RD Sharma Solutions Class 9 Rationalisation Exercise 3.1

RD Sharma Solutions Class 9 Chapter 3 Exercise 3.1

RD Sharma Class 9 Solutions Chapter 3 Ex 3.1 Free Download

1. Simplify each of the following:

(i)  \(\sqrt[3]{4}\) \(\times\) \(\sqrt[3]{16}\)

(ii) \(\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}\)

Sol:

(i) \(\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}\)

(Note: \(\sqrt[n]{a}\times \sqrt[n]{b}\) = \(\sqrt[n]{a \times b}\))

= \(\sqrt[3]{4\times 16}\)

= \(\sqrt[3]{64}\)

= \(\sqrt[3]{4^{3}}\)

= \((4^{3})^{\frac{1}{3}}\)

= \(4^({3\times \frac{1}{3}})\)

= \(4^{1}\)

= 4

(ii) \(\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}\)

(Note:  \(\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\)=  \(\sqrt[n]{\frac{a}{b}}\))

= \(\sqrt[4]{\frac{1250}{2}}\)

= \(\sqrt[4]{\frac{2\times 625}{2}}\)

= \(\sqrt[4]{625}\)

= \(\sqrt[4]{15^{4}}\)

= \(15^({4\times \frac{1}{4}})\)

= 15

2. Simplify the following expressions:

(i) (\(4 + \sqrt{7}\))( \(3 + \sqrt{2}\))

(ii) (\(3 + \sqrt{3}\))( \(5 – \sqrt{2}\))

(iii) (\(\sqrt{5} – 2\))( \(\sqrt{3} – \sqrt{5}\))

Solution:

(i) (\(4 + \sqrt{7}\))( \(3 + \sqrt{2}\))

= 12 + \(4\sqrt{2}\) + \(3\sqrt{7}\) + \(\sqrt{7\times 2}\)

= 12 + \(4\sqrt{2}\) + \(3\sqrt{7}\) + \(\sqrt{14}\)

 

(ii) (\(3 + \sqrt{3}\))( \(5 – \sqrt{2}\))

= 15 – \(3\sqrt{2}\) + \(5\sqrt{3}\)\(\sqrt{3\times 2}\)

= 15 – \(3\sqrt{2}\) + \(5\sqrt{3}\)\(\sqrt{6}\)

(iii) (\(\sqrt{5} – 2\))( \(\sqrt{3} – \sqrt{5}\))

= \(\sqrt{15} – \sqrt{25} – 2\sqrt{3} + 2\sqrt{5}\)

= \(\sqrt{15} – \sqrt{5\times 5} – 2\sqrt{3} + 2\sqrt{5}\)

= \(\sqrt{15} – 5 – 2\sqrt{3} + 2\sqrt{5}\)

3. Simplify the following expressions:

(i) (11 + \(\sqrt{11}\) )(11 – \(\sqrt{11}\))

(ii) (5 + \(\sqrt{7}\) )(5 – \(\sqrt{7}\))

(iii) (\(\sqrt{8} – \sqrt{2}\)) (\(\sqrt{8} + \sqrt{2}\))

(iv) (3 + \(\sqrt{3}\))( 3 – \(\sqrt{3}\))

(v) (\(\sqrt{5} – \sqrt{2}\))( \(\sqrt{5} + \sqrt{2}\))

Solution:

(i) (11 + \(\sqrt{11}\)) (11 – \(\sqrt{11}\))

As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\))

So, \(11^{2}\) – 11

  • 121 – 11 =110

(ii) (5 + \(\sqrt{7}\)) (5 – \(\sqrt{7}\))

As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\))

So, \(5^{2}\) – 7

  • 25 – 7 = 18

(iii) (\(\sqrt{8} – \sqrt{2}\)) (\(\sqrt{8} + \sqrt{2}\))

As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\))

  • \(\sqrt{8\times 8} – \sqrt{2\times 2}\) = 8 – 2

= 6

(iv) (3 + \(\sqrt{3}\)) ( 3 – \(\sqrt{3}\))

As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\))

= 9 – \(\sqrt{3\times 3}\)

= 6

(v) (\(\sqrt{5} – \sqrt{2}\)) (\(\sqrt{5}+\sqrt{2}\))

As we know, (a + b) (a – b) = (\(a^{2} – b^{2}\))

= \(\sqrt{5\times 5} – \sqrt{2\times 2}\)

= 5 – 2

= 3

  

4. Simplify the following expressions:

(i) \((\sqrt{3}+\sqrt{7})^{2}\)

(ii) \((\sqrt{5}-\sqrt{3})^{2}\)

(iii) \((2\sqrt{5}+3\sqrt{2})^{2}\)

Solution:

(i) \((\sqrt{3}+\sqrt{7})^{2}\)

As we know , (a + b) \(^{2}\) = (\(a^{2} + 2\times a\times b + b^{2}\))

= \(\sqrt{3^{2}} + 2\times \sqrt{3}\times \sqrt{7} + \sqrt{7^{2}}\)

= 3 + \(2\times \sqrt{3\times 7}\) + 7

= 10 + \(2\times \sqrt{21}\)

(ii) \((\sqrt{5} – \sqrt{3})^{2}\)

As we know , (a – b) \(^{2}\) = (\(a^{2} – 2\times a\times b + b^{2}\))

(iii) \((2\sqrt{5} + 3\sqrt{2})^{2}\)

As we know , (a + b) \(^{2}\) = (\(a^{2} + 2\times a\times b + b^{2}\))

= \(4\sqrt{5\times 5}+2\times 2\sqrt{5}\times 3\sqrt{2}+9\sqrt{2\times 2}\)

= 20 + \(12\sqrt{10}\) + 18

= 28 + \(12\sqrt{10}\)