Misc 6 - Solve 3x2 - 4x + 20/3 = 0 - Chapter 5 NCERT - Miscellaneous

  1. Chapter 5 Class 11 Complex Numbers
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Misc 6 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

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