Chapter 2 – Real Numbers
Practice set 2.1
1. Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
(i) \(\large \frac {13}{5}\)
Solution:
In\(\large \frac {13}{5}\), the denominator is 5 and has only 5 as the prime factor.
Thus, \(\large \frac {13}{5}\) has the terminating decimal representation.
(ii) \(\large \frac {2}{11}\)
Solution:
In \(\large \frac {2}{11}\), 11 is the denominator which has prime factors other than 2 and 5.
∴ \(\large \frac {2}{11}\) has recurring and non terminating decimal representation.
(iii) \(\large \frac {29}{16}\)
Solution:
In \(\large \frac {29}{16}\), 16 is the denominator and
16 = 2 × 2 × 2 × 2
Thus, denominator has only 2 as prime factors.
∴ \(\large \frac {29}{16}\) has the terminating decimal representation.
(iv) \(\large \frac {17}{125}\)
Solution:
In \(\large \frac {17}{125}\), 125 is the denominator and 125 = 5 × 5 × 5
Thus, the denominator has only 5 as prime factors.
∴ \(\large \frac {17}{125}\) has the terminating decimal representation.
(v) \(\large \frac {11}{6}\)
Solution:
In \(\large \frac {11}{6}\), 6 is the denominator and 6 = 2 × 3
Thus, the denominator has prime factors other than 2 and 5.
∴ \(\large \frac {11}{6}\) has recurring and non terminating decimal representation.
2. Write the following rational numbers in decimal form.
(i) \(\large \frac {127}{200}\)
Solution:
\(\large \frac {127}{200}\)
= \(\large \frac {127}{2\,×\,100}\)
= \(\large \frac {63.5}{100}\)
= 0.635
∴ \(\large \frac {127}{200}\) = 0.635
(ii) \(\large \frac {25}{99}\)
Solution:
∴ \(\large \frac {25}{99}\) = \(0.\overline{25}\)
(iii) \(\large \frac {23}{7}\)
Solution:
∴ \(\large \frac {23}{7}\) = \(3.\overline{285714}\)
(iv) \(\large \frac {4}{5}\)
Solution:
\(\large \frac {4}{5}\)
= \(\large \frac {4\,×\,2}{5\,×\,2}\)
= \(\large \frac {8}{10}\)
= 0.8
∴ \(\large \frac {4}{5}\) = 0.8
(v) \(\large \frac {17}{8}\)
Solution:
∴ \(\large \frac {17}{8}\) = 2.125
3. Write the following rational numbers in \(\large \frac {p}{q}\) form.
(i) \(0.\dot6\)
Solution:
Let x = 0.666… = \(0.\dot6\) …(i)
Multiplying both the sides by 10, we get,
10x = 6.666… = \(6.\dot6\) …(ii)
Subtracting (i) from (ii), we get,
10x – x = \(6.\dot6\) – \(0.\dot6\)
∴ 9x = 6
∴ x = \(\large \frac {6}{9}\)
∴ x = \(\large \frac {2}{3}\)
∴ \(0.\dot6\) = \(\large \frac {2}{3}\)
(ii) 0.37
Solution:
Let x = 0.3737… = \(0.\overline{37}\) …(i)
Multiplying both the sides by 100 we get,
100x = 37.3737… = \(37.\overline{37}\) …(ii)
Subtracting (i) from (ii), we get,
100x – x = \(37.\overline{37}\) – \(0.\overline{37}\)
∴ 99x = 37
∴ x = \(\large \frac {37}{99}\)
∴ \(0.\overline{37}\) = \(\large \frac {37}{99}\)
(iii) 3.17
Solution:
Let x = 3.171717… = \(3.\overline{17}\) …(i)
Multiplying both the sides by 100 we get,
100x = 317.1717… = \(317.\overline{17}\) …(ii)
Subtracting (i) from (ii), we get,
100x – x = \(317.\overline{17}\) – \(3.\overline{17}\)
∴ 99x = 314
∴ x = \(\large \frac {314}{99}\)
∴ \(3.\overline{17}\) = \(\large \frac {314}{99}\)
(iv) \(15.\overline{89}\)
Solution:
Let x = 15.8989… = \(15.\overline{89}\) …(i)
Multiplying both the sides by 100, we get,
100x = 1589.8989… = \(1589.\overline{89}\) …(ii)
Subtracting (i) from (ii), we get,
100x – x = \(1589.\overline{89}\) – \(15.\overline{89}\)
∴ 99x = 1574
∴ x = \(\large \frac {1564}{99}\)
∴ \(15.\overline{89}\) = \(\large \frac {1564}{99}\)
(v) \(2.\overline{514}\)
Solution:
Let x = 2.514514… = \(2.\overline{514}\) …(i)
Multiplying both the sides by 1000, we get,
1000x = 2514.514… = \(2514.\overline{514}\) …(ii)
Subtracting (i) from (ii)
1000x – x = \(2514.\overline{514}\) – \(2.\overline{514}\)
∴ 999x = 2512
∴ x = \(\large \frac {2512}{999}\)
∴ \(2.\overline{514}\) = \(\large \frac {2512}{999}\)
Practice set 2.2
(1) Show that 4 \(\sqrt{2}\) is an irrational number.
Solution:
Let 4 \(\sqrt{2}\) be a rational number.
∴ 4 \(\sqrt{2}\) = \(\large \frac {p}{q}\)
∴ \(\sqrt{2}\) = \(\large \frac {p}{4q}\) …(i)
Now,
\(\large \frac {p}{q}\) is a rational number then \(\large \frac {p}{4q}\) is also a rational number.
Then, \(\sqrt{2}\) is a rational number …[from (i)]
But, it contradicts the fact that \(\sqrt{2}\) is an irrational number.
∴ \(\sqrt{2}\) ≠ \(\large \frac {p}{4q}\)
Our assumption is wrong.
Hence, 4 \(\sqrt{2}\) is an irrational number.
(2) Prove that 3 + \(\sqrt{5}\) is an irrational number.
Solution:
Let 3 + \(\sqrt{5}\) be a rational number.
∴ 3 + \(\sqrt{5}\) = \(\large \frac {p}{q}\)
∴ \(\sqrt{5}\) = \(\large \frac {p}{q}\) – 3 …(i)
Now,
\(\large \frac {p}{q}\) is a rational number then \(\large \frac {p}{q}\) – 3 is also a rational number.
Then, \(\sqrt{5}\) is a rational number …[from (i)]
But, it contradicts the fact that \(\sqrt{5}\) is an irrational number.
∴ \(\sqrt{5}\) ≠ \(\large \frac {p}{q}\) – 3
Our assumption is wrong.
Hence, 3 + \(\sqrt{5}\) is an irrational number.
(3) Represent the numbers \(\sqrt{5}\) and \(\sqrt{10}\) on a number line.
Solution:
(i) To represent \(\sqrt{5}\) on the number line:
Point A corresponds to \(\sqrt{5}\) on the number line.
(ii) To represent \(\sqrt{10}\) on the number line:
Point A corresponds to \(\sqrt{10}\) on the number line.
(4) Write any three rational numbers between the two numbers given below.
(i) 0.3 and – 0.5
Solution:
We know that,
0.3 = 0.30 and – 0.5 = – 0.50
∴ 0.30 > 0.29 > 0.28 > 0.27 > …… > – 0.50
∴ Three rational numbers between 0.3 and – 0.5 are 0.25, 0.20, – 0.12.
(ii) – 2.3 and – 2.33
Solution:
We know that,
– 2.3 = – 2.300 and – 2.33 = – 2.330
∴ – 2.300 > – 2.301 > – 2.302 > – 2.303 > … > – 2.330
∴ Three rational numbers between –2.3 and – 2.33 are – 2.302, – 2.304, – 2.306.
(iii) 5.2 and 5.3
Solution:
We know that,
5.2 = 5.20 and 5.3 = 5.30
∴ 5.20 < 5.21 < 5.22 < 5.23 < … < 5.30
∴ Three rational numbers between 5.2 and 5.3 are 5.21, 5.22, 5.23.
(iv) – 4.5 and – 4.6
Solution:
Now – 4.5 = – 4.50 and – 4.6 = – 4.60
∴ – 4.50 > – 4.51 > – 4.52 > – 4.53 > … > – 4.60
∴ Three rational numbers between –4.5 and – 4.6 are – 4.51, – 4.52, – 4.53.
Practice set 2.3
(1) State the order of the surds given below.
(i) \(\sqrt[3]{2}\)
Ans: The order of the surd is 3
(ii) 5\(\sqrt{12}\)
Ans: The order of the surd is 2
(iii) \(\sqrt[4]{10}\)
Ans: The order of the surd is 4
(iv) \(\sqrt[2]{39}\)
Ans: The order of the surd is 2
(v) \(\sqrt[3]{18}\)
Ans: The order of the surd is 3
(2) State which of the following are surds. Justify.
(i) \(\sqrt[3]{51}\)
Ans: \(\sqrt[3]{51}\) is a surd
(ii) \(\sqrt[4]{16}\)
Ans: \(\sqrt[4]{16}\) = 2
Hence it is not a surd
(iii) \(\sqrt[5]{81}\)
Ans: \(\sqrt[5]{81}\) is a surd
(iv) \(\sqrt[2]{256}\)
Ans: \(\sqrt[2]{256}\) = 16
Hence it is not a surd
(v) \(\sqrt[3]{64}\)
Ans: \(\sqrt[3]{64}\) = 4
Hence it is not a surd
(vi) \(\sqrt{\large \frac {22}{7}}\)
Ans: \(\sqrt{\large \frac {22}{7}}\) is a surd
(3) Classify the given pair of surds into like surds and unlike surds.
(i) \(\sqrt{52}\) , 5 \(\sqrt{13}\)
Solution:
\(\sqrt{52}\)
= \(\sqrt{4\,×\,13}\)
= \(\sqrt{4}\) × \(\sqrt{13}\)
= 2 \(\sqrt{13}\)
2 \(\sqrt{13}\) and 5 \(\sqrt{13}\) are like surds.
∴ \(\sqrt{52}\) and 5 \(\sqrt{13}\) are like surds.
(ii) \(\sqrt{68}\) , 5 \(\sqrt{3}\)
Solution:
\(\sqrt{68}\)
= \(\sqrt{4\,×\,17}\)
= \(\sqrt{4}\) × \(\sqrt{17}\)
= 2 \(\sqrt{17}\)
Here 2 \(\sqrt{17}\) and 5 \(\sqrt{3}\) are not like surds.
∴ \(\sqrt{68}\) and 5 \(\sqrt{3}\) are not like surds.
(iii) 4 \(\sqrt{18}\) , 7 \(\sqrt{2}\)
Solution:
4 \(\sqrt{18}\)
= 4 \(\sqrt{9\,×\,2}\)
= 4 \(\sqrt{9}\) × \(\sqrt{2}\)
= 4 × 3 \(\sqrt{2}\)
= 12 \(\sqrt{2}\)
Here 12 \(\sqrt{2}\) and 7 \(\sqrt{2}\) are like surds.
∴ 4 \(\sqrt{18}\) and 7 \(\sqrt{2}\) are like surds.
(iv) 19 \(\sqrt{12}\) , 6 \(\sqrt{3}\)
Solution:
19 \(\sqrt{12}\)
= 19 \(\sqrt{4\,×\,3}\)
= 19 \(\sqrt{4}\) × \(\sqrt{3}\)
= 19 × 2 \(\sqrt{3}\)
= 38 \(\sqrt{3}\)
Here 38 \(\sqrt{3}\) and 6 \(\sqrt{3}\) are like surds.
∴ 19 \(\sqrt{12}\) and 6 \(\sqrt{3}\) are like surds.
(v) 5 \(\sqrt{22}\) , 7 \(\sqrt{33}\)
Solution:
Both surds 5 \(\sqrt{22}\) and 7 \(\sqrt{33}\) are in the simplest form
∴ 5 \(\sqrt{22}\) and 7 \(\sqrt{33}\) are not like surds.
(vi) 5 \(\sqrt{5}\) , \(\sqrt{75}\)
Solution:
\(\sqrt{75}\)
= \(\sqrt{25\,×\,3}\)
= \(\sqrt{25}\) × \(\sqrt{3}\)
= 5 \(\sqrt{3}\)
Here 5 \(\sqrt{5}\) and 5 \(\sqrt{3}\) are not like surds.
∴ 5 \(\sqrt{5}\) and \(\sqrt{75}\) are not like surds.
(4) Simplify the following surds.
(i) \(\sqrt{27}\)
Solution:
\(\sqrt{27}\)
= \(\sqrt{9\,×\,3}\)
= 3 \(\sqrt{3}\)
(ii) \(\sqrt{50}\)
Solution:
\(\sqrt{50}\)
= \(\sqrt{25\,×\,2}\)
= 5 \(\sqrt{2}\)
(iii) \(\sqrt{250}\)
Solution:
\(\sqrt{250}\)
= \(\sqrt{25\,×\,10}\)
= 5 \(\sqrt{10}\)
(iv) \(\sqrt{112}\)
Solution:
\(\sqrt{112}\)
= \(\sqrt{16\,×\,7}\)
= 4 \(\sqrt{7}\)
(v) \(\sqrt{168}\)
Solution:
\(\sqrt{168}\)
= \(\sqrt{2\,×\,2\,×\,2\,×\,3\,×\,7}\)
= 2 \(\sqrt{2\,×\,3\,×\,7}\)
= 2 \(\sqrt{42}\)
(5) Compare the following pair of surds.
