Completing the Square: When to Use It and How

Quick answer: Completing the square rewrites a quadratic like x^2 + 6x + 5 into the form (x + p)^2 + q — here, (x + 3)^2 - 4. You halve the coefficient of x to get p, then subtract p^2 to cancel the extra bit you added. Reach for it whenever you need the turning point, an exact (surd) answer, or the vertex form — not when the quadratic factorises cleanly.

The first time my teacher completed the square on the board, I copied every line down and still had no idea where the "-9" came from. It looked like a magic trick. It isn't. It's one honest move repeated, and once it clicks you get the turning point, exact solutions and even the whole quadratic formula almost for free. Here's how I finally made it stick, plus the exact sign errors that cost me marks in my Year 10 mocks.

What completing the square actually means

Multiply out (x + p)^2 and you get x^2 + 2px + p^2. Notice the middle term is 2px. So any expression like x^2 + bx is nearly a perfect square — it's just missing the p^2 on the end, where 2p = b, which means p = b/2.

The whole method is: add that missing p^2 so the front becomes a perfect square, then subtract the same p^2 straight back so you haven't changed the value. That's it. Everything else is bookkeeping.

The method when a = 1, step by step

Take x^2 + 6x + 5. This is the classic AQA and Edexcel Higher-tier phrasing: "write x^2 + 6x + 5 in the form (x + a)^2 + b." CBSE Class 10 and Cambridge IGCSE 0580 use the same idea to solve by completing the square.

• Step 1 — halve the x-coefficient. b = 6, so p = 6/2 = 3. Your bracket is going to be (x + 3).
• Step 2 — write the squared bracket. (x + 3)^2. But this expands to x^2 + 6x + 9, so it has an extra +9 that wasn't in the original.
• Step 3 — subtract what you overshot. Take off that 9: (x + 3)^2 - 9.
• Step 4 — bring down the original constant. The +5 was always there, so add it: (x + 3)^2 - 9 + 5.
• Step 5 — tidy the numbers. -9 + 5 = -4, giving (x + 3)^2 - 4. Done.

To solve x^2 + 6x + 5 = 0 from here: (x + 3)^2 = 4, so x + 3 = ±2, giving x = -1 or x = -5. (Yes, this one also factorises — I'm using easy numbers so the method is clear.)

Reading the turning point straight off

This is the payoff. From (x + 3)^2 - 4, the vertex is at (-3, -4). A squared bracket can never be negative, so the smallest the whole thing can be is when (x + 3)^2 = 0 — that is, at x = -3, where the value drops to -4.

So the graph has a minimum at (-3, -4), and the line of symmetry is x = -3. Watch the sign: the bracket says (x + 3) but the turning point x-value is -3, not +3. That single flip catches people out constantly.

When a is bigger than 1

If there's a number in front of x^2, factorise it out of the first two terms first. Express 2x^2 - 8x + 5 in the form a(x + p)^2 + q — the exact Cambridge and WAScE Further Maths phrasing.

• Step 1 — factor the 2 out of the x-terms only. 2(x^2 - 4x) + 5. Leave the +5 outside.
• Step 2 — complete the square inside the bracket. Half of -4 is -2, so x^2 - 4x = (x - 2)^2 - 4.
• Step 3 — put it back in. 2[(x - 2)^2 - 4] + 5.
• Step 4 — multiply the 2 back through. 2(x - 2)^2 - 8 + 5. The -4 becomes -8 because it's inside the bracket the 2 multiplies — miss this and everything after is wrong.
• Step 5 — simplify. 2(x - 2)^2 - 3. Minimum at (2, -3).

To solve in exact form: 2(x - 2)^2 = 3, so (x - 2)^2 = 3/2, giving x = 2 ± √6/2. If the question says "give your answer in surd form," a calculator decimal loses you the accuracy mark — this is exactly when completing the square is the intended route.

Where the quadratic formula comes from

Complete the square on the general equation and the formula falls out. Doing this once made me stop fearing it.

