarrow \sqrt[3]{4} = (4)^{\frac{1}{3}} = (4^4)^{\frac{1}{12}} = \sqrt[12]{256}$

$\Rightarrow \sqrt[3]{4} > \sqrt[4]{5}$

(ii) $\sqrt[3]{3} \ or \ \sqrt[4]{4}$

LCM of $3$ and $4$  is $12$

$\Rightarrow \sqrt[3]{3} = (3)^{\frac{1}{3}} = (3^4)^{\frac{1}{12}} = \sqrt[12]{81}$

$\Rightarrow \sqrt[4]{4} = (4)^{\frac{1}{4}} = (4^3)^{\frac{1}{12}} = \sqrt[12]{64}$

$\Rightarrow \sqrt[3]{3} > \sqrt[4]{4}$

(iii) $\sqrt{3} \ or \ \sqrt[3]{5}$

LCM of $2$ and $3$  is $6$

$\Rightarrow \sqrt{3} = (3)^{\frac{1}{2}} = (3^3)^{\frac{1}{6}} = \sqrt[6]{27}$

$\Rightarrow \sqrt[3]{5} = (5)^{\frac{1}{3}} = (5^2)^{\frac{1}{6}} = \sqrt[6]{25}$

$\Rightarrow \sqrt{3} > \sqrt[3]{5}$

(iv) $\sqrt{7}-\sqrt{3} \ or \ \sqrt{5}-1$

First simplify each of the given terms

$\sqrt{7}-\sqrt{3} =$ $\frac{(\sqrt{7}-\sqrt{3}) \times (\sqrt{7}+\sqrt{3})}{(\sqrt{7}+\sqrt{3})}$ $=$ $\frac{4}{(\sqrt{7}+\sqrt{3})}$

$\sqrt{5}-1 =$ $\frac{(\sqrt{5}-1) \times (\sqrt{5}+1)}{(\sqrt{5}+1)}$ $=$ $\frac{4}{(\sqrt{5}+1)}$

For both the terms, the numerator is the same which is 4. Therefore whichever term has a higher denominator, would be the smaller term. Let’s compare the two denominators.

$\sqrt{7} > \sqrt{5} \ and \ \sqrt{3} > \sqrt{1} \Rightarrow \sqrt{7}+\sqrt{3} > \sqrt{5}+1$

$\Rightarrow$ $\frac{4}{(\sqrt{7}+\sqrt{3})}$ $<$ $\frac{4}{(\sqrt{5}+1)}$

$\Rightarrow \sqrt{5}-1 > \sqrt{7}-\sqrt{3}$

(v) $\sqrt{17}-\sqrt{12} \ or \ \sqrt{11}-\sqrt{6}$

First simplify each of the given terms

$\sqrt{17}-\sqrt{12} =$ $\frac{(\sqrt{17}-\sqrt{12}) \times (\sqrt{17}+\sqrt{12})}{(\sqrt{17}+\sqrt{12})}$ $=$ $\frac{5}{(\sqrt{17}+\sqrt{12})}$

$\sqrt{11}-\sqrt{6} =$ $\frac{(\sqrt{11}-\sqrt{6}) \times (\sqrt{11}+\sqrt{6})}{(\sqrt{11}+\sqrt{6})}$ $=$ $\frac{5}{(\sqrt{11}+\sqrt{6})}$

For both the terms, the numerator is the same which is 5. Therefore whichever term has a higher denominator, would be the smaller term. Let’s compare the two denominators.

$\sqrt{17} > \sqrt{11} \ and \ \sqrt{12} > \sqrt{6} \Rightarrow \sqrt{17}+\sqrt{12} > \sqrt{11}+\sqrt{6}$

$\Rightarrow$ $\frac{5}{(\sqrt{17}+\sqrt{12})}$ $<$ $\frac{5}{(\sqrt{11}+\sqrt{6})}$

$\Rightarrow \sqrt{11}-\sqrt{6} > \sqrt{17}-\sqrt{12}$

Question 7: Arrange in Ascending Order:

(i) $\sqrt[4]{3}, \sqrt[6]{10}, \sqrt[12]{25}$

LCM of $4, 6 \ and \ 12 \ is \ 12$

Now convert all the above terms to order of 12

$\Rightarrow \sqrt[4]{3} = (3)^{\frac{1}{4}} = (3^3)^{\frac{1}{12}} = \sqrt[12]{27}$

$\Rightarrow \sqrt[6]{10} = (10)^{\frac{1}{6}} = (10^2)^{\frac{1}{12}} = \sqrt[12]{100}$

$\Rightarrow \sqrt[12]{25} = (25)^{\frac{1}{12}} = (25^1)^{\frac{1}{12}} = \sqrt[12]{25}$

Now comparing the number under the root sign as they are all of the same order.

