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Question 23 (OR 2
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nd
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question)
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Solve for x: 1/((a + b + x)) = 1/a + 1/b + 1/x

[a ≠ 0, b ≠ 0, x ≠ 0, x ≠ –(a + b)]

Last updated at Nov. 28, 2018 by Teachoo

**
Question 23 (OR 2
**
**
nd
**
**
question)
**

Solve for x: 1/((a + b + x)) = 1/a + 1/b + 1/x

[a ≠ 0, b ≠ 0, x ≠ 0, x ≠ –(a + b)]

Transcript

Question 23 (OR 2nd question) Solve for x: 1/((๐ + ๐ + ๐ฅ)) = 1/๐ + 1/๐ + 1/๐ฅ [a โ 0, b โ 0, x โ 0, x โ โ(a + b)] 1/((๐ + ๐ + ๐ฅ)) = 1/๐ + 1/๐ + 1/๐ฅ 1/((๐ + ๐ + ๐ฅ)) โ 1/๐ฅ = 1/๐ + 1/๐ (๐ฅ โ (๐ + ๐ + ๐ฅ))/(๐ + ๐ + ๐ฅ)๐ฅ = 1/๐ + 1/๐ (๐ฅ โ ๐ โ ๐ โ ๐ฅ)/(๐ + ๐ + ๐ฅ)๐ฅ = (๐ + ๐)/๐๐ (โ(๐ + ๐))/(๐ + ๐ + ๐ฅ)๐ฅ = ((๐ + ๐))/๐๐ (โ1)/(๐ + ๐ + ๐ฅ)๐ฅ = 1/๐๐ โab = x(a + b + x) โab = x(a + b) + x2 0 = x2 + x(a + b) + ab x2 + x(a + b) + ab = 0 Comparing with Ax2 + BX + C = 0 A = 1, B = (a + b), C = ab x = (โ๐ต ยฑ โ(๐ต^(2 )โ 4๐ด๐ถ) )/2๐ด x = (โ(๐ + ๐) ยฑ โ(ใ(๐ + ๐)ใ^(2 )โ 4ร1ร๐๐) )/(2 ร 1) x = (โ(๐ + ๐) ยฑ โ(๐^2 + ๐^2 + 2๐๐ โ 4๐๐) )/2 x = (โ(๐ + ๐) ยฑ โ(๐^2 + ๐^2 โ 2๐๐) )/2 x = (โ(๐ + ๐) ยฑ โ((๐ โ ๐)^2 ) )/2 x = (โ(๐ + ๐) ยฑ (๐ โ ๐) )/2 So, x = (โ(๐ + ๐) + (๐ โ ๐) )/2 x = (โ๐ โ ๐ + ๐ โ ๐)/2 x = (โ2๐)/2 x = โb x = (โ(๐ + ๐) โ (๐ โ ๐) )/2 x = (โ๐ โ ๐ โ ๐+ ๐)/2 x = (โ2๐)/2 x = โa

CBSE Class 10 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2 (Or 1st)

Question 2 (Or 2nd)

Question 3 (Or 1st)

Question 3 (Or 2nd)

Question 4

Question 5

Question 6

Question 7 (Or 1st)

Question 7 (Or 2nd)

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17 (Or 1st)

Question 17 (Or 2nd)

Question 18

Question 19 (Or 1st)

Question 19 (Or 2nd)

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23 (Or 1st)

Question 23 (Or 2nd) You are here

Question 24

Question 25

Question 26

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28 (Or 1st)

Question 28 (Or 2nd)

Question 29

Question 30

Class 10

Solutions of Sample Papers for Class 10 Boards

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.