Question

How do you solve the quadratic with complex numbers given `1/4x^2-1/3x+1=0`?

Answer 1

`x = 1/3(2+-4sqrt(2)i)`

Explanation:

Let `f(x) = 1/4x^2-1/3x+1`

`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)`