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

1 Answer
Oct 22, 2016

Answer:

#x=-1+-sqrt7#

Explanation:

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

Move the constant to the right side by adding 6 to both sides.

#x^2+color(magenta)2xcolor(white)(aaaaaa)=6#

Divide the coefficient #color(magenta)2# of the middle term #color(magenta)(2)x# by #2#:

#color(magenta)2/2 =color(blue)1#

Square the #color(blue)1# and add the result to both sides.

#color(blue)1^2=color(red)1#

#x^2+ 2x +color(red)1=6+color(red)1#

Factor the left side and sum the right side. Notice the #color(blue)1# in the factored form is the same #color(blue)1# you obtained by dividing the coefficient of the middle term by #2#

#(x+color(blue)1)(x+color(blue)1)=7#

Express the left side as the square of the binomial.

#(x+color(blue)1)^2=7#

Square root both sides.

#sqrt((x+1)^2)=sqrt7#

#x+1=+-sqrt7#

Subtract 1 from each side.

#x=-1+-sqrt7#