Rationalise the denominator and evaluate by taking 2 = 1.414 and 3 = 1.732 of 1 3 + 2

To solve this question, 1 ( 3 + 2 ) Multiplying and dividing the expression by ( 3 - 2 ), We get, 1 ( 3 + 2 ) × ( 3 - 2 ) ( 3 - 2 ) = ( 3 - 2 ) ( 3 + 2 ) × ( 3 - 2 ) We know that: ( a + b ) × ( a - b ) = a 2 - b 2 Using this in the above equation we get, ( 3 - 2 ) 3 - 2 = 3 - 2 Putting the given values, we get: = 1 . 732 - 1 . 414 = 0 . 318

Simplify: 6 2 + 3 + 3 6 + 3 - 4 3 6 + 2

Given, 6 2 + 3 + 3 6 + 3 - 4 3 6 + 2 Rationalising each of them and using the identity, ( a + b ) ( a - b ) = a 2 - b 2 , we get 6 2 + 3 = 6 2 + 3 × 2 - 3 2 - 3 = 12 - 18 2 2 - 3 2 = 12 - 18 - 1 3 6 + 3 = 3 6 + 3 × 6 - 3 6 - 3 = 3 6 - 3 3 6 2 - 3 2 = 3 ( 6 - 3 ) 6 - 3 = 3 ( 6 - 3 ) 3 = 6 - 3 And, 4 3 6 + 2 = 4 3 6 + 2 × 6 - 2 6 - 2 = 4 18 - 4 6 6 2 - 2 2 = 4 ( 18 - 6 ) 4 = 18 - 6 Now according to the question:- 6 2 + 3 + 3 6 + 3 - 4 3 6 + 2 = 18 - 12 + 6 - 3 - 18 + 6 = - 2 3 - 3 + 2 6 = - 3 3 + 2 6 = 2 6 - 3 3 Hence, the required answer is 2 6 - 3 3

If x = 3 + 5 , find the value of x 2 + 1 x 2

Given, x = 3 + 5 On squaring both the sides, we get, ⇒ x 2 = 3 + 5 2 = 3 2 + 5 2 + 2 × 3 × 5 = 9 + 5 + 6 5 = 14 + 6 5 Now, 1 x 2 = 1 14 + 6 5 = 1 14 + 6 5 × 14 - 6 5 14 - 6 5 = 14 - 6 5 14 2 - 6 5 2 = 14 - 6 5 196 - 180 = 14 - 6 5 16 Now according to the question:- x 2 + 1 x 2 = 14 + 6 5 + 14 - 6 5 16 ⇒ x 2 + 1 x 2 = 224 + 96 5 + 14 - 6 5 16 ⇒ x 2 + 1 x 2 = 238 + 90 5 16 = 119 + 45 5 8 Hence, the required answer is 119 + 45 5 8 .

If x = 3 - 2 2 , then find the value of x 2 - 1 x 2 .

Given, x = 3 - 2 2 Now, 1 x = 1 3 - 2 2 = 1 3 - 2 2 × 3 + 2 2 3 + 2 2 Using the identity, a 2 - b 2 = ( a + b ) ( a - b ) , we get, = 3 + 2 2 9 - 8 = 3 + 2 2 Now, x 2 - 1 x 2 = ( 3 - 2 2 ) 2 - ( 3 + 2 2 ) 2 ⇒ x 2 - 1 x 2 = ( 9 + 8 - 12 2 ) - 9 + 8 + 12 2 ⇒ x 2 - 1 x 2 = 9 + 8 - 12 2 - 9 - 8 - 12 2 ⇒ x 2 - 1 x 2 = - 24 2 Hence, the value of x 2 - 1 x 2 is - 24 2 .

If x = 4 - 15 , find the value of x 3 - 1 x 3

Given, x = 4 - 15 1 x = 1 4 - 15 = 1 4 - 15 × 4 + 15 4 + 15 Using the identity, a 2 - b 2 = ( a + b ) ( a - b ) , we get, = 4 + 15 16 - 15 = 4 + 15 Now, x - 1 x = 4 - 15 - 4 - 15 = - 2 15 Therefore, x 3 - 1 x 3 = x - 1 x 3 + 3 × x × 1 x ( x - 1 x ) = x - 1 x 3 + 3 x - 1 x ⇒ x 3 - 1 x 3 = - 2 15 3 - 3 × 2 15 ⇒ x 3 - 1 x 3 = - 8 × 15 15 - 6 15 ⇒ x 3 - 1 x 3 = - 120 15 - 6 15 = - 126 15 Hence, x 3 - 1 x 3 = - 126 15

