# Misc 5 - Class 12

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 5 Solve the equations 𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 = 0, a ≠ 0 Solving ∆ = 𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 Applying R1 → R1 + R2 + R3 = 𝑥+𝑎+𝑥+𝑥𝑥+𝑥+𝑎+𝑥𝑥+𝑥+𝑥+𝑎𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 = 𝟑𝒙+𝒂𝟑𝒙+𝒂𝟑𝒙+𝒂𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 Taking (3x + a) common from R1 = (3x + a) 111𝑥𝑥+𝑎𝑥𝑥𝑥𝑥+𝑎 Applying C2 → C2 – C1 = (3x + a) 1𝟏−𝟏1𝑥𝑥+𝑎−𝑥𝑥𝑥𝑥−𝑥𝑥+𝑎 = (3x+ a) 1𝟎1𝑥𝑎𝑥𝑥0𝑥+𝑎 Applying C3 →C3 – C1 = (3x + a) 10𝟏−𝟏𝑥𝑎𝑥−𝑥𝑥0𝑥+𝑎−𝑥 = (3x + a) 10𝟎𝑥𝑎0𝑥0𝑎 = (3x+ a) 1 𝑎00𝑎−0 𝑥0𝑥0+0 𝑥𝑎𝑥0 = (3x+ a) 1 𝑎00𝑎−0+0 = (3x+ a) (1(a2 – 0) – 0 + 0) = (3x+ a) (a2) ∴ ∆ = (3x+ a) (a2) Given ∆ = 0 ∴ (3x + a) a2 = 0

Chapter 4 Class 12 Determinants

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.