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

1 Answer
George C.
Dec 21, 2016

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)

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