1.

The quadratic equation ax^2-4ax+2a+1=0 has repeated root if a=?​

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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