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….