(i) 7 \(\sqrt{2}\) , 5 \(\sqrt{3}\)
Solution:
7 \(\sqrt{2}\)
= \(\sqrt{49}\) × \(\sqrt{2}\)
= \(\sqrt{98}\)
5 \(\sqrt{3}\)
= \(\sqrt{25}\) × \(\sqrt{3}\)
= \(\sqrt{75}\)
\(\sqrt{98}\) > \(\sqrt{75}\)
∴ 7 \(\sqrt{2}\) > 5 \(\sqrt{3}\)
(ii) \(\sqrt{247}\) , \(\sqrt{274}\)
Solution:
\(\sqrt{247}\) < \(\sqrt{274}\)
(iii) 2 \(\sqrt{7}\) , \(\sqrt{28}\)
Solution:
2 \(\sqrt{7}\)
= \(\sqrt{4}\) × \(\sqrt{7}\)
= \(\sqrt{28}\)
\(\sqrt{28}\) = \(\sqrt{28}\)
∴ 2 \(\sqrt{7}\) = \(\sqrt{28}\)
(iv) 5 \(\sqrt{5}\) , 7 \(\sqrt{2}\)
Solution:
5 \(\sqrt{5}\)
= \(\sqrt{25}\) × \(\sqrt{5}\)
= \(\sqrt{125}\)
7 \(\sqrt{2}\)
= \(\sqrt{49}\) × \(\sqrt{2}\)
= \(\sqrt{98}\)
\(\sqrt{125}\) > \(\sqrt{98}\)
∴ 5 \(\sqrt{5}\) × 7 \(\sqrt{2}\)
(v) 4 \(\sqrt{42}\) , 9 \(\sqrt{2}\)
Solution:
4 \(\sqrt{42}\)
= \(\sqrt{4}\) × \(\sqrt{42}\)
= \(\sqrt{672}\)
9 \(\sqrt{2}\)
= \(\sqrt{81}\) × \(\sqrt{2}\)
= \(\sqrt{162}\)
\(\sqrt{672}\) > \(\sqrt{162}\)
∴ 4 \(\sqrt{42}\) > 9 \(\sqrt{2}\)
(vi) 5 \(\sqrt{3}\) , 9
Solution:
5 \(\sqrt{3}\)
= \(\sqrt{25}\) × \(\sqrt{3}\)
= \(\sqrt{75}\)
9 = \(\sqrt{81}\)
\(\sqrt{75}\) < \(\sqrt{81}\)
∴ 5 \(\sqrt{3}\) < 9
(vii) 7 , 2 \(\sqrt{5}\)
Solution:
7 = \(\sqrt{49}\)
2 \(\sqrt{5}\)
= \(\sqrt{4}\) × \(\sqrt{5}\)
= \(\sqrt{20}\)
\(\sqrt{49}\) > \(\sqrt{20}\)
∴ 7 > 2 \(\sqrt{5}\)
(6) Simplify.
(i) 5 \(\sqrt{3}\) + 8 \(\sqrt{3}\)
Solution:
5 \(\sqrt{3}\) + 8 \(\sqrt{3}\)
= (5 + 8) \(\sqrt{3}\)
= 13 \(\sqrt{3}\)
(ii) 9 \(\sqrt{5}\) – 4 \(\sqrt{5}\) + \(\sqrt{5}\)
Solution:
9 \(\sqrt{5}\) – 4 \(\sqrt{5}\) + \(\sqrt{5}\)
= (9 – 4 + 1) \(\sqrt{5}\)
= 6 \(\sqrt{5}\)
(iii) 7 \(\sqrt{48}\) – \(\sqrt{27}\) – \(\sqrt{3}\)
Solution:
7 \(\sqrt{48}\)
= 7 \(\sqrt{16\,×\,3}\)
= (7 × 4) \(\sqrt{3}\)
= 28 \(\sqrt{3}\)
\(\sqrt{27}\)
= \(\sqrt{9\,×\,3}\)
= 3 \(\sqrt{3}\)
∴ 7 \(\sqrt{48}\) – \(\sqrt{27}\) – \(\sqrt{3}\)
= 28 \(\sqrt{3}\) – 3 \(\sqrt{3}\) – \(\sqrt{3}\)
= (28 – 3 – 1) \(\sqrt{3}\)
= 24 \(\sqrt{3}\)
(iv) \(\sqrt{7}\) – \(\large \frac {3}{5}\) \(\sqrt{7}\) + 2 \(\sqrt{7}\)
Solution:
\(\sqrt{7}\) – \(\large \frac {3}{5}\) \(\sqrt{7}\) + 2 \(\sqrt{7}\)
= (1 – \(\large \frac {3}{5}\) + 2) \(\sqrt{7}\)
= \(\large(\) 3 – \(\large \frac {3}{5})\) \(\sqrt{7}\)
= \(\large (\frac {3\,×\,5}{1\,×\,5}\) – \(\large \frac {3}{5})\) \(\sqrt{7}\)
= \(\large (\frac {15}{5}\) – \(\large \frac {3}{5})\) \(\sqrt{7}\)
= \(\large (\frac {15\,–\,3}{5})\) \(\sqrt{7}\)
= \(\large (\frac {12}{5})\) \(\sqrt{7}\)
(7) Multiply and write the answer in the simplest form.
(i) 3 \(\sqrt{12}\) × \(\sqrt{18}\)
Solution:
3 \(\sqrt{12}\)
= 3 \(\sqrt{4\,×\,3}\)
= (3 × 4) \(\sqrt{3}\)
= 12 \(\sqrt{3}\)
\(\sqrt{18}\)
= \(\sqrt{9\,×\,2}\)
= \(\sqrt{9}\) × \(\sqrt{2}\)
= 3 \(\sqrt{2}\)
3 \(\sqrt{12}\) × \(\sqrt{18}\)
= 12 \(\sqrt{3}\) × 3 \(\sqrt{2}\)
= (12 × 3) \(\sqrt{3}\)
= 36 \(\sqrt{3}\)
(ii) 3 \(\sqrt{12}\) × 7 \(\sqrt{15}\)
Solution:
3 \(\sqrt{12}\)
= 4 \(\sqrt{4\,×\,3}\)
= 3 \(\sqrt{4}\) × \(\sqrt{3}\)
= (3 × 2) \(\sqrt{2}\)
= 6 \(\sqrt{2}\)
3 \(\sqrt{12}\) × 7 \(\sqrt{15}\)
= (3 × 7) (\(\sqrt{12}\) × \(\sqrt{15}\))
= 21 \(\sqrt{12\,×\,15}\)
= 21 \(\sqrt{180}\)
= 21 \(\sqrt{36\,×\,5}\)
= 21 × 6 \(\sqrt{5}\)
= 126 \(\sqrt{5}\)
(iii) 3 \(\sqrt{8}\) × \(\sqrt{5}\)
Solution:
3 \(\sqrt{8}\) × \(\sqrt{5}\)
= 3 \(\sqrt{8\,×\,5}\)
= 3 \(\sqrt{4\,×\,2\,×\,5}\)
= 3 × 2 \(\sqrt{2\,×\,5}\)
= 6 \(\sqrt{10}\)
(iv) 5\(\sqrt{8}\) × 2 \(\sqrt{8}\)
Solution:
5\(\sqrt{8}\) × 2 \(\sqrt{8}\)
= (5 × 2) \(\sqrt{8}\)
= 10 \(\sqrt{4\,×\,2}\)
= 10 × 2 \(\sqrt{2}\)
= 20 \(\sqrt{2}\)
(8) Divide, and write the answer in simplest form.
