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

[a ≠ 0, b ≠ 0, x ≠ 0, x ≠ –(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

Go Ad-free

Transcript

Question 23 (OR 2nd question) Solve for x: 1/((𝑎 + 𝑏 + 𝑥)) = 1/𝑎 + 1/𝑏 + 1/𝑥 [a ≠ 0, b ≠ 0, x ≠ 0, x ≠ –(a + b)] 1/((𝑎 + 𝑏 + 𝑥)) = 1/𝑎 + 1/𝑏 + 1/𝑥 1/((𝑎 + 𝑏 + 𝑥)) – 1/𝑥 = 1/𝑎 + 1/𝑏 (𝑥 − (𝑎 + 𝑏 + 𝑥))/(𝑎 + 𝑏 + 𝑥)𝑥 = 1/𝑎 + 1/𝑏 (𝑥 − 𝑎 − 𝑏 − 𝑥)/(𝑎 + 𝑏 + 𝑥)𝑥 = (𝑏 + 𝑎)/𝑎𝑏 (−(𝑎 + 𝑏))/(𝑎 + 𝑏 + 𝑥)𝑥 = ((𝑎 + 𝑏))/𝑎𝑏 (−1)/(𝑎 + 𝑏 + 𝑥)𝑥 = 1/𝑎𝑏 –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 = (−𝐵 ± √(𝐵^(2 )− 4𝐴𝐶) )/2𝐴 x = (−(𝑎 + 𝑏) ± √(〖(𝑎 + 𝑏)〗^(2 )− 4×1×𝑎𝑏) )/(2 × 1) x = (−(𝑎 + 𝑏) ± √(𝑎^2 + 𝑏^2 + 2𝑎𝑏 − 4𝑎𝑏) )/2 x = (−(𝑎 + 𝑏) ± √(𝑎^2 + 𝑏^2 − 2𝑎𝑏) )/2 x = (−(𝑎 + 𝑏) ± √((𝑎 − 𝑏)^2 ) )/2 x = (−(𝑎 + 𝑏) ± (𝑎 − 𝑏) )/2 So, x = (−(𝑎 + 𝑏) + (𝑎 − 𝑏) )/2 x = (−𝑎 − 𝑏 + 𝑎 − 𝑏)/2 x = (−2𝑏)/2 x = –b x = (−(𝑎 + 𝑏) − (𝑎 − 𝑏) )/2 x = (−𝑎 − 𝑏 − 𝑎+ 𝑏)/2 x = (−2𝑎)/2 x = –a

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo