The quadratic formula always works. That is exactly why you should not always reach for it.
A method that never fails feels like the safe choice, and on a timed test the safe choice is often the slow one. OpenStax puts it plainly in its chapter on the quadratic formula: you can solve any quadratic this way, "but that is not always the easiest method." Factoring, when it is available, can finish the job in the time it takes to write the formula down. Completing the square looks like the worst of the three until you meet the question it was built for, at which point the other two cannot help you at all.
There is a way to know which one to use before you commit, and it costs you one subtraction.
Check the discriminant first
Take the part of the quadratic formula that lives under the square root:
b² − 4ac
That is the discriminant, and it is cheap to compute. It tells you two separate things, and most students are only taught the first one.
The famous part. Positive means two real solutions. Zero means one repeated solution. Negative means no real solutions, so if the question asked for real roots you can stop right there and write that down.
The part that decides your method. A perfect square down there means the quadratic factors. Anything else means it does not, and no amount of staring will change that. So every second you spend hunting for factor pairs on a discriminant that is not a perfect square is a second thrown away for nothing.
LibreTexts states the second half directly in its guidelines for solving quadratics: "If the discriminant is a perfect square, then we could have factored the original equation." OpenStax's College Algebra table says the same thing from the other side, splitting a positive discriminant into two rational solutions when it is a perfect square and two irrational ones when it is not.
One condition neither book spells out, because both are working with whole numbers throughout: this test behaves as advertised when a, b and c are integers. That covers essentially every quadratic you will be handed in a course or on an exam, but it is the reason the rule works, so it is worth knowing.
That second fact converts factoring from a guessing game into a decision. You are no longer wondering whether this thing factors. You know.
The same equation, three ways
Take x² − 7x + 12 = 0. The discriminant is 49 − 48 = 1, which is a perfect square, so this one factors.
- Factoring: (x − 3)(x − 4) = 0, so x = 3 or x = 4.
- Formula: x = [7 ± √1]/2 = (7 ± 1)/2, so x = 4 or x = 3.
- Completing the square: (x − 7/2)² = 49/4 − 12 = 1/4, so x − 7/2 = ±1/2, so x = 4 or x = 3.
Three routes, one answer, wildly different costs. Factoring took a few seconds. The formula took a bit longer but never risked anything. Completing the square dragged you through quarters and halves to reach a result you could have seen by inspection.
That last one is the important lesson. Completing the square is not a bad method. It was simply the wrong tool for this particular question, which is a different criticism.
When factoring is not available
Now for one that behaves differently: x² − 4x − 3 = 0, with a discriminant of 16 + 12 = 28.
Twenty-eight is not a perfect square. So this one does not factor over the rationals, and the useful part is that you knew that before writing down a single line of working.
The formula gives x = [4 ± √28]/2, which simplifies to 2 ± √7.
But watch what completing the square does here. The leading coefficient is 1 and the middle coefficient is even, which is the situation this method was made for:
- x² − 4x = 3
- (x − 2)² − 4 = 3
- (x − 2)² = 7
- x = 2 ± √7
No fractions anywhere, and arguably faster than the formula. The method that embarrassed itself on the previous problem just won this one. Nothing about the method changed. The numbers changed.
The case that catches everyone
Here is where the simple rule "if it factors, factor it" falls apart. Take 6x² + 11x − 35 = 0.
The discriminant is 121 + 840 = 961, and 961 is 31 squared. Perfect square, so it factors. The factorisation is (3x − 5)(2x + 7).
Now be honest about how long finding that would have taken you. With a leading coefficient of 6 and a constant of −35, you are searching through pairs from two different factor lists with signs to juggle. It is findable. It is not fast.
The formula, meanwhile, does it mechanically: x = (−11 ± 31)/12, giving 5/3 and −7/2. Same answers, no searching.
So there is a gap between "it factors" and "factoring is the fastest route here". Confusing those two is where students bleed time. The rule that actually works has a clock built into it. Perfect-square discriminant, and you can see the factorisation inside about ten seconds? Factor it. Ten seconds gone and you are still listing pairs? You have already spent longer than the formula would have taken you. Switch.
That clock is not something I invented. Paul's Online Math Notes, in its summary of solving quadratics, gives the same advice in words: "you shouldn't spend a lot of time trying to determine if the quadratic equation factors. Look at the equation and if you can quickly determine that it factors then go with that. If you can't quickly determine that it factors then don't worry about it."
Two textbooks, two different orders
Worth knowing that the standard references do not agree on how to start, because it explains why students get contradictory advice.
OpenStax and Paul's both say: try factoring first, and give up quickly if it does not come. LibreTexts says the opposite, that you should compute the discriminant first and let it tell you which method to use.
Both are defensible, and which one suits you depends on something personal. If you factor quickly by eye, the try-it-first order is faster, because on the quadratics that factor easily you will be finished before you would have computed anything. If factoring is not automatic for you, the discriminant-first order is better, because it stops you sinking two minutes into a hunt that was never going to succeed.
The ten-second clock is just a way of getting both. Glance for the factors the way OpenStax suggests, and if nothing appears, fall back on the discriminant to tell you whether to keep looking or move on.
The question only one method answers
Everything above treats these three as competitors for a single job. They are not, and this is the part that makes completing the square worth learning properly rather than tolerating.
Paul's notes are blunt that as a way of solving equations it "should always be your last choice", and then immediately add the caveat almost everyone skips: "This doesn't mean however that it isn't an important method. We will see the completing the square process arise in several sections in later chapters. Interestingly enough when we do see this process in later sections we won't be solving equations!" It comes back for vertex form, for conic sections, for certain integrals, and it is what the quadratic formula was derived from in the first place.
Ask for the vertex of y = x² − 6x + 5, and the formula has nothing useful to say. It finds where a parabola crosses the axis, not where it turns.
Completing the square rewrites the whole thing as (x − 3)² − 4, and the vertex is sitting there in the open at (3, −4). Minimum value −4, occurring at x = 3, read straight off the page.
Any question about a maximum, a minimum, a turning point, or the range of a quadratic is a completing-the-square question. Profit models, projectile heights, and optimisation problems are all this question wearing a costume. Roots are not what is being asked for, so root-finding tools do not apply. If converting to vertex form is the step that keeps slipping, drill that on its own rather than inside a longer word problem.
The rule, compressed
- Asked for a vertex, maximum, minimum, or range? Complete the square. Skip the rest of this list.
- Otherwise compute b² − 4ac first.
- Negative, and the question wants real roots? You are already done.
- Perfect square, and you see the factors quickly? Factor.
- Perfect square, but ten seconds have gone and you are still hunting? Formula.
- Not a perfect square, leading coefficient 1, middle coefficient even? Completing the square is clean here.
- Anything else? Formula, without guilt.
Notice how much of this is decided before you write a single line of working. That is the point. Choosing the method is a separate skill from executing it, and it is the one that saves time under pressure, in the same way that reading a function's structure before differentiating saves more time than getting faster at the rules themselves.
Practise it the cheap way. Take twenty quadratics, compute nothing but the discriminant for each, and write down which method you would use. No solving. Ten minutes of that will do more for your speed than another page of grinding, and when you do want the full working for one of them, a step-by-step run through the roots and discriminant will show you whether your instinct about the method was right.