(i) \(\sqrt{98}\) ÷ \(\sqrt{2}\)
Solution:
\(\sqrt{98}\) ÷ \(\sqrt{2}\)
= \(\large \frac {\sqrt{98}}{\sqrt{2}}\)
= \(\large \sqrt \frac {49\,×\,2}{2}\)
= \(\sqrt{49}\)
= 7
(ii) \(\sqrt{125}\) ÷ \(\sqrt{50}\)
Solution:
\(\sqrt{125}\) ÷ \(\sqrt{50}\)
= \(\large \frac {\sqrt{125}}{\sqrt{50}}\)
= \(\large \sqrt \frac {25\,×\,5}{25\,×\,2}\)
= \(\large \sqrt \frac {5}{2}\)
(iii) \(\sqrt{54}\) ÷ \(\sqrt{27}\)
Solution:
\(\sqrt{54}\) ÷ \(\sqrt{27}\)
= \(\large \frac {\sqrt{54}}{\sqrt{27}}\)
= \(\large \sqrt \frac {9\,×\,6}{9\,×\,3}\)
= \(\sqrt {2}\)
(iv) \(\sqrt{310}\) ÷ \(\sqrt{5}\)
Solution:
\(\sqrt{310}\) ÷ \(\sqrt{5}\)
= \(\large \frac {\sqrt{310}}{\sqrt{5}}\)
= \(\large \sqrt \frac {62\,×\,5}{5}\)
= \(\large \sqrt {62}\)
(9) Rationalize the denominator.
(i) \(\large \frac {3}{\sqrt{5}}\)
Solution:
Rationalizing factor of \(\sqrt{5}\) is \(\sqrt{5}\)
∴ \(\large \frac {3}{\sqrt{5}}\)
= \(\large \frac {3}{\sqrt{5}}\) × \(\large \frac {\sqrt{5}}{\sqrt{5}}\)
= \(\large \frac {3\,×\,\sqrt{5}}{\sqrt{5}\,×\,\sqrt{5}}\)
= \(\large \frac {3\sqrt{5}}{5}\)
(ii) \(\large \frac {1}{\sqrt{14}}\)
Solution:
Rationalizing factor of \(\sqrt{14}\) is \(\sqrt{14}\)
∴ \(\large \frac {1}{\sqrt{14}}\)
= \(\large \frac {1}{\sqrt{14}}\) × \(\large \frac {\sqrt{14}}{\sqrt{14}}\)
= \(\large \frac {1\,×\,\sqrt{14}}{\sqrt{14}\,×\,\sqrt{14}}\)
= \(\large \frac {\sqrt{14}}{14}\)
(iii) \(\large \frac {5}{\sqrt{7}}\)
Solution:
Rationalizing factor of \(\sqrt{7}\) is \(\sqrt{7}\)
∴ \(\large \frac {5}{\sqrt{7}}\)
= \(\large \frac {5}{\sqrt{7}}\) × \(\large \frac {\sqrt{7}}{\sqrt{7}}\)
= \(\large \frac {5\,×\,\sqrt{7}}{\sqrt{7}\,×\,\sqrt{7}}\)
= \(\large \frac {5\sqrt{7}}{7}\)
(iv) \(\large \frac {6}{9\sqrt{3}}\)
Solution:
Rationalizing factor of 9\(\sqrt{3}\) is \(\sqrt{3}\)
∴ \(\large \frac {6}{9\sqrt{3}}\)
= \(\large \frac {6}{9\sqrt{3}}\) × \(\large \frac {\sqrt{3}}{\sqrt{3}}\)
= \(\large \frac {6\,×\,\sqrt{3}}{9\sqrt{3}\,×\,\sqrt{3}}\)
= \(\large \frac {2\sqrt{7}}{3\,×\,3}\)
= \(\large \frac {2\sqrt{7}}{9}\)
(v) \(\large \frac {11}{\sqrt{3}}\)
Solution:
Rationalizing factor of \(\sqrt{3}\) is \(\sqrt{3}\)
∴ \(\large \frac {11}{\sqrt{3}}\)
= \(\large \frac {11}{\sqrt{3}}\) × \(\large \frac {\sqrt{3}}{\sqrt{3}}\)
= \(\large \frac {11\,×\,\sqrt{3}}{\sqrt{3}\,×\,\sqrt{3}}\)
= \(\large \frac {11\sqrt{3}}{3}\)
Practice set 2.4
(1) Multiply.
(i) \(\sqrt{3}\) (\(\sqrt{7}\) – \(\sqrt{3}\))
Solution:
\(\sqrt{3}\) ( \(\sqrt{7}\) – \(\sqrt{3}\) )
= \(\sqrt{3}\) × \(\sqrt{7}\) – \(\sqrt{3}\) × \(\sqrt{3}\)
= \(\sqrt{3\,×\,7}\) – 3
= \(\sqrt{21}\) – 3
= – 3 + \(\sqrt{21}\)
(ii) (\(\sqrt{5}\) – \(\sqrt{7}\)) \(\sqrt{2}\)
Solution:
(\(\sqrt{5}\) – \(\sqrt{7}\)) \(\sqrt{2}\)
= \(\sqrt{5}\) × \(\sqrt{2}\) – \(\sqrt{7}\) × \(\sqrt{2}\)
= \(\sqrt{5\,×\,2}\) – \(\sqrt{7\,×\,2}\)
= \(\sqrt{10}\) – \(\sqrt{14}\)
(iii) (3 \(\sqrt{2}\) – \(\sqrt{3}\) )(4 \(\sqrt{3}\) – \(\sqrt{2}\) )
Solution:
(3 \(\sqrt{2}\) – \(\sqrt{3}\) )(4 \(\sqrt{3}\) – \(\sqrt{2}\) )
= 3\(\sqrt{2}\) × 4\(\sqrt{3}\) – 3\(\sqrt{2}\) × \(\sqrt{2}\) – \(\sqrt{3}\) × 4\(\sqrt{3}\) + \(\sqrt{3}\) × \(\sqrt{2}\)
= 12\(\sqrt{6}\) – (3 × 2) – (3 × 4) + \(\sqrt{6}\)
= 12\(\sqrt{6}\) – 6 – 12 + \(\sqrt{6}\)
= 12\(\sqrt{6}\) + \(\sqrt{6}\) – 18
= 13 \(\sqrt{6}\) – 18
= – 18 + 13 \(\sqrt{6}\)
(2) Rationalize the denominator.
