How do you solve #x^2 + 8x + 2 = 0# by completing the square?

We will use the difference of squares identity, which can be written:

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

with #a=(x+4)# and #b=sqrt(14)# as follows:

#0 = x^2+8x+2#

#=(x+4)^2-16+2#

#=(x+4)^2-(sqrt(14))^2#

#=((x+4)-sqrt(14))((x+4)+sqrt(14))#

#=(x+4-sqrt(14))((x+4+sqrt(14))#

Hence:

#x = -4+-sqrt(14)#

Completing the square is based on the consistency of the answers to the square of a binomial.

#(x - 3)^2 = x^2 - 6x + 9#
#(x - 5)^2 = x^2 - 10x + 25#
#(x + 6)^2 = x^2 + 12x + 36#
In all of the products above, #ax^2+ bx + c# we see the following:

#a = 1#
The first and last terms, #a and c# are perfect squares.'
There is a specific relationship between 'b' - the coefficient of the #x# term and 'c'. Half of b, squared equals c.

Knowing this, it is always possible to add in a missing value for #c# to have the square of a binomial, which can then be written as #(x+?)^2#

In #x^2 + 8x + 2 = 0#, 2 is obviously not the correct value of #c#.
It is therefore moved to the right hand side and the wanted value of #c# is added to BOTH sides of the equation.

#x^2 + 8x + color(red)(16)# = #-2 + color(red)(16) rArr [16 = (8÷2)^2]#
#(x + 4)^2 = 14 rArr# where 4 is either from #b÷2 or sqrt16#
#x + 4 = +-sqrt14 rArr# take the square root of both sides

This gives 2 possible answers for #x#.

#x = sqrt14 -4# OR #x = -sqrt14 -4#