How do you solve #k^2+ 8k + 12 = 0# by completing the square?

1 Answer
Shwetank Mauria
May 27, 2016

#k=-6# or #k=-2#

Explanation:

To solve #k^2+8k+12=0#, by completing the square method

remember #(k+a)^2=k^2+2ka+a^2#

Now here we have #k^2+8k# in place of #k^2+2ka#, hence #2a=8# or #a=4# and we should have #a^2# or #4^2=16# in addition to #k^2+8k# to make it a square.

Hence we can write #k^2+8k+12=0# as

#k^2+8k+16-16+12=0#

or #(k+4)^2-4=0#

or #(k+4)^2-2^2=0#

As now this is in the form #a^2-b^2# which can be factorized as #(a+b)(a-b)#, we will have

#(k+4+2)(k+4-2)=0# i.e, #(k+6)(k+2)=0#

Hence either #k+6=0# or #k+2=0# or

#k=-6# or #k=-2#

Related questions
See all questions in Completing the Square
Impact of this question
3927 views around the world
You can reuse this answer
Creative Commons License