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

Answer:

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

Answer:

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}

Answer:

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}

Answer:

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

Answer:

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

Answer:

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}