• Start with ax^2 + bx + c = 0.
• Divide everything by a: x^2 + (b/a)x + c/a = 0.
• Move the constant across: x^2 + (b/a)x = -c/a.
• Complete the square on the left (half of b/a is b/2a): (x + b/2a)^2 - b^2/4a^2 = -c/a.
• Rearrange: (x + b/2a)^2 = (b^2 - 4ac) / 4a^2.
• Square-root both sides: x + b/2a = ±√(b^2 - 4ac) / 2a.
• Subtract b/2a: x = [-b ± √(b^2 - 4ac)] / 2a.

That's the quadratic formula. It's just completing the square done once, in general.

The sign errors that cost real marks

• Forgetting to subtract p^2. Writing (x + 3)^2 + 5 instead of (x + 3)^2 - 9 + 5. The bracket always overshoots — you must take it back off.
• Reading the vertex x with the wrong sign. (x + 3)^2 gives x = -3, not +3.
• Not multiplying the subtracted square back by a. With 2(x - 2)^2, the inner -4 must become -8. Leaving it as -4 is the most common a-greater-than-1 mistake.
• Rounding when b is odd. x^2 + 5x becomes (x + 5/2)^2 - 25/4. Keep the fractions — mark schemes want exact values, not 2.5 turned into a decimal that drifts.
• Sign of the bracket following b. x^2 - 4x gives (x - 2)^2 - 4; the inside sign matches b's sign.

When to actually use it

Use completing the square when the question:

• says "write in the form (x + a)^2 + b" or "a(x + p)^2 + q" — that's a direct instruction;
• asks for the turning point, vertex, minimum or maximum value, or line of symmetry;
• says "solve, giving your answer in surd form" or "exact form";
• asks you to prove a quadratic has no real roots, or to find the range of a function (Cambridge Additional Maths 0606, AP Precalculus vertex form).

Don't bother when it factorises in a couple of seconds — factorise instead. And if a decimal answer is fine, the quadratic formula is faster.

Test yourself

1. Write x^2 - 8x + 11 in the form (x + a)^2 + b, and state the turning point.
2. Solve x^2 + 4x - 6 = 0 by completing the square, giving your answer in surd form.
3. Express 3x^2 + 12x + 5 in the form a(x + p)^2 + q, and state the minimum value.

Quick answers:

1. (x - 4)^2 - 5, turning point (4, -5).
2. (x + 2)^2 - 10 = 0, so x = -2 ± √10.
3. 3(x + 2)^2 - 7, minimum value -7.

Want each line marked the way an examiner would? Paste your working into Math solver to check every step, or run a few more through Mark my answer to see where a method mark would slip. Do five and the whole method locks in.

FAQ

When should I complete the square instead of factorising? When the quadratic doesn't factorise with whole numbers, when you need the turning point, or when the paper demands an exact surd answer. If it factorises fast, just factorise.

How do I complete the square when b is odd? Halve it and keep the fraction. x^2 + 5x becomes (x + 5/2)^2 - 25/4. Don't round — the mark scheme expects the exact fraction, and rounding early loses accuracy marks.

How do you find the turning point from completed-square form? From (x + p)^2 + q, the vertex is at (-p, q). Flip the sign of the number inside the bracket for the x-coordinate; the q is the y-coordinate and the minimum (or maximum) value. Drill this with quick recall cards in Flashcards.

Why do we halve the coefficient of x? Because (x + p)^2 expands to x^2 + 2px. To match the bx in your quadratic you need 2p = b, so p = b/2. Halving isn't a rule to memorise — it's forced by the expansion. Still stuck on a specific line? Drop it into Math solver and watch it work through.

In short: completing the square turns a quadratic into (x + p)^2 + q by halving b and subtracting p^2. That one form hands you the turning point, exact surd solutions, and even the quadratic formula. Watch the vertex sign flip, and with a in front, multiply the subtracted square back through. Nail those and this becomes some of the most reliable marks on the paper.