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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