How do you solve the quadratic with complex numbers given #1/4x^2-1/3x+1=0#?
1 Answer
Aug 13, 2016
Explanation:
Let
#0 = 36 f(x) = 9x^2-12x+36#
#= (3x-2)^2-4+36#
#= (3x-2)^2+32#
#= (3x-2)^2-(4sqrt(2)i)^2#
#= ((3x-2)-4sqrt(2)i))((3x-2)+4sqrt(2)i)#
#= (3x-(2+4sqrt(2)i))(3x-(2-4sqrt(2)i))#
Hence zeros:
#x = 1/3(2+-4sqrt(2)i)#
