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Transcript

Question 2 Solve the equation 3๐‘ฅ2 โ€“ 4๐‘ฅ + 20/3 = 0 3๐‘ฅ2 โ€“ 4๐‘ฅ + 20/3 = 0 Multiplying both sides by 3 3 ร— (3๐‘ฅ2 โ€“ 4๐‘ฅ "+ " 20/3) = 3 ร— 0 3 ร— 3x2 โ€“ 3 ร— 4x + 3 ร— 20/3 = 0 9๐‘ฅ2 โ€“ 12๐‘ฅ + 20 = 0 The above equation is of the form ๐‘Ž๐‘ฅ2 + ๐‘๐‘ฅ + ๐‘ = 0 Where a = 9, b = โ€“12, and c = 20 x = (โˆ’๐‘ยฑโˆš( ๐‘^2 โˆ’4๐‘Ž๐‘ ))/2๐‘Ž = (โˆ’(โˆ’12) ยฑ โˆš((โˆ’12)^2 โˆ’ 4 ร— 9 ร— 20))/(2 ร— 9) = (12 ยฑ โˆš(144 โˆ’720))/18 = (12 ยฑ โˆš(โˆ’576))/18 = (12 ยฑ โˆš(โˆ’1 ร— 576))/18 = (12 ยฑ โˆš(โˆ’1 ) ร— โˆš( 576))/18 = (12 ยฑ ๐‘– ร— โˆš( 576))/18 = (12 ยฑ ๐‘– ร— โˆš( 242))/18 = (12 ยฑ ๐‘– ร— 24)/18 = (6(2 ยฑ ๐‘– ร— 4))/18 = ((2 ยฑ ๐‘– ร— 4))/3 = ((2 ยฑ ๐‘–4))/3 Thus, ๐‘ฅ = ((2 ยฑ ๐‘–4))/3

  1. Chapter 4 Class 11 Complex Numbers
  2. Serial order wise

About the Author

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