Check sibling questions

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

[a β‰  0, b β‰  0, x β‰  0, x β‰  –(a + b)]

Solve for x: 1/(a + b + x) = 1/a + 1/b + 1/x [with Video] - Teachoo

Question 23 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 2
Question 23 (Or 2nd) - CBSE Class 10 Sample Paper for 2019 Boards - Part 3

This video is only available for Teachoo black users


Transcript

Question 23 (OR 2nd question) Solve for x: 1/((π‘Ž + 𝑏 + π‘₯)) = 1/π‘Ž + 1/𝑏 + 1/π‘₯ [a β‰  0, b β‰  0, x β‰  0, x β‰  –(a + b)] 1/((π‘Ž + 𝑏 + π‘₯)) = 1/π‘Ž + 1/𝑏 + 1/π‘₯ 1/((π‘Ž + 𝑏 + π‘₯)) – 1/π‘₯ = 1/π‘Ž + 1/𝑏 (π‘₯ βˆ’ (π‘Ž + 𝑏 + π‘₯))/(π‘Ž + 𝑏 + π‘₯)π‘₯ = 1/π‘Ž + 1/𝑏 (π‘₯ βˆ’ π‘Ž βˆ’ 𝑏 βˆ’ π‘₯)/(π‘Ž + 𝑏 + π‘₯)π‘₯ = (𝑏 + π‘Ž)/π‘Žπ‘ (βˆ’(π‘Ž + 𝑏))/(π‘Ž + 𝑏 + π‘₯)π‘₯ = ((π‘Ž + 𝑏))/π‘Žπ‘ (βˆ’1)/(π‘Ž + 𝑏 + π‘₯)π‘₯ = 1/π‘Žπ‘ –ab = x(a + b + x) –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 = (βˆ’π΅ Β± √(𝐡^(2 )βˆ’ 4𝐴𝐢) )/2𝐴 x = (βˆ’(π‘Ž + 𝑏) Β± √(γ€–(π‘Ž + 𝑏)γ€—^(2 )βˆ’ 4Γ—1Γ—π‘Žπ‘) )/(2 Γ— 1) x = (βˆ’(π‘Ž + 𝑏) Β± √(π‘Ž^2 + 𝑏^2 + 2π‘Žπ‘ βˆ’ 4π‘Žπ‘) )/2 x = (βˆ’(π‘Ž + 𝑏) Β± √(π‘Ž^2 + 𝑏^2 βˆ’ 2π‘Žπ‘) )/2 x = (βˆ’(π‘Ž + 𝑏) Β± √((π‘Ž βˆ’ 𝑏)^2 ) )/2 x = (βˆ’(π‘Ž + 𝑏) Β± (π‘Ž βˆ’ 𝑏) )/2 So, x = (βˆ’(π‘Ž + 𝑏) + (π‘Ž βˆ’ 𝑏) )/2 x = (βˆ’π‘Ž βˆ’ 𝑏 + π‘Ž βˆ’ 𝑏)/2 x = (βˆ’2𝑏)/2 x = –b x = (βˆ’(π‘Ž + 𝑏) βˆ’ (π‘Ž βˆ’ 𝑏) )/2 x = (βˆ’π‘Ž βˆ’ 𝑏 βˆ’ π‘Ž+ 𝑏)/2 x = (βˆ’2π‘Ž)/2 x = –a

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.