Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you determine what solving method to use, for example what is the reason you solve by the quadratic formula, or by factoring, or by completing the square, or by taking a square root.

share|cite|improve this question

To find $x$, each approach can be considered more specific to a certain form of problem, or less general, than the next:

Taking the square root is only really useful when you have $ax^2 = n$.

Completing the square works well when you have $(x+c)^2 = n$, or a statement that can easily be arranged to form something similar. All quadratics can be rearranged so you can solve them by completing the square, but the effort in rearranging often makes it easier to just use the quadratic formula:

Using the quadratic formula is the most general approach, though it is also the most laborious. It works for all expressions of form $ax^2+bx+c=n$.

So you use the quadratic formula only when you can't conveniently do it by taking a square root or completing a square.

share|cite|improve this answer
Every quadratic can be done by completing the square -- it is the equivalent of the quadratic formula I believe. – Eric Thoma Dec 19 '13 at 1:51
@Newb, you contradict yourself in the second paragraph. You claim that completing the square only "works" in certain situations and then go on to say it can always be done. I would recommend rephrasing the first sentence since it seems that perhaps wasn't your intended meaning. – Spencer Dec 19 '13 at 2:39
@Spencer replaced "only works" w/ "works well". – Newb Dec 19 '13 at 3:29

Sounds like you are trying to solve a quadratic equation. If you have something of the form $x^2$ = a, take the square root of both sides. This would also work well if you have something like $(x-1)^2$ = a.

If that is not your canse and you can see quickly how to factor it, do so.

If you cannot see quickly how to factor it, use the quadratic formula. "Quickly" means you can factor it faster than you can compute the quadratic formula.

I personally would never bother to complete the square. The quadratic formula is already the generalization of completing the square, so you don't have to consider how to go about the completion -- you just plug the formula in.

share|cite|improve this answer

I think the only consideration here is how hard the problem is. If it's already set up as $x^2=a$, take the square root of both sides (don't forget the negative root too of course). If you can factor it easily, do so. If factors don't come easily (may not be real or rational), completing the square can work well, especially if the middle term is even. If all of your coefficients are decimals (I remember having to solve quadratics in chemistry class...), then your best bet is quadratic formula. And a calculator.

share|cite|improve this answer

When solving a quadratic equation, all the methods that you described work. However, sometimes some of them are easier and faster than others.

The quadratic formula acts as a general approach to solving a quadratic and can be useful if you are unsure of how to solve an equation. However, it will often not be the simplest or easiest way to solve the problem.

When the roots of a quadratic equation are integers, factoring tends to be a quick method of solving the equations because it does not taking as long as going through the mechanics of the quadratic formula. Completing the square can be useful to solve more complex equations by breaking them down into a more simple factoring problem. Taking the square root of both sides of the equation will generally only be useful in solving equations of the form $ax^2 = n$.

For more help with solving quadratic equations, visit:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.