Question

How do you solve using the completing the square method `x^2+2x+4=0`?

Answer 1

`x = -1+-sqrt(3)i`

Explanation:

Given:

`x^2+2x+4 = 0`

Completing the square we get:

`0 = x^2+2x+4`

`color(white)(0) = x^2+2x+1+3`

`color(white)(0) = (x+1)^2+3`

Note that for any Real value of `x`, we have:

`(x+1)^2 >= 0`

and so:

`(x+1)^2+3 >= 3`

To solve this quadratic we need to use Complex numbers.

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this with `a=(x+1)` and `b=sqrt(3)i` as follows:

`0 = (x+1)^2+3`

`color(white)(0) = (x+1)^2-(sqrt(3)i)^2`

`color(white)(0) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)`

`color(white)(0) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)`

Hence:

`x = -1+-sqrt(3)i`