Find rational numbers a and b such that 2 - 5 2 + 5 = a 5 + b

Given, 2 - 5 2 + 5 = a 5 + b ........... ( 1 ) To rationalize the denominator, we multiply the numerator and denominator by its rationalizing factor. Here, 2 - 5 , the conjugate of the denominator, is the rationalizing factor of 2 - 5 2 + 5 . Then, multiplying the numerator and the denominator of the fraction by the rational factor (conjugate), we get, 2 - 5 2 + 5 = ( 2 - 5 ) ( 2 + 5 ) × ( 2 - 5 ) ( 2 - 5 ) = ( 2 - 5 ) ( 2 - 5 ) ( 2 + 5 ) ( 2 - 5 ) = ( 2 - 5 ) 2 2 2 - ( 5 ) 2 = 2 2 - 2 × 2 × 5 + ( 5 ) 2 4 - 5 = 4 - 4 5 + 5 - 1 = 9 - 4 5 - 1 = 4 5 - 9 , whose denominator is rational. From equation ( 1 ) , 2 - 5 2 + 5 = a 5 + b ⇒ 4 5 - 9 = a 5 + b . On comparing both sides, we get, a = 4 and b = - 9 .

Find rational numbers a and b such that 3 + 2 3 - 2 = a + b 6

Given, 3 + 2 3 - 2 = a + b 6 → ( 1 ) . To rationalize the denominator, we multiply the numerator and denominator by its rationalizing factor. Here, 3 + 2 , the conjugate of the denominator, is the rationalizing factor of 3 + 2 3 - 2 . Then, multiplying the numerator and the denominator of the fraction by the rational factor (conjugate), we get, 3 + 2 3 - 2 = ( 3 + 2 ) ( 3 - 2 ) × ( 3 + 2 ) ( 3 + 2 ) = ( 3 + 2 ) ( 3 + 2 ) ( 3 - 2 ) ( 3 + 2 ) = ( 3 + 2 ) 2 ( 3 ) 2 - ( 2 ) 2 = ( 3 ) 2 + ( 2 × 3 × 2 ) + ( 2 ) 2 3 - 2 = 3 + 2 6 + 2 3 - 2 = 5 + 2 6 1 = 5 + 2 6 , whose denominator is rational. From equation ( 1 ) , 3 + 2 3 - 2 = a + b 6 ⇒ 5 + 2 6 = a + b 6 . On comparing both sides, we get, a = 5 and b = 2 .

Find the values of a and b if 7 + 3 5 3 + 5 - 7 - 3 5 3 - 5 = a + b 5

LHS = 7 + 3 5 3 + 5 - 7 - 3 5 3 - 5 On rationalising the denominator, 7 + 3 5 3 + 5 × 3 - 5 3 - 5 - 7 - 3 5 3 - 5 × 3 + 5 3 + 5 [According to the formula a 2 - b 2 = ( a + b ) ( a - b ) ] = 7 × 3 - 7 5 + 3 5 × 3 - 3 5 × 5 3 2 - 5 2 - 7 × 3 + 7 5 - 3 5 × 3 - 3 5 × 5 3 2 - 5 2 = 21 - 7 5 + 9 5 - 15 9 - 5 - 21 + 7 5 - 9 5 - 15 9 - 5 = 6 + 2 5 4 - 6 - 2 5 4 = 6 + 2 5 - 6 + 2 5 4 = 0 + 4 5 4 = 0 + 5 Thus, we see, 7 + 3 5 3 + 5 - 7 - 3 5 3 - 5 = a + b 5 ⇒ 0 + 5 = a + b 5 ⇒ a = 0 and b = 1 .

E M B I B E

Secondary School Mathematics FOR CLASS 9>Number Systems>Chapter 1 - Number Systems>EXERCISE 1F

R S Aggarwal and Veena Aggarwal Solutions for Exercise 6: EXERCISE 1F

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 6: EXERCISE 1F

Attempt the free practice questions from Exercise 6: EXERCISE 1F with hints and solutions to strengthen your understanding. Secondary School Mathematics FOR CLASS 9 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 6: EXERCISE 1F with Hints & Solutions

MEDIUM

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 1 - Number Systems>EXERCISE 1F>Q 1

If the rationalizing factor of the denominator $\frac{1}{\sqrt{2} + \sqrt{3}}$ is in the form of $\sqrt{k} - \sqrt{(k + 1)}$ , then find the value of $k$ .

HARD

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 1 - Number Systems>EXERCISE 1F>Q 2

Rationalising the denominator of $\frac{1 + \sqrt{2}}{2 - \sqrt{2}}$ , we get, $\frac{k + 3 \sqrt{3}}{2}$ . Find $k$ .

MEDIUM

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 1 - Number Systems>EXERCISE 1F>Q 3

It being given that $\sqrt{2} = 1.414$ , $\sqrt{3} = 1.732$ , $\sqrt{5} = 2.236$ and $\sqrt{10} = 3.162$ , find the value to three places of decimals, of the following.

$\frac{2}{\sqrt{5}}$

MEDIUM

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 1 - Number Systems>EXERCISE 1F>Q 3

It being given that $\sqrt{2} = 1.414$ , $\sqrt{3} = 1.732$ , $\sqrt{5} = 2.236$ and $\sqrt{10} = 3.162$ , find the value to three places of decimals, of the following.

$\frac{2 - \sqrt{3}}{\sqrt{3}}$

MEDIUM

9th CBSE

IMPORTANT

Secondary School Mathematics FOR CLASS 9>Chapter 1 - Number Systems>EXERCISE 1F>Q 3

It being given that $\sqrt{2} = 1.414$ , $\sqrt{3} = 1.732$ , $\sqrt{5} = 2.236$ and $\sqrt{10} = 3.162$ , find the value to three places of decimals of the following.