How do you solve the equation #x^2+2x-6=0# by completing the square?

1 Answer
Dec 24, 2016

Answer:

#x = -1+-sqrt(7)#

Explanation:

The difference of squares identity can be written:

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

Use this with #a=(x+1)# and #b=sqrt(7)# as follows:

#0 = x^2+2x-6#

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

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

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

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

Hence:

#x = -1+-sqrt(7)#