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

Exercise 3(a) continued….

 

 

 

 

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