How do you solve #3x^2+10x+2=0#?

1 Answer
May 12, 2016

Answer:

#x = -5/3+-sqrt(19)/3#

Explanation:

One way involves completing the square. To avoid fractions a little, premultiply by #3# first, then use the difference of squares identity:

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

with #a=(3x+5)# and #b=sqrt(19)# as follows:

#0 = 3(3x^2+10x+2)#

#=9x^2+30x+6#

#=(3x)^2+2(3x)(5)+6#

#=(3x+5)^2-25+6#

#=(3x+5)^2-19#

#=(3x+5)-(sqrt(19))^2#

#=((3x+5)-sqrt(19))((3x+5)+sqrt(19))#

#=(3x+5-sqrt(19))(3x+5+sqrt(19))#

#=9(x+5/3-sqrt(19)/3)(x+5/3+sqrt(19)/3)#

Hence:

#x = -5/3+-sqrt(19)/3#