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The quadratic equation ax^2-4ax+2a+1=0 has repeated root if a=? |
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Answer» Step-by-step explanation: Ax square - 4ax + 2a+1 = 0 by applying B square - 4AC = 0 (-4a) square - 4.a.(2a+1) =0 16a square - 4a ( 2a+1) =0 16a.a - 8a.a -4a =0 by taking 4a common 4a (2a-1) =0 (2a-1) = 0 a = 1/2 or a = 0 hope it HELPS.. |
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