How do you solve using the completing the square method #x^2+2x+4=0#?

1 Answer
Apr 5, 2017

Answer:

#x = -1+-sqrt(3)i#

Explanation:

Given:

#x^2+2x+4 = 0#

Completing the square we get:

#0 = x^2+2x+4#

#color(white)(0) = x^2+2x+1+3#

#color(white)(0) = (x+1)^2+3#

Note that for any Real value of #x#, we have:

#(x+1)^2 >= 0#

and so:

#(x+1)^2+3 >= 3#

To solve this quadratic we need to use Complex numbers.

The difference of squares identity can be written:

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

We use this with #a=(x+1)# and #b=sqrt(3)i# as follows:

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

#color(white)(0) = (x+1)^2-(sqrt(3)i)^2#

#color(white)(0) = ((x+1)-sqrt(3)i)((x+1)+sqrt(3)i)#

#color(white)(0) = (x+1-sqrt(3)i)(x+1+sqrt(3)i)#

Hence:

#x = -1+-sqrt(3)i#