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

1 Answer
Oct 7, 2016

Answer:

#x=-5/6+-sqrt23/6i#

Explanation:

#3x^2+5x+4 color(white)(aaa)=color(white)(aa) 0#

#color(white)(aaaaaa^1a)-4color(white)(1aaa1)-4color(white)(aaa)# Subtract 4 from both sides

#3x^2+5xcolor(white)(aaaaaa)=-4#

#3(x^2+5/3xcolor(white)(aaaa))=-4color(white)(aaa)#Factor out the leading coefficient 3

Divide the coefficient of the #x# term by 2: #color(white)(aa)5/3-:2=5/6#

Square the result: #(5/6)^2=25/36# and add it to both sides:

#3(x^2+5/3x+25/36)=-4+3(25/36)color(white)(aa)#Note: add #3*25/36# on the#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)# right to account for the #color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#3 factored out on the left.

#3(x+5/6)(x+5/6)=-23/12color(white)(aaa)# Factor the left side and #color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaa)# simplify the right

Note the * #5/6# in the binomial #(x+5/6)#is the same as the coefficient of the x term divided by 2 above.*

#(x+5/6)(x+5/6)=(x+5/6)^2#...thus...

#3(x+5/6)^2=-23/12#

#3/3(x+5/6)^2=(-23/12)/3color(white)(aaa)#Divide both sides by 3

#(x+5/6)^2=-23/36#

#sqrt((x+5/6)^2)=sqrt(-23/36)color(white)(aaa)#Square root both sides

#x+5/6=+-sqrt23/6i#

Subtracting #5/6# from both side gives...

#x=-5/6+-sqrt23/6i#