Question

How do you solve `x^2 + 3x +6 =0` by completing the square?

Answer 1

`x = ± sqrt3 - 3`

Explanation:

By adding 9 to both sides of the equation to obtain:

`( x^2 + 3x + 9 )+ 6 = 9`

now `(x + 3 )^ 2 = 9-6 = 3`

`( x + 3 )^2 color(black)( " is a perfect square ")`

Taking the 'square root' of both sides :

`sqrt((x+ 3 )^2 )= sqrt3`

hence x + 3 = ± `sqrt3`

so x = ±`sqrt3 - 3`

Answer 2

there are two `x` values

`x_1=(-3+sqrt(15)i)/2`

`x_2=(-3-sqrt(15)i)/2`

Explanation:

Completing the square method:
Do this only when the numerical coefficient of `x^2` is `1`.
Start with the numerical coefficient of `x` which is the number `3`.
Divide this number by 2 then square the result. That is

`(3/2)^2=9/4`

Add `9/4` to both sides of the equation

`x^2+3x+9/4+6=0+9/4`

the first three terms now become one group which is a PST-Perfect Square Trinomial

`(x^2+3x+9/4)+6=9/4`

`(x+3/2)^2+6=9/4`

`(x+3/2)^2=9/4-6` after transposing the `6` to the right side

`(x+3/2)^2=(9-24)/4`

`sqrt((x+3/2)^2)=+-sqrt((9-24)/4)`

`x+3/2=+-sqrt((-15)/4)`

`x+3/2=+-sqrt(-15)/sqrt(4)`

`x+3/2=+-sqrt(-15)/2`

Finally, transpose the `3/2` to the right side of the equation

`x=-3/2+-sqrt(-15)/2`

take note: `sqrt(-15)=sqrt(15)*sqrt(-1)=sqrt(15)i`

therefore

`x=-3/2+-(sqrt(15)i)/2`

there are two `x` values

`x_1=(-3+sqrt(15)i)/2`

`x_2=(-3-sqrt(15)i)/2`