How do you solve #m^2 + 2m + 6# by completing the square?

1 Answer
Mar 9, 2016

Also look at http://socratic.org/s/asBgAcAP

#(m+1)^2+5#

Explanation:

Given: #" "m^2+2m+6#

Using short cuts - By sight we have:

#(m+1)^2+5#

'~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 1")#

Add brackets#->(m^2+2m)+6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

Take the power outside the bracket#->(m+2m)^2+6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Remove the #m" from "2m ->(m+2)^2+6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

Multiply the constant inside the brackets
by #1/2" "->(m+[1/2xx2])^2+6 #

#= (m+1)^2+6#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")#

When squaring the brackets you get #m^2+2m+color(red)(1^2)+6#

The #color(red)(1^2)# is not in the original equation so it is an error. It has to be removed by including #color(magenta)(-1)#

# (m+1)^2color(magenta)(-1)+6#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Step 6")#

Answer is:#" "color(green)((m+1)^2+5)#