(i) \(\large \frac {1}{\sqrt{7} \,+\, \sqrt{2}}\)
Solution:
Conjugate of \(\sqrt{7} \,+\, \sqrt{2}\) is \(\sqrt{7} \,–\, \sqrt{2}\)
\(\large \frac {1}{\sqrt{7} \,+\, \sqrt{2}}\)
= \(\large \frac {1}{\sqrt{7} \,+\, \sqrt{2}}\) × \(\large \frac {\sqrt{7} \,–\, \sqrt{2}}{\sqrt{7} \,–\, \sqrt{2}}\)
= \(\large \frac {1\,×\,(\sqrt{7} \,–\, \sqrt{2})}{(\sqrt{7} \,+\, \sqrt{2})(\sqrt{7} \,–\, \sqrt{2})}\)
= \(\large \frac {\sqrt{7}\,–\,\sqrt{2}}{(\sqrt{7})^2 \,–\, (\sqrt{2})^2}\)
= \(\large \frac {\sqrt{7}\,–\, \sqrt{2}}{7\,–\,2}\)
= \(\large \frac {\sqrt{7}\,–\, \sqrt{2}}{5}\)
(ii) \(\large \frac {3}{2 \sqrt{5} \,–\, 3 \sqrt{2}}\)
Solution:
Conjugate of \(2\sqrt{5} \,–\, 3\sqrt{3}\) is \(2\sqrt{5} \,+\, 3\sqrt{3}\)
\(\large \frac {3}{2 \sqrt{5} \,–\, 3 \sqrt{2}}\)
= \(\large \frac {3}{2 \sqrt{5} \,–\, 3 \sqrt{2}}\) × \(\large \frac {2\sqrt{5} \,+\, 3\sqrt{3}}{2\sqrt{5} \,+\, 3\sqrt{3}}\)
= \(\large \frac {3\,×\,(2\sqrt{5} \,+\, 3\sqrt{3})}{(2\sqrt{5} \,–\, 3\sqrt{3})(2\sqrt{5} \,+\, 3\sqrt{3})}\)
= \(\large \frac {6\sqrt{5} \,+\, 9\sqrt{3}}{(2\sqrt{5})^2 \,–\,(3\sqrt{3})^2}\)
= \(\large \frac {6\sqrt{5} \,+\, 9\sqrt{3}}{(4\,×\,5)\,–\,(9\,×\,3)}\)
= \(\large \frac {6\sqrt{5} \,+\, 9\sqrt{3}}{20\,–\,18}\)
= \(\large \frac {6\sqrt{5} \,+\, 9\sqrt{3}}{2}\)
(iii) \(\large \frac {4}{7 \,+\, 4 \sqrt{3}}\)
Solution:
Conjugate of \(7\,+\, 4\sqrt{3}\) is \(7 \,–\, 4\sqrt{3}\)
∴ \(\large \frac {4}{7 \,+\, 4 \sqrt{3}}\)
= \(\large \frac {4}{7 \,+\, 4 \sqrt{3}}\) × \(\large \frac {7 \,–\, 4\sqrt{3}}{7 \,–\, 4\sqrt{3}}\)
= \(\large \frac {4\,×\,(7 \,–\, 4\sqrt{3})}{(7 \,+\, 4\sqrt{3})(7 \,–\, 4\sqrt{3})}\)
= \(\large \frac {28\,–\,16\sqrt{3}}{{7}^2 \,–\,(4\sqrt{2})^2}\)
= \(\large \frac {28\,–\,16\sqrt{3}}{49\,–\,(16\,×\,2)}\)
= \(\large \frac {28\,–\,16\sqrt{3}}{49\,–\,48}\)
= \(\large \frac {28\,–\,16\sqrt{3}}{1}\)
= 28 –\(16\sqrt{3}\)
(iv) \(\large \frac {\sqrt{5} \,–\, \sqrt{3}}{\sqrt{5} \,+\, \sqrt{3}}\)
Solution:
Conjugate of \(\sqrt{5} \,+\, \sqrt{3}\) is \(\sqrt{5} \,–\, \sqrt{3}\)
\(\large \frac {\sqrt{5} \,–\, \sqrt{3}}{\sqrt{5} \,+\, \sqrt{3}}\)
= \(\large \frac {\sqrt{5} \,–\, \sqrt{3}}{\sqrt{5} \,+\, \sqrt{3}}\) × \(\large \frac {\sqrt{5} \,–\, \sqrt{3}}{\sqrt{5} \,–\, \sqrt{3}}\)
= \(\large \frac {(\sqrt{5} \,–\, \sqrt{3})(\sqrt{5} \,–\, \sqrt{3})}{(\sqrt{5} \,+\, \sqrt{3})(\sqrt{5} \,–\, \sqrt{3})}\)
= \(\large \frac {(\sqrt{5} \,–\, \sqrt{3})^2}{(\sqrt{5})^2 \,+\, (\sqrt{3})^2}\)
= \(\large \frac {(\sqrt{5})^2\,–\, 2\,×\,\sqrt{5}\,×\,\sqrt{3}\,+\,(\sqrt{3})^2}{5\,–\,3}\)
= \(\large \frac {5\,–\,2\sqrt{15}\,+\,3}{2}\)
= \(\large \frac {8\,–\,2\sqrt{15}}{2}\)
= \(\large \frac {2(4\,–\,\sqrt{15})}{2}\)
= 4 – \(\sqrt{15}\)
Practice set 2.5
(1) Find the value.
(i) | 15 – 2 |
Solution:
| 15 – 2 |
= | 13 |
= 13
(ii) | 4 – 9 |
Solution:
| 4 – 9 |
= | – 5 |
= 5
(iii) | 7 | × | – 4 |
Solution:
| 7 | × | – 4 |
= | 7 × – 4 |
= | – 28 |
= 28
(2) Solve.
(i) | 3x – 5 | = 1
Solution:
| 3x – 5 | = 1
∴ 3x – 5 = 1
∴ 3x = 1 + 5
∴ 3x = 6
∴ x = \(\large \frac {6}{3}\)
∴ x = 2
∴ 3x – 5 = – 1
∴ 3x = – 1 + 5
∴ 3x = 4
∴ x = \(\large \frac {4}{3}\)
(ii) | 7 – 2x | = 5
Solution:
| 7 – 2x | = 5
∴ 7 – 2x = 5
∴ – 2x = 5 – 7
∴ – 2x = – 2
∴ x = \(\large \frac {–\,2}{–\,2}\)
∴ x = 1
∴ 7 – 2x = – 5
∴ – 2x = – 5 – 7
∴ – 2x = – 12
∴ x = \(\large \frac {– \,12}{–\, 2}\)
∴ x = 6
(iii) | \(\large \frac {8\,–\,x}{2}\) | = 5
Solution:
| \(\large \frac {8\,–\,x}{2}\) | = 5
∴ \(\large \frac {8\,–\,x}{2}\) = 5
∴ 8 – x = 5 × 2
∴ 8 – x = 10
∴ – x = 10 – 8
∴ – x = 2
∴ x = – 2
∴ \(\large \frac {8\,–\,x}{2}\) = – 5
∴ 8 – x = – 5 × 2
∴ 8 – x = – 10
∴ – x = – 10 – 8
∴ – x = – 18
∴ x = 18
(iv) | 5 + \(\large \frac {x}{4}\) | = 5
Solution:
| 5 + \(\large \frac {x}{4}\) | = 5
∴ 5 + \(\large \frac {x}{4}\) = 5
∴ \(\large \frac {x}{4}\) = 5 – 5
∴ \(\large \frac {x}{4}\) = 0
∴ x = 0
∴ 5 + \(\large \frac {x}{4}\) = – 5
∴ \(\large \frac {x}{4}\) = – 5 – 5
∴ \(\large \frac {x}{4}\) = – 10
∴ x = – 10 × 4
∴ x = – 40
Problem Set 2
(1) Choose the correct alternative answer for the questions given below.