$\Rightarrow \sqrt[12]{25} < \sqrt[12]{27} < \sqrt[12]{100}$

or $\Rightarrow \sqrt[12]{25} < \sqrt[4]{3} < \sqrt[6]{10}$

(ii) $\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$

LCM of $6, 3 \ and \ 4 \ is \ 12$

Now convert all the above terms to order of 12

$\Rightarrow \sqrt[6]{6} = (6)^{\frac{1}{6}} = (6^2)^{\frac{1}{12}} = \sqrt[12]{36}$

$\Rightarrow \sqrt[3]{7} = (7)^{\frac{1}{3}} = (7^4)^{\frac{1}{12}} = \sqrt[12]{2401}$

$\Rightarrow \sqrt[4]{8} = (8)^{\frac{1}{4}} = (8^3)^{\frac{1}{12}} = \sqrt[12]{512}$

Now comparing the number under the root sign as they are all of the same order.

$\Rightarrow \sqrt[12]{36} < \sqrt[12]{512} < \sqrt[12]{2401}$

or $\Rightarrow \sqrt[6]{6} < \sqrt[4]{8} < \sqrt[3]{7}$

Question 8: Arrange in Descending Order:

(i) $2\sqrt{3}, 3\sqrt{2}, 5\sqrt{7}$

Convert $2\sqrt{3}, 3\sqrt{2}, 5\sqrt{7}$ into simple surds

$\Rightarrow \sqrt{12}, \sqrt{18}, \sqrt{175}$

Since the order of all the terms is the same, just compare the terms inside the square root. Hence, the descending order is

$\sqrt{175} > \sqrt{18} > \sqrt{12}$

(ii) $\sqrt[3]{10}, \sqrt[3]{36}, \sqrt {3}, \sqrt[6]{5}, \sqrt[8]{60}$

LCM of $3, 2, 6 \ and \ 8 \ is \ 24$

Now convert all the above terms to order of 24

$\Rightarrow \sqrt[3]{10} = (10)^{\frac{1}{3}} = (10^8)^{\frac{1}{24}} = \sqrt[24]{100000000}$

$\Rightarrow \sqrt[3]{36} = (36)^{\frac{1}{3}} = (36^8)^{\frac{1}{24}} = \sqrt[24]{36^8}$

$\Rightarrow \sqrt[2]{3} = (3)^{\frac{1}{2}} = (3^12)^{\frac{1}{12}} = \sqrt[12]{531441}$

$\Rightarrow \sqrt[6]{5} = (5)^{\frac{1}{6}} = (5^4)^{\frac{1}{24}} = \sqrt[24]{625}$

$\Rightarrow \sqrt[8]{60} = (60)^{\frac{1}{8}} = (60^3)^{\frac{1}{24}} = \sqrt[24]{216000}$

Now comparing the number under the root sign as they are all of the same order.

$\Rightarrow \sqrt[24]{36^8} > \sqrt[24]{100000000} > \sqrt[12]{531441} > \sqrt[24]{216000} > \sqrt[24]{625}$

or $\Rightarrow \sqrt[3]{36} > \sqrt[3]{10} > \sqrt[2]{3} > \sqrt[8]{60} > \sqrt[6]{5}$

Question 9: Simplify:

(i) $\sqrt{63}+\sqrt{28}-\sqrt{175}+\sqrt{162}-\sqrt{32}$

$= \sqrt{9 \times 7}+\sqrt{4 \times 7} - \sqrt{25 \times 7} + \sqrt{81 \times 2} - \sqrt{16 \times 2}$

