How is quadratic formula different than completing the square?

1 Answer
Oct 25, 2015


They are closely related, but you might choose to use one over the other for a specific purpose, e.g. finding the vertex of a parabola rather than its #x# intercepts.


Given #ax^2+bx+c = 0#, we can rewrite by completing the square as:

#a(x+b/(2a))^2 + (c - b^2/(4a)) = 0#

Then subtract #(c-b^2/(4a))# from both sides to get:

#a(x+b/(2a))^2 = b^2/(4a) - c = (b^2-4ac)/(4a)#

Divide both ends by #a# to get:

#(x+b/(2a))^2 = (b^2-4ac)/(4a^2) = (b^2-4ac)/(2a)^2#

Taking square roots we get:

#x+b/(2a) = +-sqrt(b^2-4ac)/(2a)#

Finally subtract #b/(2a)# from both sides to get:

#x = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b+-sqrt(b^2-4ac))/(2a)#

If we just want to get straight to the roots then the quadratic formula is probably the best choice.

If we want to get to a vertex form to find where the vertex of the parabola is, then completing the square is more appropriate.

Of course, one advantage of completing the square over the quadratic formula is less 'magic'. It is much clearer how you are getting to the solution and it does not entail much more work. However, now you can derive the quadratic formula yourself you have every reason to use it if it's more convenient.