Question

How do you solve `x^2-2x+9=0`?

Answer 1

`x=1+i2sqrt2` or `x=1-i2sqrt2`

Explanation:

In the equation `x^2-2x+9=0`, the coefficients are rational but as the discriminant is `(-2)^2-4×1×9=4-36=-32` is negative, the roots of the equation are pair of complex conjugate numbers. Hence, we can use quadratic formula to get roots.

Quadratic formula gives the roots of `ax^2+bx+c=0` as `x=(-b+-sqrt(b^2-4ac))/(2a)`. Note that `b^2-4ac` is the discriminant.

Hence solution of `x^2-2x+9=0` is given by
`x=(-(-2)+-sqrt((-2)^2-4×1×9))/(2×1)` or

`x=(2+-sqrt(-32))/2` i.e.
`x=(2+i4sqrt2)/2` or `x=(2-i4sqrt2)/2` i.e.

`x=1+i2sqrt2` or `x=1-i2sqrt2`