Last updated at March 9, 2017 by Teachoo

Transcript

Ex9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) :"(3 − 2 " √2) Introduction Componendo dividendo If 𝑥/𝑦 = 𝑎/𝑏 Applying componendo dividendo (𝑥 + 𝑦)/(𝑥 − 𝑦) = (𝑎 + 𝑏)/(𝑎 − 𝑏) Eg: Taking 1/2 = 4/8 (1+ 2)/(1 − 2) = (4 + 8)/(4 − 8) 3/(−1) = 12/(−4) -3 = -3 Ex 9.3, 28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2 √2) :"(3 − 2 " √2) Let a & b be the numbers We know that Geometric mean of two numbers a & b is √𝑎𝑏 i.e. GM of a & b = √𝑎𝑏 According to the question Sum of two numbers a and b is 6 times of their GM a + b = 6√𝑎𝑏 Solving, (𝑎 + 𝑏 )/(2√𝑎𝑏) = 3/1 Applying componendo & dividendo (𝑎+𝑏+2√𝑎𝑏)/(𝑎+𝑏 −2√𝑎𝑏) = (3 + 1)/(3 − 1 ) ((√𝑎)2+(√𝑏)2+2(√𝑎×√𝑏))/((√𝑎)2+(√𝑏)2−2(√𝑎×√𝑏) ) = 4/2 Using (x + y)2 = x2 + y2 + 2xy (x - y)2 = x2 + y2 - 2xy (√𝑎 + √𝑏)2/(√𝑎 − √𝑏)2 = 2/1 ((√𝑎 + √𝑏)/(√𝑎 − √𝑏))^2 = 2/1 (√𝑎 + √𝑏)/(√𝑎 − √𝑏) = √2/( 1) Again applying componendo & dividendo ((√𝑎+ √𝑏)+(√𝑎 − √𝑏))/((√𝑎 + √𝑏) −(√𝑎 − √𝑏) ) = (√2 + 1)/(√2 − 1) (√𝑎 +√𝑎 + √𝑏 − √𝑏)/(√𝑎 + √𝑏 − √𝑎 + √𝑏) = (√2 + 1)/(√2 − 1) (2√𝑎 + 0)/(√𝑏 + √𝑏 + √𝑎 − √𝑎) = (√2 + 1)/(√2 − 1) (2√𝑎)/(2√𝑏 + 0) = (√2 + 1)/(√2 − 1) (2√𝑎)/(2√𝑏 ) = (√2 + 1)/(√2 − 1) √(𝑎/𝑏) = (√2 + 1)/(√2 − 1) Squaring both sides (√(𝑎/𝑏))^2 = ((√2 + 1)/(√2 − 1))^2 𝑎/𝑏 = ((√2 + 1)2)/((√2 − 1)2) 𝑎/𝑏 = ((√2)2 + (1)2 + 2√2 × 1)/((√2)2 + (1)2 − 2√2 × 1) 𝑎/𝑏 = (2 + 1 + 2√2)/(2 + 1 − 2√2) 𝑎/𝑏 = (3 + 2√2)/(3 − 2√2) Thus the ratio of a & b is 3 + 2√3: 3 – 2√2 Hence proved

Ex 9.3, 1

Ex 9.3, 2

Ex 9.3, 3 Important

Ex 9.3, 4

Ex 9.3, 5

Ex 9.3, 6

Ex 9.3, 7

Ex 9.3, 8

Ex 9.3, 9

Ex 9.3, 10

Ex 9.3, 11 Important

Ex 9.3, 12

Ex 9.3, 13

Ex 9.3, 14

Ex 9.3, 15

Ex 9.3, 16

Ex 9.3, 17 Important

Ex 9.3, 18 Important

Ex 9.3, 19

Ex 9.3, 20

Ex 9.3, 21

Ex 9.3, 22 Important

Ex 9.3, 23

Ex 9.3, 24

Ex 9.3, 25

Ex 9.3, 26

Ex 9.3, 27 Important

Ex 9.3, 28 Important You are here

Ex 9.3, 29 Important

Ex 9.3, 30

Ex 9.3, 31

Ex 9.3, 32

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.