$= 3\sqrt{7} + 2 \sqrt{7} - 5 \sqrt{7} + 9\sqrt{2}-4 \sqrt{2}$

$= 5\sqrt{2}$

(ii) $5\sqrt{3}+ 2 \sqrt{27}+ 4$ $\sqrt{\frac{1}{3}}$

$= 5\sqrt{3} + 2 \sqrt{9 \times 3} +$ $\frac{4}{3}$ $\sqrt{3}$

$= 5\sqrt{3}+ 6\sqrt{3}+$ $\frac{4}{3}$ $\sqrt{3}$

$= (5 + 6 +$ $\frac{4}{3}$ $)\sqrt{3}$

$=$ $\frac{37}{3}$ $\sqrt{3}$

(iii) $2$ $\sqrt{\frac{8}{9}}$ $- 3$ $\sqrt{\frac{1}{2}}$ $+ 5$ $\sqrt{\frac{9}{8}}$

$= 2$ $\sqrt{\frac{4 \times 2}{3 \times 3}}$ $- 3$ $\sqrt{\frac{2}{2 \times 2}}$ $+ 5$ $\sqrt{\frac{3\times 3 \times 2}{4 \times 2 \times 2}}$

$=$ $\frac{4}{3}$ $\sqrt{2}-$ $\frac{3}{2}$ $\sqrt{2} +$ $\frac{15}{4}$ $\sqrt{2}$

$= ($ $\frac{4}{3}$ $-$ $\frac{3}{2}$ $+$ $\frac{15}{4})$ $\sqrt{2}$

$=$ $\frac{43}{12}$ $\sqrt{2}$

(iv) $\sqrt[2]{147}-5$ $\sqrt{\frac{1}{3}}$ $+ 9$ $\sqrt{\frac{1}{3}}$

$= \sqrt[2]{3 \times 7 \times 7}-5$ $\sqrt{\frac{3}{3 \times 3}}$ $+ 9$ $\sqrt{\frac{3}{3 \times 3}}$

$= 7\sqrt{3}-$ $\frac{5}{3}$ $\sqrt{3}+3\sqrt{3}$

$= (7 -$ $\frac{5}{3}$ $+ 3) \sqrt{3}$

$=$ $\frac{25}{3}$ $\sqrt{3}$

(v) $4\sqrt{2} - 2\sqrt{8}+$ $\frac{3}{\sqrt{2}}$

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

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

$= \frac{3}{2}$ $\sqrt{2}$

Question 10: Multiply

(i) $4\sqrt{12} \times 7\sqrt{16}$

$= 4\sqrt{4 \times 3} \times 7\sqrt{4 \times 4}$

$= 4 \times 2 \times \sqrt{3} \times 7 \times 4$

$= 224 \sqrt{3}$

(ii) $\sqrt[3]{7} \times \sqrt{2}$

$LCM of 3 \ and \ 2 \ is \ 6$

$= \sqrt[6]{7^2} \times \sqrt[6]{2^3}$

$= \sqrt[6]{48 \times 8}$

$= \sqrt[6]{392}$

(iii) $\sqrt[4]{2} \times \sqrt[3]{3} \times \sqrt[3]{4}$

LCM of $4, 3, 3, \ is \ 12$

$= \sqrt[12]{2^3} \times \sqrt[12]{3^4} \times \sqrt[12]{4^4}$

$= \sqrt[12]{2^3 \times 3^4 \times 4^4}$

$= \sqrt[12]{2^11 \times 3^4}$

(iv) $3\sqrt{5} \times 2\sqrt{7} \times 5\sqrt{2}$

$= \sqrt{45} \times \sqrt{28} \times \sqrt{50}$

$= \sqrt{45 \times 28 \times 50}$

$= \sqrt{63000} = \sqrt{70 \times 900} = 30 \sqrt{70}$

(v) $3 \times \sqrt[3]{32} \times 3 \sqrt[3]{4}$

$= \sqrt[3]{27} \times \sqrt[3]{32} \times \sqrt[3]{27 \times 4}$

$= \sqrt[3]{3^3 \times 2^5 \times 2^2 \times 3^3}$

$= 36 \sqrt[3]{2}$

Question 11: Divide

(i) $\sqrt{98} \div \sqrt{2}$

$=$ $\sqrt{\frac{98}{2}}$ $= \sqrt{49} = 7$

(ii) $25 \sqrt[4]{33} \div 5 \sqrt[4]{11}$

$=$ $\frac{25 \sqrt[4]{33}}{5 \sqrt[4]{11}}$ $= 5$ $\sqrt[4]{\frac{33}{11}}$ $= 5 \sqrt[4]{3}$