(i) Which one of the following is an irrational number?
(A) \(\large \frac {16}{25}\)
(B) \(\sqrt{5}\)
(C) \(\large \frac {3}{9}\)
(D) \(\sqrt{196}\)
Ans: Option (B) : \(\sqrt{5}\)
(ii) Which of the following is an irrational number?
(A) 0.17
(B) \(1.\overline{513}\)
(C) \(1.\overline{2746}\)
(D) 0.101001000…..
Ans: Option (D) : 0.101001000…..
(iii) Decimal expansion of which of the following is non-terminating recurring ?
(A) \(\large \frac {2}{5}\)
(B) \(\large \frac {3}{16}\)
(C) \(\large \frac {3}{11}\)
(D) \(\large \frac {137}{25}\)
Ans: Option (C) : \(\large \frac {3}{11}\)
(iv) Every point on the number line represent, which of the following numbers?
(A) Natural numbers
(B) Irrational numbers
(C) Rational numbers
(D) Real numbers
Ans: Option (D) : Real numbers
(v) The number \(0.\dot4\) in \(\large \frac {p}{q}\) form is _____
(A) \(\large \frac {4}{9}\)
(B) \(\large \frac {40}{9}\)
(C) \(\large \frac {3.6}{9}\)
(D) \(\large \frac {36}{9}\)
Ans: Option (A) : \(\large \frac {4}{9}\)
(vi) What is \(\sqrt{n}\), if n is not a perfect square number?
(A) Natural number
(B) Rational number
(C) Irrational number
(D) Options A, B, C all are correct.
Ans: Option (C) : Irrational number
(vii) Which of the following is not a surd ?
(A) \(\sqrt{7}\)
(B) \(\sqrt[3]{17}\)
(C) \(\sqrt[3]{64}\)
(D) \(\sqrt{193}\)
Ans: Option (C) : \(\sqrt[3]{64}\)
(viii) What is the order of the surd \(\sqrt[3]{\sqrt{5}}\) ?
(A) 3
(B) 2
(C) 6
(D) 5
Ans: Option (C) : 6
(ix) Which one is the conjugate pair of 2 \(\sqrt{5}\) + \(\sqrt{3}\) ?
(A) – 2 \(\sqrt{5}\) + \(\sqrt{3}\)
(B) – 2 \(\sqrt{5}\) – \(\sqrt{3}\)
(C) 2 \(\sqrt{3}\) – \(\sqrt{5}\)
(D) \(\sqrt{3}\) + 2 \(\sqrt{5}\)
Ans: Option (C) : 2 \(\sqrt{3}\) – \(\sqrt{5}\)
(x) The value of |12 – (13 + 7) × 4| is ______
(A) – 68
(B) 68
(C) – 32
(D) 32
Ans: Option (B) : 68
(2) Write the following numbers in \(\large \frac {p}{q}\) form.
(i) 0.555
Solution:
0.555
= \(\large \frac {555}{1000}\)
= \(\large \frac {5\,×\,111}{5\,×\,200}\)
= \(\large \frac {111}{200}\)
∴ 0.555 = \(\large \frac {111}{200}\)
(ii) \(29.\overline{568}\)
Solution:
Let x = 29.568568….. = \(29.\overline{568}\) …(i)
Multiplying both the sides by 1000, we get,
∴ 1000x = 29568.568…. = \(29568.\overline{568}\) …(ii)
Subtracting (i) from (ii), we get,
∴ 1000x – x = \(29568.\overline{568}\) – \(29.\overline{568}\)
∴ 999x = 29539
∴ x = \(\large \frac {29539}{999}\)
∴ 29.568 = \(\large \frac {29539}{999}\)
(iii) 9.315315 …
Solution:
Let x = 9.315315…. = \(9.\overline{315}\) …(i)
Multiplying both the sides by 1000, we get,
∴ 1000x = 9315.315315…. = \(9315.\overline{315}\) …(ii)
Subtracting (i) from (ii), we get,
∴ 1000x – x = \(9315.\overline{315}\) – \(9.\overline{315}\)
∴ 999x = 9306
∴ x = \(\large \frac {9306}{999}\)
∴ x = \(\large \frac {1034}{111}\)
∴ 9.315315… = \(\large \frac {1034}{111}\)
(iv) 357.417417
Solution:
Let x = 357.417417…. = \(357.\overline{417}\) (i)
Multiplying both the sides by 1000, we get,
∴ 1000x = 357417.417417…. = \(357417.\overline{417}\) …(ii)
Subtracting (i) from (ii)
∴ 1000x – x = \(357417.\overline{417}\) – \(357.\overline{417}\)
∴ 999x = 357060
∴ x = \(\large \frac {357060}{999}\)
∴ 357.417417 = \(\large \frac {357060}{999}\)
(v) 30.219
Solution:
Let x = 30.219219….. = \(30.\overline{219}\) …(i)
Multiplying both the sides by 1000, we get,
∴ 1000x = 30219.219219…. = \(30219.\overline{219}\) …(ii)
Subtracting (i) from (ii)
∴ 1000x – x = \(30219.\overline{219}\) – \(30.\overline{219}\)
∴ 999x = 30189
∴ x = \(\large \frac {30189}{999}\)
∴ x = \(\large \frac {10063}{333}\)
∴ 30.219 = \(\large \frac {10063}{333}\)
(3) Write the following numbers in its decimal form.
(i) \(\large \frac {–\,5}{7}\)
Solution:
∴ \(\large \frac {5}{7}\) = \(0.\overline{714285}\)
∴ \(\large \frac {–\,5}{7}\) = \(–\,0.\overline{714285}\)
(ii) \(\large \frac {9}{11}\)
Solution:
∴ \(\large \frac {9}{11}\) = \(0.\overline{81}\)
(iii) \(\sqrt{5}\)
Solution:
∴ \(\sqrt{5}\) = 2.2360…
(iv) \(\large \frac {121}{13}\)
Solution:
∴ \(\large \frac {121}{13}\) = \(9.\overline{307692}\)
(v) \(\large \frac {29}{8}\)
Solution:
∴ \(\large \frac {29}{8}\) = 3.625
(4) Show that 5 + \(\sqrt{7}\) is an irrational number.
