Question

What is the solution set for `x^2 - 2x + 5 = 0`?

Answer 1

`x =(2+ sqrt( - 16))/ (2)`
`x =(2- sqrt( - 16))/ (2)`

Explanation:

Since ------- `(-2)^2` - `(4 xx1 xx 5)` < 0, x has imaginary roots

`x =(-b+- sqrt(b^2 - (4ac)))/ (2a)`
`x =(-(-2)+- sqrt((-2)^2 - (4xx1xx5)))/ (2xx1)`
`x =(-(-2)+- sqrt(4 - 20))/ (2)`
`x =(2+ sqrt( - 16))/ (2)`
`x =(2- sqrt( - 16))/ (2)`