How do you complete the square for #x^2-4x#?

1 Answer
Nov 26, 2017

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

Explanation:

To complete the square, we need to find some binomial, #a+b#, such that #(a+b)^2# only differs by a constant to the expression we want.

Since #(a+b)^2=a^2+2ab+b^2#, #a# must be equal to #x#. Now we only need to find #b#:
#(x+b)^2=x^2+2bx+b^2#.

Since we want the middle term to be #-4x#, we know that #2b=-4#. Dividing both sides by two, we get:
#b=-4/2=-2#.

Now we have a square that will get us the first two terms, but we also will have an extra bit at the end, which isn't right:
#(x-2)^2=x^2-4x+4#

Since we want no constant at all, we just have to subtract by #4#:
#(x-2)^2-4#

We can expand it just to check:
#(x-2)^2-4=x^2-4xcancel(+4-4)=x^2-4x#