Question

How do you solve `x^2 + 6x + 10 = 0` by completing the square?

Answer 1

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

Explanation:

Completing the square is a method of getting as close as you can with one set of brackets and adding or subtracting the rest.

It ends up in a form like

`x^2 + ax + b = (x + c)^2 + d`

`c` is always half of `a`, so that when you expand out the brackets you have the right coefficients for `x^2` and `x`.

`a = 6 -> c = 3`
`(x + 3)^2 = x^2 + 6x + 9`

As you can see this is pretty close to the original quadratic. All you need to do is add `1` and it equates perfectly.

`x^2 + 6x + 10 = (x+3)^2 + 1`

To then solve, rearrange like so

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

Answer 2

There is no solution for the given equation for any `x` in the set of 'Real Numbers'.

`color(blue)(x=-3+-i)`

Explanation:

Consider the standard form of `y=ax^2+bx+c`

The by completing the square we have:

`y=a(x+b/(2a))^2 +c - [(b/(2))^2]`

In your case `a=1` so we have:

`y=(x+b/2)^2+c-[(b/2)^2]`

`=>y=(x+3)^2+10 -(3^2)`

`=> y= (x+3)^2+1`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`x_("vertex") = (-1)xx(+3) = -3`
`y_("vertex") =+1`

The coefficient of `x^2` = +1 so the graph is of shape type `uu`

Thus the vertex is a minimum and above the x-axis

Thus there is no solution for `x` at `y=0` where `x in RR`

However their will be a solution for `x in CC` (Complex numbers)

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

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

`=>+-(x+3)= i`

Updated: The `+-` is for the `x" and " 3`. Not for the `i`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Suppose we had "-(x+3)= i`

This becomes: `-x-3=i`

`=> -x=+3+i`
Multiply by (-1)

`color(blue)(=>+x=-3-i)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(brown)("Suppose we had "+(x+3)= i)`

This becomes: `x=-3+i`

`color(blue)(=>+x=-3+i)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Solution for `x` at `y=0` is

`color(magenta)(x=-3+-i)`