(iii) $6 \sqrt[3]{25} \div \sqrt[2]{5}$

$= \frac{6 \sqrt[3]{25}}{\sqrt[2]{5}}$

$= \frac{6 \sqrt[6] {25 \times 25}}{\sqrt[3]{5 \times 5 \times 5}}$

$= 6$ $\sqrt[6]{\frac{25 \times 25}{5 \times 5 \times 5}}$

$= 6 \sqrt[6]{5}$

(iv) $\sqrt{a^3b^4} \div \sqrt[3]{a^4b^3}$

$= \frac{\sqrt{a^3b^4}}{\sqrt[3]{a^4b^3}}$

$= \frac{\sqrt[6]{(a^3b^4)^3}}{\sqrt[6]{(a^4b^3)2}}$

$= \sqrt[6]{\frac{a^9b^12}{a^8b^6}}$

$= b\sqrt{a}$

(v) $\sqrt{m^2n^2} \times \sqrt[6]{m^2n^2} \times \sqrt[3]{m^2n^2}$

$= \sqrt[6]{(m^2n^2)^3} \times \sqrt[6]{m^2n^2} \times \sqrt[6]{(m^2n^2)^2}$

$= \sqrt[6] {m^6n^6 \times m^2n^2 \times m^4n^4}$

$= \sqrt[6]{m^{12} n^{12}}$

$= m^2n^2$

$\\$

Question 12: Find the rationalising factors of the following:

(i) $\sqrt{108}$

$\sqrt{108} = \sqrt{2^2 \times 3^3} = 6 \sqrt{3}$.

We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt{3} =$ $3^{\frac{1}{2}}$ the rationalizing factor should be $3^{(1-\frac{1}{2})} = 3^{\frac{1}{2}} = \sqrt{3}$

(ii) $\sqrt[5]{486}$

$\sqrt[5]{486} = \sqrt[5]{2 \times 3^5} = 3 \sqrt[5]{2}$.

We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt[5]{2} =$ $2^{\frac{1}{5}}$ the rationalizing factor should be $2^{(1-\frac{1}{5})} = 2^{\frac{4}{5}} = \sqrt[5]{16}$

(iii) $\sqrt{2}+\sqrt{3}+\sqrt{5}$

We find that

$( \sqrt{2}+\sqrt{3}+\sqrt{5}) ( \sqrt{2}+\sqrt{3}-\sqrt{5}) = ( \sqrt{2}+\sqrt{3})^2 - (\sqrt{5})^2 = 2 + 3 + 2 \sqrt{6} - 5 = 2 \sqrt{6}$

Rationalizing factor of $\sqrt{6}$ is $\sqrt{6}$.

Hence $\sqrt{6}( \sqrt{2}+\sqrt{3}-\sqrt{5})$ is the rationalizing factor of  $\sqrt{2}+\sqrt{3}+\sqrt{5}$

(iv) $\sqrt{3}+ \sqrt{8+2\sqrt{15}}$

We have $\sqrt{3}+ \sqrt{8+2\sqrt{15}} = \sqrt{3}+ \sqrt{5 + 3+2\sqrt{5 \times 3}} = \sqrt{3}+ \sqrt{(\sqrt{5} + \sqrt{3})^2}$

$= \sqrt{3} +\sqrt{5}+\sqrt{3} = 2 \sqrt{3}+ \sqrt{5}$

The conjugate of $(2 \sqrt{3}+ \sqrt{5})$ is $(2 \sqrt{3}- \sqrt{5})$

Therefore the rationalizing factor of $\sqrt{3}+ \sqrt{8+2\sqrt{15}}$ is $(2\sqrt{3}-\sqrt{5})$.

(v) $\sqrt[3]{5}$

We know that the rationalizing factor of monomial $a^{\frac{1}{n}}$ is $a^{(1-\frac{1}{n})}$. Therefore the monomial $\sqrt[3]{5} =$ $5^{\frac{1}{3}}$ the rationalizing factor should be $5^{(1-\frac{1}{3})} = 5^{\frac{2}{3}} = \sqrt[3]{25}$

Question 13: Rationalize the denominator and simplify

(i)    $\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$

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

$=$ $\frac{36 + 32 -48\sqrt{2}}{36-32}$

$=$ $\frac{78-48\sqrt{2}}{4}$

$= 17 - 12\sqrt{2}$

(ii) $\frac{b^2}{\sqrt{a^2+b^2} +a}$

$=$ $\frac{b^2}{\sqrt{a^2+b^2} +a}$ $\times$ $\frac{\sqrt{a^2+b^2} -a}{\sqrt{a^2+b^2} -a}$

