Miscellaneous
Last updated at December 16, 2024 by Teachoo
Transcript
Question 3 Solve the equations š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·Æļ·Æ = 0, a ā 0 Solving ā = š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·Æļ·Æ Applying R1 ā R1 + R2 + R3 = š„+š+š„+š„ļ·®š„+š„+š+š„ļ·®š„+š„+š„+šļ·®š„ļ·®š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·Æļ·Æ = šš+šļ·®šš+šļ·®šš+šļ·®š„ļ·®š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·Æļ·Æ Taking (3x + a) common from R1 = (3x + a) 1ļ·®1ļ·®1ļ·®š„ļ·®š„+šļ·®š„ļ·®š„ļ·®š„ļ·®š„+šļ·Æļ·Æ Applying C2 ā C2 ā C1 = (3x + a) 1ļ·®šāšļ·®1ļ·®š„ļ·®š„+šāš„ļ·®š„ļ·®š„ļ·®š„āš„ļ·®š„+šļ·Æļ·Æ = (3x+ a) 1ļ·®šļ·®1ļ·®š„ļ·®šļ·®š„ļ·®š„ļ·®0ļ·®š„+šļ·Æļ·Æ Applying C3 āC3 ā C1 = (3x + a) 1ļ·®0ļ·®šāšļ·®š„ļ·®šļ·®š„āš„ļ·®š„ļ·®0ļ·®š„+šāš„ļ·Æļ·Æ = (3x + a) 1ļ·®0ļ·®šļ·®š„ļ·®šļ·®0ļ·®š„ļ·®0ļ·®šļ·Æļ·Æ = (3x+ a) 1 šļ·®0ļ·®0ļ·®šļ·Æļ·Æā0 š„ļ·®0ļ·®š„ļ·®0ļ·Æļ·Æ+0 š„ļ·®šļ·®š„ļ·®0ļ·Æļ·Æļ·Æ = (3x+ a) 1 šļ·®0ļ·®0ļ·®šļ·Æļ·Æā0+0ļ·Æ = (3x+ a) (1(a2 ā 0) ā 0 + 0) = (3x+ a) (a2) ā“ ā = (3x+ a) (a2) Given ā = 0 ā“ (3x + a) a2 = 0