IB Maths AA SL Topic 2 — Functions Paper 1 & 2 🎯 Skill ~4 min practice

AA SL Completing the Square skills

Rewrite any quadratic in (x − h)² + k form. The trick is one move: half the coefficient of x, square it, then balance. The result hands you the turning point on a plate and turns “exact form” questions into one-liners.

The Method

x² + bx + c → (x + b2)² − (b2)² + c half the b · square it · balance the same amount

Half the coefficient of x — that’s b/2. Use it inside the bracket as (x + b/2)².
Subtract (b/2)² outside the bracket — this cancels the extra term you just created.
Add the constant from the original quadratic, then simplify.

Half the b — the only move that matters

Example: complete the square on x² + 8x + 3

Step 1 — half b 8 ÷ 2 = 4

→

Step 2 — square it 4² = 16

→

Step 3 — balance (x + 4)² − 16 + 3

Final answer: (x + 4)² − 13 · turning point at (−4, −13)

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Why this form is so useful

Turning point shows up directly: (x − h)² + k has minimum/maximum at (h, k).
Solving in exact form takes one square-root and a rearrange — much cleaner than the formula.
Sketching the parabola is easy: vertex first, then the shape.

Worked examples

WE 1 EASY

Write x² + 6x + 1 in the form (x − h)² + k.

step 1 — half the b half of 6 is 3 → bracket: (x + 3)²step 2 — square and subtract 3² = 9 → subtract 9 outside x² + 6x = (x + 3)² − 9step 3 — add the original constant (x + 3)² − 9 + 1 = (x + 3)² − 8(x + 3)² − 8 turning point: (−3, −8) — read it straight off the form!

WE 2 MEDIUM

Write x² − 5x + 2 in completed-square form, then state the turning point.

step 1 — half the b half of −5 is −5/2 bracket: (x − 5/2)²step 2 — square and subtract (−5/2)² = 25/4 → subtract x² − 5x = (x − 5/2)² − 25/4step 3 — add the original constant (x − 5/2)² − 25/4 + 2 = (x − 5/2)² − 25/4 + 8/4 = (x − 5/2)² − 17/4(x − 5/2)² − 17/4 · min at (5/2, −17/4) odd b → fractions. just keep them tidy and convert to common denominators!

WE 3 HARD

Write 2x² − 12x + 7 in the form a(x − h)² + k.

step 1 — factor out 2 from x terms only 2(x² − 6x) + 7step 2 — complete the square inside half of −6 is −3, square is 9 2((x − 3)² − 9) + 7step 3 — distribute the 2 2(x − 3)² − 18 + 7 = 2(x − 3)² − 112(x − 3)² − 11 · min at (3, −11) when a ≠ 1, factor a out of the x terms FIRST — leave the constant alone!

Practice questions

Try each one yourself first, then click the question to reveal the worked answer. Always half the b first — that’s the move.

Q1 EASY Complete the square: x² + 4x + 7 Show answer ▼Hide answer ▲

half of 4 is 2, 2² = 4 (x + 2)² − 4 + 7 (x + 2)² + 3

Q2 EASY Complete the square: x² − 10x + 30 Show answer ▼Hide answer ▲

half of −10 is −5, (−5)² = 25 (x − 5)² − 25 + 30 (x − 5)² + 5

Q3 MEDIUM Complete the square and state the turning point: x² + 7x − 4 Show answer ▼Hide answer ▲

half of 7 is 7/2, (7/2)² = 49/4 (x + 7/2)² − 49/4 − 4 = (x + 7/2)² − 49/4 − 16/4 (x + 7/2)² − 65/4 · min at (−7/2, −65/4)

Q4 MEDIUM Solve x² − 8x + 5 = 0 by completing the square. Give exact answers. Show answer ▼Hide answer ▲

complete the square first (x − 4)² − 16 + 5 = 0 (x − 4)² = 11 x − 4 = ±√11 x = 4 ± √11

Q5 HARD Write 3x² + 12x − 1 in the form a(x − h)² + k. Show answer ▼Hide answer ▲

step 1 — factor 3 from x terms 3(x² + 4x) − 1 step 2 — complete the square inside half of 4 is 2, 2² = 4 3((x + 2)² − 4) − 1 step 3 — distribute the 3 3(x + 2)² − 12 − 1 3(x + 2)² − 13 · min at (−2, −13) don’t include the constant in the bracket — only the x² and x terms!

⚠ Common mistakes

Not subtracting (b/2)² to balance. The bracket adds an extra (b/2)², so you must subtract the same amount outside — otherwise you’ve changed the expression.
Halving the wrong number. You half the coefficient of x, not the constant or x².
Sign errors with negative b. If b = −5, half is −5/2 — keep the minus sign through every step.
Including the constant when factoring out a leading coefficient. 2x² − 12x + 7 → 2(x² − 6x) + 7 — leave the +7 outside, don’t divide it by 2.
Misreading the turning point. (x − h)² + k means the vertex is at (h, k) — the sign of h flips inside the bracket.
Forgetting to distribute when a ≠ 1. After completing the square inside the bracket, you must multiply the −(b/2)² piece by a too.

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Want the theory?

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