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Ex 5.3, 9 - Class 11 CBSE - Solve x2 + x + 1/root2 - Ex 5.3

  1. Chapter 5 Class 11 Complex Numbers
  2. Serial order wise
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Ex5.3, 9 Solve the equation ๐‘ฅ2 + ๐‘ฅ + 1/โˆš2 = 0 ๐‘ฅ2 + ๐‘ฅ + 1/โˆš2 = 0 Multiplying both sides by โˆš2 โˆš2x2 + โˆš2x + โˆš2 ร— 1/โˆš2 = 0 ร— โˆš2 โˆš2x2 + โˆš2x + 1 = 0 The above equation is of the form ๐‘Ž๐‘ฅ2 + ๐‘๐‘ฅ + ๐‘ = 0 Where a = โˆš2 , b = โˆš2, and c = 1 ๐‘ฅ = (โˆ’๐‘ ยฑ โˆš( ๐‘2 โˆ’ 4๐‘Ž๐‘ ))/2๐‘Ž = (โˆ’โˆš2 ยฑ โˆš((โˆš2)^2 โˆ’ 4 ร— โˆš(2 )ร— 1))/(2 ร— โˆš2) = (โˆ’โˆš2 ยฑ โˆš((โˆš2)^2โˆ’ 4 ร—โˆš2 ร—1))/(2โˆš2) = (โˆ’โˆš2 ยฑ (โˆš(2 โˆ’ 4โˆš2) ))/(2โˆš2) = (โˆ’โˆš(2 )ยฑ โˆš(2 ) (โˆš(1 โˆ’ 2โˆš2) ))/(2โˆš2) = (โˆš(2 ) ( โˆ’1ยฑ โˆš(1 โˆ’ 2โˆš2) ))/(2โˆš2) = (โˆ’1 ยฑ โˆš(1 โˆ’ 2โˆš2) )/2 = (โˆ’1 ยฑโˆš(โˆ’(2โˆš2 โˆ’1) ))/2 = (โˆ’1 ยฑโˆš(โˆ’1 ร— (2โˆš2 โˆ’1) ))/2 = (โˆ’1 ยฑ โˆš(โˆ’1) ร— โˆš( (2โˆš2 โˆ’1) ))/2 = (โˆ’1 ยฑ ๐‘– ร— โˆš( (2โˆš2 โˆ’1) ) )/2 = (โˆ’1 ยฑ ๐‘–โˆš( (2โˆš2 โˆ’1) ) )/2 Thus, ๐‘ฅ = (โˆ’1 ยฑ ๐‘–โˆš( (2โˆš2 โˆ’1) ) )/2

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