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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  1. Class 10
  2. Sample Papers, Previous Year Papers and Other Questions

Transcript

Question 23 (OR 2nd question) Solve for x: 1/((๐‘Ž + ๐‘ + ๐‘ฅ)) = 1/๐‘Ž + 1/๐‘ + 1/๐‘ฅ [a โ‰  0, b โ‰  0, x โ‰  0, x โ‰  โ€“(a + b)] 1/((๐‘Ž + ๐‘ + ๐‘ฅ)) = 1/๐‘Ž + 1/๐‘ + 1/๐‘ฅ 1/((๐‘Ž + ๐‘ + ๐‘ฅ)) โ€“ 1/๐‘ฅ = 1/๐‘Ž + 1/๐‘ (๐‘ฅ โˆ’ (๐‘Ž + ๐‘ + ๐‘ฅ))/(๐‘Ž + ๐‘ + ๐‘ฅ)๐‘ฅ = 1/๐‘Ž + 1/๐‘ (๐‘ฅ โˆ’ ๐‘Ž โˆ’ ๐‘ โˆ’ ๐‘ฅ)/(๐‘Ž + ๐‘ + ๐‘ฅ)๐‘ฅ = (๐‘ + ๐‘Ž)/๐‘Ž๐‘ (โˆ’(๐‘Ž + ๐‘))/(๐‘Ž + ๐‘ + ๐‘ฅ)๐‘ฅ = ((๐‘Ž + ๐‘))/๐‘Ž๐‘ (โˆ’1)/(๐‘Ž + ๐‘ + ๐‘ฅ)๐‘ฅ = 1/๐‘Ž๐‘ โ€“ab = x(a + b + x) โ€“ab = 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 = (โˆ’๐ต ยฑ โˆš(๐ต^(2 )โˆ’ 4๐ด๐ถ) )/2๐ด x = (โˆ’(๐‘Ž + ๐‘) ยฑ โˆš(ใ€–(๐‘Ž + ๐‘)ใ€—^(2 )โˆ’ 4ร—1ร—๐‘Ž๐‘) )/(2 ร— 1) x = (โˆ’(๐‘Ž + ๐‘) ยฑ โˆš(๐‘Ž^2 + ๐‘^2 + 2๐‘Ž๐‘ โˆ’ 4๐‘Ž๐‘) )/2 x = (โˆ’(๐‘Ž + ๐‘) ยฑ โˆš(๐‘Ž^2 + ๐‘^2 โˆ’ 2๐‘Ž๐‘) )/2 x = (โˆ’(๐‘Ž + ๐‘) ยฑ โˆš((๐‘Ž โˆ’ ๐‘)^2 ) )/2 x = (โˆ’(๐‘Ž + ๐‘) ยฑ (๐‘Ž โˆ’ ๐‘) )/2 So, x = (โˆ’(๐‘Ž + ๐‘) + (๐‘Ž โˆ’ ๐‘) )/2 x = (โˆ’๐‘Ž โˆ’ ๐‘ + ๐‘Ž โˆ’ ๐‘)/2 x = (โˆ’2๐‘)/2 x = โ€“b x = (โˆ’(๐‘Ž + ๐‘) โˆ’ (๐‘Ž โˆ’ ๐‘) )/2 x = (โˆ’๐‘Ž โˆ’ ๐‘ โˆ’ ๐‘Ž+ ๐‘)/2 x = (โˆ’2๐‘Ž)/2 x = โ€“a

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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.