How do you solve #x^2+4x+1=0# by completing the square?
1 Answer
May 6, 2016
Explanation:
In addition to completing the square, use the difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
with
#0 = x^2 + 4x + 1#
#=(x+2)^2-4+1#
#=(x+2)^2-3#
#=(x+2)^2-(sqrt(3))^2#
#=((x+2)-sqrt(3))((x+2)+sqrt(3))#
#=(x+2-sqrt(3))(x+2+sqrt(3))#
Hence:
#x = -2+-sqrt(3)#