Solution:
Let 5 + \(\sqrt{7}\) = x, where ‘a’ is any rational number.
∴ \(\sqrt{7}\) = x – 5 … (i)
Now,
x is a rational number then x – 5 is also a rational number.
Then, \(\sqrt{7}\) is a rational number …[from (i)]
But, it contradicts the fact that \(\sqrt{7}\) is an irrational number.
∴ \(\sqrt{7}\) ≠ x – 5
Our assumption is wrong.
Hence, 5 + \(\sqrt{7}\) is an irrational number.
(5) Write the following surds in simplest form.
(i) \(\large \frac {3}{4}\) \(\sqrt{8}\)
Solution:
\(\large \frac {3}{4}\) \(\sqrt{8}\)
= \(\large \frac {3}{4}\) \(\sqrt{4\,×\,2}\)
= \(\large \frac {3}{4}\) × 2 \(\sqrt{2}\)
= \(\large \frac {3}{2}\) \(\sqrt{2}\)
(ii) – \(\large \frac {5}{9}\) \(\sqrt{45}\)
Solution:
– \(\large \frac {5}{9}\) \(\sqrt{45}\)
= – \(\large \frac {5}{9}\) \(\sqrt{9\,×\,5}\)
= – \(\large \frac {5}{9}\) × 3\(\sqrt{5}\)
(6) Write the simplest form of rationalising factor for the given surds.
(i) \(\sqrt{32}\)
Solution:
\(\sqrt{32}\)
= \(\sqrt{16\,×\,2}\)
= 4 \(\sqrt{2}\)
∴ Simplest rationalization factor is \(\sqrt{2}\)
(ii) \(\sqrt{50}\)
Solution:
\(\sqrt{50}\)
= \(\sqrt{25\,×\,2}\)
= 5 \(\sqrt{2}\)
∴ Simplest rationalization factor is \(\sqrt{2}\)
(iii) \(\sqrt{27}\)
Solution:
\(\sqrt{27}\)
= \(\sqrt{9\,×\,3}\)
= 3 \(\sqrt{3}\)
∴ Simplest rationalization factor is \(\sqrt{3}\)
(iv) \(\large \frac {3}{5}\)\(\sqrt{10}\)
Solution:
Simplest rationalization factor is \(\sqrt{10}\)
(v) 3 \(\sqrt{72}\)
Solution:
3 \(\sqrt{72}\)
= 3 \(\sqrt{36\,×\,2}\)
= 3 × 6 \(\sqrt{2}\)
= 18 \(\sqrt{2}\)
∴ Simplest rationalization factor is \(\sqrt{2}\)
(vi) 4 \(\sqrt{11}\)
Solution:
Simplest rationalization factor is \(\sqrt{11}\)
(7) Simplify.
(i) \(\large \frac {4}{7}\) \(\sqrt{147}\) + \(\large \frac {3}{8}\) \(\sqrt{192}\) – \(\large \frac {1}{5}\) \(\sqrt{75}\)
Solution:
\(\large \frac {4}{7}\) \(\sqrt{147}\) + \(\large \frac {3}{8}\) \(\sqrt{192}\) – \(\large \frac {1}{5}\) \(\sqrt{75}\)
= \(\large \frac {4}{7}\) \(\sqrt{49\,×\,3}\) + \(\large \frac {3}{8}\) \(\sqrt{64\,×\,3}\) – \(\large \frac {1}{5}\) \(\sqrt{25\,×\,3}\)
= \(\large \frac {4}{7}\) × 7 \(\sqrt{3}\) + \(\large \frac {3}{8}\) × 8\(\sqrt{3}\) – \(\large \frac {1}{5}\) × 5\(\sqrt{3}\)
= 4 \(\sqrt{3}\) + 3 \(\sqrt{3}\) – \(\sqrt{3}\)
= (4 + 3 – 1) \(\sqrt{3}\)
= 6 \(\sqrt{3}\)
(ii) 5 \(\sqrt{3}\) + 2 \(\sqrt{27}\) + \(\large \frac {1}{\sqrt3}\)
Solution:
5 \(\sqrt{3}\) + 2 \(\sqrt{27}\) + \(\large \frac {1}{\sqrt3}\)
= 5 \(\sqrt{3}\) + 2 \(\sqrt{9\,×\,3}\) + \(\large \frac {1}{\sqrt3}\) × \(\large \frac {\sqrt3}{\sqrt3}\)
= 5 \(\sqrt{3}\) + 2 × 3\(\sqrt{3}\) + \(\large \frac {\sqrt3}{(\sqrt3)(\sqrt3)}\)
= 5 \(\sqrt{3}\) + 6 \(\sqrt{3}\) + \(\large \frac {\sqrt3}{3}\)
= (5 + 6 + \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= (11 + \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= (\(\large \frac {11\,×\,3}{3}\) + \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= (\(\large \frac {33}{3}\) + \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= \(\large \frac {33\,+\,1}{3}\) \(\sqrt{3}\)
= \(\large \frac {34}{3}\) \(\sqrt{3}\)
(iii) \(\sqrt{216}\) – 5 \(\sqrt{6}\) + \(\sqrt{294}\) – \(\large \frac {3}{\sqrt6}\)
Solution:
\(\sqrt{216}\) – 5 \(\sqrt{6}\) + \(\sqrt{294}\) – \(\large \frac {3}{\sqrt6}\)
= \(\sqrt{36\,×\,6}\) – 5 \(\sqrt{6}\) + \(\sqrt{49\,×\,6}\) – \(\large \frac{3}{\sqrt6}\) × \(\large \frac {\sqrt6}{\sqrt6}\)
= 6 \(\sqrt{6}\) – 5 \(\sqrt{6}\) + 7 \(\sqrt{6}\) – \(\large \frac{3\,×\,\sqrt6}{(\sqrt6)(\sqrt6)}\)
= 6 \(\sqrt{6}\) – 5 \(\sqrt{6}\) + 7 \(\sqrt{6}\) – \(\large \frac{\sqrt6}{2}\)
= (6 – 5 + 7 – \(\large \frac {1}{2}\)) \(\sqrt{6}\)
= 8 – \(\large \frac {1}{2}\) \(\sqrt{6}\)
= \(\large \frac {8\,×\,2}{2}\) – \(\large \frac {1}{2}\) \(\sqrt{6}\)
= \(\large \frac {16}{2}\) – \(\large \frac {1}{2}\) \(\sqrt{6}\)
= \(\large \frac {16\,–\,1}{2}\) \(\sqrt{6}\)
= \(\large \frac {15}{2}\) \(\sqrt{6}\)
(iv) 4\(\sqrt{12}\) – \(\sqrt{75}\) – 7 \(\sqrt{48}\)
Solution:
4\(\sqrt{12}\) – \(\sqrt{75}\) – 7 \(\sqrt{48}\)
= 4\(\sqrt{4\,×\,3}\) – \(\sqrt{25\,×\,3}\) – 7 \(\sqrt{16\,×\,3}\)
= 4 × 2 \(\sqrt{3}\) – 5 \(\sqrt{3}\) – 7 × 4 \(\sqrt{3}\)
= 8 \(\sqrt{3}\) – 5 \(\sqrt{3}\) – 28 \(\sqrt{3}\)
= (8 – 5 – 28) \(\sqrt{3}\)
= – 25 \(\sqrt{3}\)
(v*) 2 \(\sqrt{48}\) – \(\sqrt{75}\) – \(\large \frac {1}{\sqrt3}\)
Solution:
2 \(\sqrt{48}\) – \(\sqrt{75}\) – \(\large \frac {1}{\sqrt3}\)
= 2 \(\sqrt{16\,×\,3}\) – \(\sqrt{25\,×\,3}\) – \(\large \frac {1}{\sqrt3}\) × \(\large \frac {\sqrt3}{\sqrt3}\)
= 2 × 4 \(\sqrt{3}\) – 5 \(\sqrt{3}\) – \(\large \frac {\sqrt3}{(\sqrt3)(\sqrt3)}\)
= 8 \(\sqrt{3}\) – 5 \(\sqrt{3}\) – \(\large \frac {\sqrt3}{3}\)
= (8 – 5 – \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= (3 – \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= (\(\large \frac {3\,×\,3}{3}\) – \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= (\(\large \frac {9}{3}\) – \(\large \frac {1}{3}\)) \(\sqrt{3}\)
= \(\large \frac {9\,–\,1}{3}\) \(\sqrt{3}\)
= \(\large \frac {8}{3}\) \(\sqrt{3}\)
(8) Rationalize the denominator.