$=$ $\frac{b^2(\sqrt{a^2+b^2)}}{a^2+b^2-a^2}$

$=$ $\sqrt{a^2+b^2} - a$

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

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

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

$=$ $5 - 2\sqrt{6}$

(iv) $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$

$=$ $\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$ $\times$ $\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}$

$=$ $\frac{6\sqrt{30} - 15 + 24 - 2 \sqrt{30}}{45-24}$

$=$ $\frac{4\sqrt{3}+9}{21}$

(v)  $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$

$=$ $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$ $\times$ $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a-b}}$

$=$ $\frac{a+b+\sqrt{a^2-b^2} + \sqrt{a^2-b^2}+a-b}{a+b-a+b}$

$=$ $\frac{2(a+\sqrt{a^2-b^2})}{2b}$

$=$ $\frac{a+\sqrt{a^2-b^2}}{b}$

Question 14: Simplify

(i) $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$ $+$ $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$

$=$ $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$ $\times$ $\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}$ $+$ $\frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$ $\times$ $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=$ $\frac{18-6\sqrt{6} - 6\sqrt{6}+12}{18-12}$ $+$ $\frac{\sqrt{12}(\sqrt{3}+\sqrt{2})}{1}$

$=$ $\frac{30-12\sqrt{6}}{6}$ $+$ $\sqrt{12} (\sqrt{3}+\sqrt{2})$

$=$ $5 - 2\sqrt{6} + 6 + 2\sqrt{6}$

$=$ $11$

(ii) $\frac{1}{2+\sqrt{3}}$ $+$ $\frac{2}{\sqrt{5} - \sqrt{3}}$ $+$ $\frac{1}{2-\sqrt{5}}$

$=$ $\frac{1}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $+$ $\frac{2}{\sqrt{5} - \sqrt{3}}$ $\times$ $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ $+$ $\frac{1}{2-\sqrt{5}}$ $\times$ $\frac{2+\sqrt{5}}{2+\sqrt{5}}$

$=$ $\frac{2-\sqrt{3}}{4-3}$ $+$ $\frac{2(\sqrt{5}+\sqrt{3})}{5-3}$ $+$ $\frac{2+\sqrt{5}}{-1}$

$= 2 -\sqrt{3}+\sqrt{5}+\sqrt{3}-2-\sqrt{5}$

$= 0$

(iii) $\frac{2}{\sqrt{5}+\sqrt{3}}$ $+$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $-$ $\frac{3} {\sqrt{5}+\sqrt{2}}$

$=$ $\frac{2}{\sqrt{5}+\sqrt{3}}$ $\times$ $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ $+$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+$ $\frac{3} {\sqrt{5}+\sqrt{2}}$ $\times$ $\frac{\sqrt{5}-\sqrt{2}} {\sqrt{5}-\sqrt{2}}$

$=$ $\frac{2(\sqrt{5}-\sqrt{3})}{2}$ $+$ $\frac{\sqrt{3}-\sqrt{2}}{1}$ $-$ $\frac{3(\sqrt{5}-\sqrt{2})}{3}$

$=$ $\sqrt{5} - \sqrt{3} + \sqrt{3} - \sqrt{2} - \sqrt{5} + \sqrt{2}$

$= 0$

(iv) $\frac{1}{2-\sqrt{3}}$ $-$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $+$ $\frac{5}{3-\sqrt{2}}$

$=$ $\frac{1}{2-\sqrt{3}}$ $\times$ $\frac{2+\sqrt{3}}{2+\sqrt{3}}$ $-$ $\frac{1}{\sqrt{3}+\sqrt{2}}$ $\times$ $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ $+$ $\frac{5}{3-\sqrt{2}}$ $\times$ $\frac{3+\sqrt{2}}{3+\sqrt{2}}$

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

$=$ $2 + \sqrt{3}-\sqrt{3}+\sqrt{2}$ $+$ $\frac{15+5\sqrt{2}}{7}$

$=$ $\frac{14+7\sqrt{2}+15+5\sqrt{2}}{7}$

$=$ $\frac{29+12\sqrt{2}}{7}$

(v)  $\frac{4\sqrt{3}}{2+\sqrt{3}}$ $-$ $\frac{30}{4\sqrt{3}-3\sqrt{2}}$ $-$ $\frac{3\sqrt{2}}{3+2\sqrt{2}}$

