Class 9: Surds – Exercise 3.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

$\displaystyle \text{(i) } \sqrt{18}$ $\displaystyle \text{(ii) }$ $\displaystyle \text{(iii) }$ $\displaystyle \text{(iv) } \sqrt[3]{\pi}$ $\displaystyle \text{(v) } \sqrt{\frac{3}{16}}$

Answer:

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

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

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

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

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

$\displaystyle \\$

Question 2: Simplify the following:

$\displaystyle \text{(i) } (\sqrt[4]{5})^4$ $\displaystyle \text{(ii) } \sqrt[3]{54}$ $\displaystyle \text{(iii) } \sqrt[4]{1875}$ $\displaystyle \text{(iv) } \sqrt[5]{\sqrt[3]{12}}$ $\displaystyle \text{(v) } 4\sqrt[5]{64}$

Answer:

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

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

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

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

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

$\displaystyle \\$

Question 3: Express the following as pure surds:

$\displaystyle \text{(i) } 5\sqrt[3]{4}$ $\displaystyle \text{(ii) } \frac{3}{2} \sqrt{\frac{3}{2}}$ $\displaystyle \text{(iii) } \frac{1}{4} \sqrt[3]{128}$ $\displaystyle \text{(iv) } \frac{1}{7} \sqrt[3]{1029}$ $\displaystyle \text{(v) } \frac{a}{b} \sqrt[3]{\frac{b^4}{a^4}}$

Answer:

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{(i) } \frac{3}{4} \sqrt[3]{128}$ $\displaystyle \text{(ii) } 6 \sqrt[7]{384}$ $\displaystyle \text{(iv) } \sqrt{112}$ $\displaystyle \text{(v) } \sqrt[4]{243}$

Answer:

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

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

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

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

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

$\displaystyle \\$

Question 5: Convert:

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

LCM of $\displaystyle 4$ and $\displaystyle 8$ is $\displaystyle 8$

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

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

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

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

$\displaystyle \\$

Question 6: Which is greater?

$\displaystyle \text{(i)} \sqrt[4]{5} \ or \ \sqrt[3]{4}$ $\displaystyle \text{(ii)} \sqrt[3]{3} \ or \ \sqrt[4]{4}$ $\displaystyle \text{(iii)} \sqrt{3} \ or \ \sqrt[3]{5}$

$\displaystyle \text{(iv)} \sqrt{7}-\sqrt{3} \ or \ \sqrt{5}-1$ $\displaystyle \text{(v)} \sqrt{17}-\sqrt{12} \ or \ \sqrt{11}-\sqrt{6}$

Answer:

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

LCM of $\displaystyle 4$ and $\displaystyle 3$ is $\displaystyle 12$

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

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

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

$\displaystyle \text{(ii)} \sqrt[3]{3} \ or \ \sqrt[4]{4}$

LCM of $\displaystyle 3$ and $\displaystyle 4$ is $\displaystyle 12$

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

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

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

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

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

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

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

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

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

First simplify each of the given terms

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

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

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.

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

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

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

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

First simplify each of the given terms

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

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

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.

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

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

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

$\displaystyle \\$

Question 7: Arrange in Ascending Order:

$\displaystyle \text{(i) } \sqrt[4]{3}, \sqrt[6]{10}, \sqrt[12]{25}$ $\displaystyle \text{(ii) } \sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$

Answer:

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

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

Now convert all the above terms to order of 12

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

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

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

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

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

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

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

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

Now convert all the above terms to order of 12

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

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

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

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

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

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

$\displaystyle \\$

Question 8: Arrange in Descending Order:

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

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

Answer:

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

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

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

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

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

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

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

Now convert all the above terms to order of 24

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 9: Simplify:

$\displaystyle \text{(i) } \sqrt{63}+\sqrt{28}-\sqrt{175}+\sqrt{162}-\sqrt{32}$ $\displaystyle \text{(ii) } 5\sqrt{3}+ 2 \sqrt{27}+ 4 \sqrt{\frac{1}{3}}$

$\displaystyle \text{(iii) } 2 \sqrt{\frac{8}{9}} - 3 \sqrt{\frac{1}{2}} + 5 \sqrt{\frac{9}{8}}$ $\displaystyle \text{(iv) } \sqrt[2]{147}-5 \sqrt{\frac{1}{3}} + 9 \sqrt{\frac{1}{3}}$

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

Answer:

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

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

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

$\displaystyle = 5\sqrt{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 10: Multiply:

