How do you factor and solve #x^2+3x+1=0#?

1 Answer
Jun 5, 2016

Answer:

#x = -3/2+sqrt(5)/2# or #x = -3/2-sqrt(5)/2#

Explanation:

Use the difference of squares identity:

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

with #a=(x+3/2)# and #b=sqrt(5)/2# as follows:

#0 = x^2+3x+1#

#= (x+3/2)^2-3^2/2^2+1#

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

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

#= ((x+3/2)-sqrt(5)/2)((x+3/2)+sqrt(5)/2)#

#= (x+3/2-sqrt(5)/2)(x+3/2+sqrt(5)/2)#

Hence:

#x = -3/2+sqrt(5)/2# or #x = -3/2-sqrt(5)/2#