$=$ $\frac{4\sqrt{3}}{2+\sqrt{3}}$ $\times$ $\frac{2-\sqrt{3}}{2-\sqrt{3}}$ $-$ $\frac{30}{4\sqrt{3}-3\sqrt{2}}$ $\times$ $\frac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$ $-$ $\frac{3\sqrt{2}}{3+2\sqrt{2}}$ $\times$ $\frac{3-2\sqrt{2}}{3-2\sqrt{2}}$

$=8\sqrt{3} - 12 - 4\sqrt{3}-3\sqrt{2}-9\sqrt{2}+12$

$= 4\sqrt{3}-12\sqrt{2}$

$= 4(\sqrt{3}-3\sqrt{2})$

Question 15: Determine rational numbers $a$ and $b$

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

$\Rightarrow$ $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ $\times$ $\frac{3+\sqrt{2}}{3+\sqrt{2}}$ $=$ $a + b\sqrt{2}$

$\Rightarrow$ $\frac{9+2+6\sqrt{2}}{9-2}$ $=$ $a + b\sqrt{2}$

$\Rightarrow$ $\frac{11}{7} + \frac{6}{11}$ $\sqrt{2}$ $=$ $a + b\sqrt{2}$

$\Rightarrow$ $a =$ $\frac{11}{7}$ and $b =$ $\frac{6}{11}$

(ii) $\frac{5+3\sqrt{3}}{7+4\sqrt{3}}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $\frac{5+3\sqrt{2}}{7+3\sqrt{3}}$ $\times$ $\frac{7-4\sqrt{3}}{7-4\sqrt{3}}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $\frac{35+21\sqrt{3}-20\sqrt{3}-36}{49-48}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $-1+\sqrt{3}$ $=$ $a + b\sqrt{3}$

$\Rightarrow$ $a = -1$ and $b = 1$

(iii) $\frac{1+\sqrt{48}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{72} - \sqrt{108}+\sqrt{8}+2}$ $=$ $a + b\sqrt{3}$

Let’s first simplify

$\frac{1+\sqrt{48}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{72} - \sqrt{108}+\sqrt{8}+2}$

$=$ $\frac{1+\sqrt{2^4 \times 3}}{5\sqrt{3} + 4\sqrt{2} - \sqrt{2^3 \times 3^2} - \sqrt{2^2 \times 3^3}+\sqrt{2^3}+2}$

$=$ $\frac{1+3\sqrt{3}}{5\sqrt{3} + 4\sqrt{2} - 6\sqrt{2} - 6\sqrt{3}+2\sqrt{2}+2}$

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

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

$=$ $\frac{2+8\sqrt{3}+\sqrt{3}+12}{4-3}$

$= 14+9\sqrt{3}$

Now comparing,  $14+9\sqrt{3} = a + b\sqrt{3}$

$\Rightarrow a = 14 \ and \ b = 9$

Question 16: If $x$ $=$ $\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $y$ $=$ $\frac{\sqrt{2}-1}{\sqrt{2}+1}$, find $x^2+xy+y^2$

$x^2+xy+y^2$

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

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

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

$=$ $\frac{9+9+12\sqrt{2} + 9 + 8 - 12\sqrt{2}}{9-8}$ $+1$

$= 34+1 = 35$

Question 17: If $x$ $=$ $\frac{1}{3-2\sqrt{2}}$ and $y$ $=$ $\frac{1}{3+2\sqrt{2}}$, find $xy^2+x^2y$

$xy^2+x^2y$

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

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

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

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

$= 6$

Question 18: If $x = 2 + \sqrt{3}$, find the value of $x^3 +$ $\frac{1}{x^3}$

$x = 2 + \sqrt{3}$

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

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

$x^3 +$ $\frac{1}{x^3}$ $=$ $($ $x+$ $\frac{1}{x})^3$ $- 3(x+$ $\frac{1}{x}$ $) = 4^3 - 3 \times 4 = 52$

Question 19: If $x = 3 + \sqrt{8}$, find the value of $x^2 +$ $\frac{1}{x^2}$

$x = 3 + \sqrt{8}$

Therefore $\frac{1}{x}$ $=$ $\frac{1}{3 + \sqrt{8}}$ $\times$ $\frac{3 - \sqrt{8}}{3 - \sqrt{8}}$ $=$ $3-\sqrt{8}$

