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

1 Answer
Alan P.
Dec 20, 2016

There are no Real solutions to this equation
but see below for method of completing the square
and for Complex value solutions.

Explanation:

Remember that a squared binomial takes the form
#color(white)("XXX")(x=color(red)a)^2=x^2+2color(red)ax+color(red)a^2#

So if #x^2+2x# are the first 2 terms of an expanded squared binomial:
#color(white)("XXX")color(red)a=1#
and the third term must be #color(red)(a)^2=1^2=1#
and the complete (expanded) square needs to be #color(blue)(x^2+2x+1)#

Rewriting the given equation: #x^2+2x+6=0#
to include this complete (expanded) square:
#color(white)("XXX")color(blue)(x^2+2x+1)+5=0#

or
#color(white)("XXX")color(blue)(x^2+2x+1) = -5#

or
#color(white)("XXX")color(blue)(""(x+1)^2)=-5#

We could note at this point that since any Real number squared is greater than or equal to #0# there can be no Real solutions.

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If you have advanced to Complex numbers:
#color(white)("XXX")color(blue)(x+1)=+-sqrt(-5) =+-isqrt(5)#
and
#color(white)("XXX")x=-1+-isqrt(5)#

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