If a = 7 + 40 , find the value of a + 1 a

Given, a = 7 + 40 ⇒ a = 7 + 2 × 2 × 10 ⇒ a = 7 + 2 10 ⇒ a = 5 + 2 + 2 × 5 × 2 ⇒ a = 5 2 + 2 2 + 2 × 5 × 2 Using the identity, a + b 2 = a 2 + 2 a b + b 2 , we get, ⇒ a = 5 + 2 2 ⇒ a = 5 + 2 Now, 1 a = 1 5 + 2 = 1 5 + 2 × 5 - 2 5 - 2 = 5 - 2 5 2 - 2 2 Using the identity, a 2 - b 2 = ( a + b ) ( a - b ) , we get, = 5 - 2 5 - 2 = 5 - 2 3 According to the question:- a + 1 a = 5 + 2 + 5 - 2 3 ⇒ a + 1 a = 3 5 + 2 + 5 - 2 3 ⇒ a + 1 a = 3 5 + 3 2 + 5 - 2 3 ⇒ a + 1 a = 4 5 + 2 2 3 Hence, the required answer is 4 5 + 2 2 3 .

Express a 2 - b 2 + a a 2 + b 2 + b ÷ a 2 + b 2 - b a - a 2 - b 2 in the simplest form.

Given, a 2 - b 2 + a a 2 + b 2 + b ÷ a 2 + b 2 - b a - a 2 - b 2 Now, a 2 - b 2 + a a 2 + b 2 + b ÷ a 2 + b 2 - b a - a 2 - b 2 = a 2 - b 2 + a a 2 + b 2 + b × a - a 2 - b 2 a 2 + b 2 - b Using the identity, a 2 - b 2 = ( a + b ) ( a - b ) , we get, = a 2 - a 2 - b 2 2 a 2 + b 2 2 - b 2 = a 2 - a 2 + b 2 a 2 + b 2 - b 2 = b 2 a 2 Hence, the required answer is b 2 a 2 .

Find the value of 1 3 - 8 - 1 8 - 7 + 1 7 - 6 - 1 6 - 5 + 1 5 - 2

Given, 1 3 - 8 - 1 8 - 7 + 1 7 - 6 - 1 6 - 5 + 1 5 - 2 Rationalising each one of them as:- 1 3 - 8 = 1 3 - 8 × 3 + 8 3 + 8 = 3 + 8 ( 9 - 8 ) = 3 + 8 1 8 - 7 = 1 8 - 7 × 8 + 7 8 + 7 = 8 + 7 8 - 7 = 8 + 7 1 7 - 6 = 1 7 - 6 × 7 + 6 7 + 6 = 7 + 6 7 - 6 = 7 + 6 1 6 - 5 = 1 6 - 5 × 6 + 5 6 + 5 = 6 + 5 6 - 5 = 6 + 5 1 5 - 2 = 1 5 - 2 × 5 + 2 5 + 2 = 5 - 2 5 - 4 = 5 + 2 Now according to the question:- 1 3 - 8 - 1 8 - 7 + 1 7 - 6 - 1 6 - 5 + 1 5 - 2 = 3 + 8 - 8 + 7 + 7 + 6 - 6 + 5 + ( 5 + 2 ) = 3 + 8 - 8 - 7 + 7 + 6 - 6 - 5 + 5 + 2 = 3 + 2 = 5 Hence, the required answer is 5 .

E M B I B E

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3

A K Upadhyay and S Upadhyay Solutions for Chapter: Number System, Exercise 3: EXERCISE 1.3

Author:A K Upadhyay & S Upadhyay

A K Upadhyay Mathematics Solutions for Exercise - A K Upadhyay and S Upadhyay Solutions for Chapter: Number System, Exercise 3: EXERCISE 1.3

Attempt the free practice questions on Chapter 1: Number System, Exercise 3: EXERCISE 1.3 with hints and solutions to strengthen your understanding. MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from A K Upadhyay and S Upadhyay Solutions for Chapter: Number System, Exercise 3: EXERCISE 1.3 with Hints & Solutions

EASY

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 17

Rationalise the denominator and evaluate by taking $\sqrt{2} = 1.414$ and $\sqrt{3} = 1.732$ of $\frac{1}{\sqrt{3} + \sqrt{2}}$

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 18

Simplify: $\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{3}{\sqrt{6} + \sqrt{3}} - \frac{4 \sqrt{3}}{\sqrt{6} + \sqrt{2}}$

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 22

If $x = 3 + \sqrt{5}$ , find the value of $x^{2} + \frac{1}{x^{2}}$

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 26

If $x = 3 - 2 \sqrt{2},$ then find the value of $x^{2} - \frac{1}{x^{2}}$ .

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 28

If $x = 4 - \sqrt{15}$ , find the value of $x^{3} - \frac{1}{x^{3}}$

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 30

If $a = 7 + \sqrt{40}$ , find the value of $\sqrt{a} + \frac{1}{\sqrt{a}}$

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 33

Express $\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}$ in the simplest form.

MEDIUM

9th CBSE

IMPORTANT

MATHEMATICS>Chapter 1 - Number System>EXERCISE 1.3>Q 34

Find the value of $\sqrt{\frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2}}$

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