Hence $x +$ $\frac{1}{x}$ $= 3 + \sqrt{8} + 3-\sqrt{8} = 6$

$x^2 +$ $\frac{1}{x^2}$ $=$ $($ $x+$ $\frac{1}{x})^2$ $- 2$ $= 6^2 - 2 = 34$

Question 20: If $x =$ $\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $y =$ $\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ , find the value of $3x^2+4xy-3y^2$

$3x^2+4xy-3y^2$

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

$= 3 ($ $\frac{7+\sqrt{2}}{7-\sqrt{2}}$ $) + 4 - 3 ($ $\frac{7-\sqrt{2}}{7+\sqrt{2}}$ $)$

$= 3 ($ $\frac{49+40+28\sqrt{10}-49-40+28\sqrt{10}}{49-40}$ $)$

$=$ $\frac{56\sqrt{10}+12}{3}$

Question 21: If $x =$ $\frac{\sqrt{3}+1}{2}$, find the value of $4x^3+2x^2-8x+7$

$4x^3+2x^2-8x+7$

$= 4 \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^3 + 2 \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^2 - 8\Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big] + 7$

$= \Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big]^2 \Big[ ($ $\frac{4\sqrt{3}+4}{2}$ $) + 2 \Big] - 8\Big[$ $\frac{\sqrt{3}+1}{2}$ $\Big] + 7$

$=$ $(\frac{3+1+2\sqrt{3}}{2})(\frac{4\sqrt{3}+4+4}{2})$ $- 8$ $(\frac{\sqrt{3}+1}{2})$ $+ 7$

$= (4+2\sqrt{3})$ $(\frac{\sqrt{3}+2}{2})$ $- 8$ $(\frac{\sqrt{3}+1}{2})$ $+ 7$

$=$ $\frac{4\sqrt{3}+6+8+4\sqrt{3}-8\sqrt{3}-8+14}{2}$

$= 10$

Question 22: Given $\sqrt{3}= 1.732, \sqrt{5}= 2.236, \sqrt{2}=1.4142, \sqrt{6}=2.4495$ and $\sqrt{10}=3.162$, find

(i)   $\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}}$ $+$ $\frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$

$=$ $\frac{\sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} + \sqrt{5} + \sqrt{3} -\sqrt{10} -\sqrt{6}}{5-3}$

$=$ $\frac{2\sqrt{5}-2\sqrt{6}}{2}$

$= \sqrt{5}-\sqrt{6}$

$= 2.236 - 2.4495 = -0.213$

(ii) $\frac{6}{\sqrt{5}-\sqrt{3}}$

$=$ $\frac{6}{\sqrt{5}-\sqrt{3}}$ $\times$ $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

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

$= 3 (\sqrt{5}+\sqrt{3})$

$= 3(2.236+1.732) = 11.904$

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

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

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

$= 14 + (2-\sqrt{3})$

$= 14 + (2-1.732) = 14.268$

Question 23: Rationalize and simplify:

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

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

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

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

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

(ii) $\frac{1}{1+\sqrt{5}+\sqrt{3}}$

$=$ $\frac{1}{1+\sqrt{5}+\sqrt{3}}$ $\times$ $\frac{1+\sqrt{5} - \sqrt{3}}{1+\sqrt{5} -\sqrt{3}}$

$=$ $\frac{1+\sqrt{5}-\sqrt{3}}{1+5+2\sqrt{5}-3}$

$=$ $\frac{1+\sqrt{5}-\sqrt{3}}{3+2\sqrt{5}}$ $\times$ $\frac{3-2\sqrt{5}}{3-2\sqrt{5}}$

$=$ $\frac{3+3\sqrt{5}-3\sqrt{3}-2\sqrt{5}-10+2\sqrt{15} }{9-20}$

$=$ $\frac{7-\sqrt{5}+3\sqrt{3}-2\sqrt{15}}{11}$

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

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

$=$ $\frac{2+\sqrt{6}+\sqrt{10}}{2+3+2\sqrt{6}-5}$

$=$ $\frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}$

$=$ $\frac{2\sqrt{6}+6+\sqrt{60}}{12}$

$=$ $\frac{\sqrt{6}+3+\sqrt{15}}{6}$