How do you solve by completing the square #x^2- 4x-11=0#?

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Answer 1 · 2015-04-01T17:38:53

#x=2-sqrt(15), 2+sqrt(15)#

We should reform this equation by a full square. So we have:
#x^2-4x+c=0#

I don't want to go deeper in this, lets conclude this part with #c=4#

So:

#(x-2)^2=x^2 -4x +4#

Now, lets put #(x-2)^2# in the given equation.

#Given:# #1)# # x^2 -4x-11=0#
#(x-2)^2+n=x^2-4x-11#
#n=-15#

So:

#(x-2)^2-15=x^2-4x-11#
#(x-2)^2-15=0#
#sqrt((x-2)^2) = 15#
#abs(x-2)=sqrt(15)#
#x-2=-sqrt(15), sqrt(15)#
#x=2-sqrt(15),2+sqrt(15)#