You see a quadratic on your paper: ax² + bx + c = 0. You have three tools in your bag — factoring, completing the square, and the quadratic formula. All three give the same answer when the answer exists. So which one do you pick?

Factoring is usually the fastest when it works. The quadratic formula always works, no matter how ugly the numbers get. Completing the square sits in the middle — instructive, but rarely the fastest in an exam. The skill is not knowing all three; it's knowing, in five seconds, which one to reach for.

The fast test — the discriminant

There is a single number that decides this for you. It is the discriminant:

D = b² - 4ac

You compute D first. Then you read off what kind of roots the quadratic has:

That is the whole rule. The next thirty seconds of your life depend on which bucket D falls into.

Worked example — factoring wins (D is a perfect square)

Take x² - 5x + 6 = 0.

Compute D: 25 - 24 = 1. Perfect square. Roots are rational.

Try factoring. You want two numbers that multiply to 6 and add to -5. That is -2 and -3.

(x - 2)(x - 3) = 0, so x = 2 or x = 3.

Total time: about 15 seconds. The formula would have given the same answer in 40 seconds with more arithmetic and more chances to slip.

Worked example — formula wins (D not a perfect square)

Take 2x² - 5x + 1 = 0.

Compute D: 25 - 8 = 17. Not a perfect square.

If you try to factor with integers, you will fail — there are no integer pairs that work. Every second you spend hunting is wasted. Just use the formula:

x = (5 ± √17) / 4.

Done in 30 seconds. The "factoring attempt first, then formula" student burns a minute on a dead end before even starting the right method.

Worked example — D < 0 (complex roots)

Take x² + x + 1 = 0.

Compute D: 1 - 4 = -3. Negative.

Over the real numbers, there are no roots — and no real factorisation either. If your syllabus accepts complex roots, the formula gives:

x = (-1 ± i√3) / 2.

If your syllabus is real numbers only, you write "no real roots" and move on. Either way, factoring over reals is impossible here, so do not even start.

The decision tree

This is the algorithm to run on every quadratic, every time:

  1. Compute D = b² - 4ac.
  2. If D ≥ 0 and D is a perfect square → try factoring.
  3. If D < 0 → use the formula (complex roots), or state "no real roots".
  4. Otherwise (D > 0, not a perfect square) → use the formula.

You can run this in your head in under five seconds once you have practised it.

Common perfect squares to recognise

You should know these on sight, no calculation:

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400.

Memorise up to 400 (that is up to 20²). When D pops out as one of these, factoring is on. When it pops out as 17 or 23 or 50, factoring is off.

When factoring via split-the-middle is still appropriate

For quadratics like ax² + bx + c where the leading coefficient is not 1, the trick is splitting the middle term.

You want two integers whose product is ac and whose sum is b. If those integers exist, you can split bx into two terms and factor by grouping.

If no such integers exist, factoring will not give you clean roots. Switch to the formula.

The discriminant test tells you in advance whether those integers exist. If D is a perfect square, the integers exist. If not, they do not. The two tests are equivalent.

Factoring by inspection — when the structure is obvious

Some quadratics announce themselves. Do not even compute D for these:

If you can see the structure in one glance, factor by inspection. The discriminant test is for when the structure is not obvious.

Completing the square as a middle ground

Take x² + 6x + 5 = 0. You can complete the square:

(x + 3)² - 9 + 5 = 0, so (x + 3)² = 4, so x + 3 = ±2, so x = 1 or x = -5.

This works always, just like the formula. In fact, the quadratic formula is what you get when you complete the square on a general ax² + bx + c. They are the same technique wearing different clothes.

For simple problems, completing the square is rarely the fastest path. Use it when a question explicitly asks you to (sometimes JEE problems ask you to write a quadratic in vertex form), or when you need to find the maximum/minimum of a parabola.

Worked drill — pick the method

Run the discriminant test on each. Do not solve yet — just decide.

Six problems, six clean decisions. No wasted hunting.

Exam-time strategy

In a JEE or board exam, every second matters. Build this into a reflex:

  1. Write D = b² - 4ac in the margin.
  2. Compute it.
  3. If small and a perfect square → factor.
  4. Otherwise → formula.

You will save 30-60 seconds on every quadratic. Across a paper that is several minutes — enough time for one extra problem.

Common confusions

"The quadratic formula always works, so I'll just use it every time." It works, but it is slower than factoring when factoring is available. Speed matters in exams.

"Factoring always works if I try hard enough." It does not. When roots are irrational or complex, no amount of hunting for integer pairs will help. The discriminant tells you in advance.

"Completing the square is a separate technique from the formula." It is not — the formula is derived from completing the square on the general quadratic. They are the same idea. Pick whichever is cleaner for the problem at hand.

What to check when the quadratic isn't monic

Non-monic just means the leading coefficient is not 1. The discriminant rule still applies, but the splitting method needs care.

Take 3x² + 7x + 2 = 0.

Compute D: 49 - 24 = 25. Perfect square. Factoring is on.

Split the middle: find two numbers with product ac = 3 × 2 = 6 and sum b = 7. That is 6 and 1.

3x² + 6x + x + 2 = 3x(x + 2) + 1(x + 2) = (3x + 1)(x + 2) = 0.

Roots: x = -1/3 and x = -2.

Notice you used ac, not just c. That is the only twist for non-monic quadratics. The discriminant rule itself is unchanged.

Closing

Discriminant first. Perfect square → factor. Otherwise → formula. Negative → no real roots (or formula for complex).

You do not pick a method by habit. You pick it by reading the problem. Five seconds with the discriminant tells you which tool fits, and you save the rest of your time for the question's actual idea — not the arithmetic of getting to the roots.