Last updated at Dec. 16, 2024 by Teachoo
Question 2 Solve the equation 3๐ฅ2 โ 4๐ฅ + 20/3 = 0 3๐ฅ2 โ 4๐ฅ + 20/3 = 0 Multiplying both sides by 3 3 ร (3๐ฅ2 โ 4๐ฅ "+ " 20/3) = 3 ร 0 3 ร 3x2 โ 3 ร 4x + 3 ร 20/3 = 0 9๐ฅ2 โ 12๐ฅ + 20 = 0 The above equation is of the form ๐๐ฅ2 + ๐๐ฅ + ๐ = 0 Where a = 9, b = โ12, and c = 20 x = (โ๐ยฑโ( ๐^2 โ4๐๐ ))/2๐ = (โ(โ12) ยฑ โ((โ12)^2 โ 4 ร 9 ร 20))/(2 ร 9) = (12 ยฑ โ(144 โ720))/18 = (12 ยฑ โ(โ576))/18 = (12 ยฑ โ(โ1 ร 576))/18 = (12 ยฑ โ(โ1 ) ร โ( 576))/18 = (12 ยฑ ๐ ร โ( 576))/18 = (12 ยฑ ๐ ร โ( 242))/18 = (12 ยฑ ๐ ร 24)/18 = (6(2 ยฑ ๐ ร 4))/18 = ((2 ยฑ ๐ ร 4))/3 = ((2 ยฑ ๐4))/3 Thus, ๐ฅ = ((2 ยฑ ๐4))/3
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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