How do you simplify #x^2+x+2=0# ?

1 Answer
Feb 1, 2017

Answer:

This will only "simplify" into linear factors with the help of Complex coefficients as:

#(x+1/2-sqrt(7)/2i)(x+1/2+sqrt(7)/2i) = 0#

Explanation:

Given:

#x^2+x+2 = 0#

Note that this is in the form:

#ax^2+bx+c = 0#

with #a=1#, #b=1# and #c=2#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = color(blue)(1)^2-4(color(blue)(1))(color(blue)(2)) = 1-8 = -7#

Since #Delta < 0# this quadratic has no simpler factors with Real coefficients.

We can simplify it with Complex coefficients by completing the square and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with #A=(x+1/2)# and #B=sqrt(7)/2i# as follows:

#0 = x^2+x+2#

#color(white)(0) = x^2+2(x)(1/2)+(1/2)^2+7/4#

#color(white)(0) = (x+1/2)^2-(sqrt(7)/2i)^2#

#color(white)(0) = ((x+1/2)-sqrt(7)/2i)((x+1/2)+sqrt(7)/2i)#

#color(white)(0) = (x+1/2-sqrt(7)/2i)(x+1/2+sqrt(7)/2i)#

Hence:

#x = -1/2+-sqrt(7)/2i#