How do you solve #2x ^ { 2} + 12x + 9= 0#?

1 Answer
Feb 2, 2017

Answer:

#x = -3+-(3sqrt(2))/2#

Explanation:

Complete the square then use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=2x+6# and #b=3sqrt(2)# as follows:

#0 = 2(2x^2+12x+9)#

#color(white)(0) = 4x^2+24x+18#

#color(white)(0) = (2x)^2+2(2x)(6)+(6)^2-18#

#color(white)(0) = (2x+6)^2-(3sqrt(2))^2#

#color(white)(0) = ((2x+6)-3sqrt(2))((2x+6)+3sqrt(2))#

#color(white)(0) = (2x+6-3sqrt(2))(2x+6+3sqrt(2))#

Hence:

#x = 1/2(-6+-3sqrt(2)) = -3+-(3sqrt(2))/2#