Question

How do you solve the quadratic `8x^2-4x+5=0` with complex numbers?

Answer 1

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

Explanation:

`8x^2-4x+5=0` is a quadratic equation with the form `ax^2+bx+c`, where `a=8`, `b=-4`, and `c=5`. This equation can be solved using the quadratic formula.

Quadratic Formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Plug in the given values.

`x=(-(-4)+-sqrt((-4)^2-4*8*5))/(2*8)`

Simplify.

`x=(4+-sqrt(16-160))/16`

`x=(4+-sqrt(-144))/16`

`x=(4+-12i)/16`

`x=(4+12i)/16,` `(4-12i)/16`

Reduce.

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