$\displaystyle \text{(i) } 4\sqrt{12} \times 7\sqrt{16}$ $\displaystyle \text{(ii) } \sqrt[3]{7} \times \sqrt{2}$ $\displaystyle \text{(iii) } \sqrt[4]{2} \times \sqrt[3]{3} \times \sqrt[3]{4}$

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

Answer:

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

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

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

$\displaystyle = 224 \sqrt{3}$

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

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

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

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

$\displaystyle = \sqrt[6]{392}$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 11: Divide:

$\displaystyle \text{(i)} \sqrt{98} \div \sqrt{2}$ $\displaystyle \text{(ii)} 25 \sqrt[4]{33} \div 5 \sqrt[4]{11}$ $\displaystyle \text{(iii)} 6 \sqrt[3]{25} \div \sqrt[2]{5}$

$\displaystyle \text{(iv)} \sqrt{a^3b^4} \div \sqrt[3]{a^4b^3}$ $\displaystyle \text{(v)} \sqrt{m^2n^2} \times \sqrt[6]{m^2n^2} \times \sqrt[3]{m^2n^2}$

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle = b\sqrt{a}$

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

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

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

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

$\displaystyle = m^2n^2$

$\displaystyle \\$

Question 12: Find the rationalising factors of the following:

$\displaystyle \text{(i)} \sqrt{108}$ $\displaystyle \text{(ii)} \sqrt[5]{486}$ $\displaystyle \text{(iii)} \sqrt{2}+\sqrt{3}+\sqrt{5}$

$\displaystyle \text{(iv)} \sqrt{3}+ \sqrt{8+2\sqrt{15}}$ $\displaystyle \text{(v)} \sqrt[3]{5}$

Answer:

$\displaystyle \text{(i) } \sqrt{108}$

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

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

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

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

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

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

We find that

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 13: Rationalize the denominator and simplify:

$\displaystyle \text{(i) } \frac{6-4\sqrt{2}}{6+4\sqrt{2}}$ $\displaystyle \text{(ii) } \frac{b^2}{\sqrt{a^2+b^2} +a}$ $\displaystyle \text{(iii) } \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+ \sqrt{2}}$

$\displaystyle \text{(iv) } \frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$ $\displaystyle \text{(v) } \frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$

Answer:

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

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

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

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

$\displaystyle = 17 - 12\sqrt{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 14: Simplify:

$\displaystyle \text{(i)} \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} + \frac{\sqrt{12}}{\sqrt{3}-\sqrt{2}}$ $\displaystyle \text{(ii)} \frac{1}{2+\sqrt{3}} + \frac{2}{\sqrt{5} - \sqrt{3}} + \frac{1}{2-\sqrt{5}}$

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

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

Answer:

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

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

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

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

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

$\displaystyle = 11$

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

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

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

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

$\displaystyle = 0$

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

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

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

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

$\displaystyle = 0$

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{(i) } \frac{3+\sqrt{2}}{3-\sqrt{2}} = a + b\sqrt{2}$ $\displaystyle \text{(ii) } \frac{5+3\sqrt{3}}{7+4\sqrt{3}} = a + b\sqrt{3}$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

Let’s first simplify

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

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

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

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

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

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

$\displaystyle = 14+9\sqrt{3}$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle x^2+xy+y^2$

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

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

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

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

$\displaystyle = 34+1 = 35$

$\displaystyle \\$

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

Answer:

$\displaystyle xy^2+x^2y$

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

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

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

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

$\displaystyle = 6$

$\displaystyle \\$

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

Answer:

$\displaystyle x = 2 + \sqrt{3}$

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle x = 3 + \sqrt{8}$

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle = 10$

$\displaystyle \\$

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

$\displaystyle \text{(i)} \frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}} + \frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}$ $\displaystyle \text{(ii)} \frac{6}{\sqrt{5}-\sqrt{3}}$ $\displaystyle \text{(iii)} \frac{2+\sqrt{3}}{2-\sqrt{3}} + \frac{2-\sqrt{3}}{2+\sqrt{3}} + \frac{\sqrt{3}-1}{\sqrt{3}+1}$

Answer:

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

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

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

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

$\displaystyle = 2.236 - 2.4495 = -0.213$

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

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

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

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

$\displaystyle = 3(2.236+1.732) = 11.904$

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

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

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

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

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

$\displaystyle \\$

Question 23: Rationalize and simplify:

$\displaystyle \text{(i)} \frac{1}{1+\sqrt{2}-\sqrt{3}}$ $\displaystyle \text{(ii)} \frac{1}{1+\sqrt{5}+\sqrt{3}}$ $\displaystyle \text{(iii)} \frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$