Question

How do you solve the equation `3x^2+5x+4=0` by completing the square?

Answer 1

`x = 1/6(-5+-sqrt(23)i)`

Explanation:

Given:

`f(x) = 3x^2+5x+4`

I would first premultiply by `3*2^2 = 12` to avoid having to deal with fractions very much...

`0 = 12f(x)`

`color(white)(0) = 12(3x^2+5x+4)`

`color(white)(0) = 36x^2+60x+48`

`color(white)(0) = 36x^2+60x+25+23`

`color(white)(0) = (6x+5)^2+23`

`color(white)(0) = (6x+5)^2-(sqrt(23)i)^2`

`color(white)(0) = ((6x+5)-sqrt(23)i)((6x+5)+sqrt(23)i)`

`color(white)(0) = (6x+5-sqrt(23)i)(6x+5+sqrt(23)i)`

Hence:

`x = 1/6(-5+-sqrt(23)i)`