(i) \(\large \frac {1}{\sqrt5}\)
Solution:
Rationalizing factor of \(\sqrt{5}\) is \(\sqrt{5}\)
∴ \(\large \frac {1}{\sqrt5}\)
= \(\large \frac {1}{\sqrt5}\) × \(\large \frac {\sqrt5}{\sqrt5}\)
= \(\large \frac {\sqrt5}{(\sqrt5)(\sqrt5)}\)
= \(\large \frac {\sqrt5}{5}\)
(ii) \(\large \frac {2}{3\sqrt7}\)
Solution:
Rationalizing factor of 3 \(\sqrt{7}\) is \(\sqrt{7}\)
∴ \(\large \frac {2}{3\sqrt7}\)
= \(\large \frac {2}{3\sqrt7}\) × \(\large \frac {\sqrt7}{\sqrt7}\)
= \(\large \frac {2\sqrt7}{(3\sqrt7)(\sqrt7)}\)
= \(\large \frac {2\sqrt7}{3\,×\,7}\)
= \(\large \frac {2\sqrt7}{21}\)
(iii) \(\large \frac {1}{\sqrt3\,–\,\sqrt2}\)
Solution:
Conjugate of \(\sqrt{3} \,–\, \sqrt{2}\) is \(\sqrt{3} \,+\, \sqrt{2}\)
\(\large \frac {1}{\sqrt3\,–\,\sqrt2}\)
= \(\large \frac {1}{\sqrt3\,–\,\sqrt2}\) × \(\large \frac {\sqrt{3} \,+\, \sqrt{2}}{\sqrt{3} \,+\, \sqrt{2}}\)
= \(\large \frac {1\,×\,(\sqrt{3} \,+\, \sqrt{2})}{(\sqrt{3} \,–\, \sqrt{2})(\sqrt{3} \,+\, \sqrt{2})}\)
= \(\large \frac {\sqrt{3}\,+\, \sqrt{2}}{(\sqrt{3})^2 \,–\,(\sqrt{2})^2}\)
= \(\large \frac {\sqrt{3}\,+\, \sqrt{2}}{3\,–\,2}\)
= \(\large \frac {\sqrt{3}\,+\, \sqrt{2}}{1}\)
= \(\sqrt{3}\) + \(\sqrt{2}\)
(iv) \(\large \frac {1}{3\sqrt5\,+\,2\sqrt2}\)
Solution:
Conjugate of \(3\sqrt{5} \,+\, 2\sqrt{2}\) is \(3\sqrt{5} \,–\, 2\sqrt{2}\)
\(\large \frac {1}{3\sqrt5\,+\,2\sqrt2}\)
= \(\large \frac {1}{3\sqrt5\,+\,2\sqrt2}\) × \(\large \frac {3\sqrt{5} \,–\, 2\sqrt{2}}{3\sqrt{5} \,–\, 2\sqrt{2}}\)
= \(\large \frac {1\,×\,(3\sqrt{5} \,–\, 2\sqrt{2})}{(3\sqrt{5} \,+\, 2\sqrt{2})(3\sqrt{5} \,–\, 2\sqrt{2})}\)
= \(\large \frac {3\sqrt{5} \,–\, \sqrt{2}}{(3\sqrt{5})^2 \,–\,(2\sqrt{2})^2}\)
= \(\large \frac {3\sqrt{5} \,–\, 2\sqrt{2}}{(9 × 5)\,–\, (4 × 2)}\)
= \(\large \frac {3\sqrt{5} \,–\, 2\sqrt{2}}{45\,–\,8}\)
= \(\large \frac {3\sqrt{5} \,–\, 2\sqrt{2}}{37}\)
(v) \(\large \frac {12}{4\sqrt3\,–\,\sqrt2}\)
Solution:
Conjugate of \(4\sqrt3\,–\,\sqrt2\) is \(4\sqrt3\,+\,\sqrt2\)
\(\large \frac {12}{4\sqrt3\,–\,\sqrt2}\)
= \(\large \frac {12}{4\sqrt3\,–\,\sqrt2}\) × \(\large \frac {4\sqrt3\,+\,\sqrt2}{4\sqrt3\,+\,\sqrt2}\)
= \(\large \frac {12\,×\,(4\sqrt3\,+\,\sqrt2)}{(4\sqrt3\,–\,\sqrt2)(4\sqrt3\,+\,\sqrt2)}\)
= \(\large \frac {12\,×\,4\sqrt3\,+\,12\,\sqrt2}{(4\sqrt3)^2 \,–\,(\sqrt{2})^2}\)
= \(\large \frac {48\sqrt3\,+\,12\,\sqrt2}{(16\,×\,3)\,–\,2}\)
= \(\large \frac {48\sqrt3\,+\,12\,\sqrt2}{48\,–\,2}\)
= \(\large \frac {48\sqrt3\,+\,12\,\sqrt2}{46}\)
= \(\large \frac {6\,×\,2(4\sqrt3\,+\,\sqrt2)}{2\,×\,23}\)
= \(\large \frac {6(4\sqrt3\,+\,\sqrt2)}{23}\)