How do you solve the equation #3x^2+5x+4=0# by completing the square?
1 Answer
Dec 21, 2016
Explanation:
Given:
#f(x) = 3x^2+5x+4#
I would first premultiply by
#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)#