How do you solve #x^2 - 4x = -3# by completing the square?

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Answer 1 · 2017-10-11T00:20:00

#x=3# OR #x=1#

We want to add a number on both sides of the equation so that the left side becomes a perfect square:

To do that, we look at #b# (the number next to #x#) and see that it is #-4#

If you add #(b/2)^2# to both sides, the equation on the left will be a perfect square or:

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

So:

#x^2-4xcolor(red)(+4)=-3color(red)(+4)#

#(x-2)^2=1#

#sqrt((x-2)^2)=sqrt(1)#

#x-2=1# OR #x-2=-1#

#x=3# OR #x=1#