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

1 Answer
Apr 1, 2015

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)