How do you solve `x^2 + 8x + 2 = 0` by completing the square?

We will use the difference of squares identity, which can be written:

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

with `a=(x+4)` and `b=sqrt(14)` as follows:

`0 = x^2+8x+2`

`=(x+4)^2-16+2`

`=(x+4)^2-(sqrt(14))^2`

`=((x+4)-sqrt(14))((x+4)+sqrt(14))`

`=(x+4-sqrt(14))((x+4+sqrt(14))`

Hence:

`x = -4+-sqrt(14)`

Completing the square is based on the consistency of the answers to the square of a binomial.

`(x - 3)^2 = x^2 - 6x + 9`
`(x - 5)^2 = x^2 - 10x + 25`
`(x + 6)^2 = x^2 + 12x + 36`
In all of the products above, `ax^2+ bx + c` we see the following:

`a = 1`
The first and last terms, `a and c` are perfect squares.'
There is a specific relationship between 'b' - the coefficient of the `x` term and 'c'. Half of b, squared equals c.

Knowing this, it is always possible to add in a missing value for `c` to have the square of a binomial, which can then be written as `(x+?)^2`

In `x^2 + 8x + 2 = 0`, 2 is obviously not the correct value of `c`.
It is therefore moved to the right hand side and the wanted value of `c` is added to BOTH sides of the equation.

`x^2 + 8x + color(red)(16)` = `-2 + color(red)(16) rArr [16 = (8÷2)^2]`
`(x + 4)^2 = 14 rArr` where 4 is either from `b÷2 or sqrt16`
`x + 4 = +-sqrt14 rArr` take the square root of both sides

This gives 2 possible answers for `x`.

`x = sqrt14 -4` OR `x = -sqrt14 -4`