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)]


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Last updated at Oct. 5, 2020 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)]
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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
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
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