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)]
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)]
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
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Question 7 (Or 1st)
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Question 23 (Or 1st)
Question 23 (Or 2nd) You are here
Question 24
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Question 28 (Or 2nd)
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CBSE Class 10 Sample Paper for 2019 Boards
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