IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~8 min read

Solving Quadratic Equations

A quadratic equation is just ax² + bx + c = 0. You’ve now got three tools for solving it: factorising, the quadratic formula, and completing the square. They all work — the trick is picking the right one quickly. Factorising is fastest when the numbers are nice. The quadratic formula is the safety net that always works. Completing the square is best when you need exact answers in surd form. On Paper 2, your GDC handles all three. Here’s how to choose.

📘 What you need to know

Every quadratic equation can be written as ax² + bx + c = 0. Always rearrange to this form first.
Factorising: write a(x − p)(x − q) = 0, then x = p or q. Use when factors are obvious.
Quadratic formula: x = −b ± √(b² − 4ac)2a. In the formula booklet. Always works.
Completing the square: rewrite as a(x − h)² + k = 0, isolate the squared term, take ±√, solve for x.
GDC on Paper 2: the polynomial root finder gives all real roots instantly. Always sketch the graph as part of your working.
Hidden quadratics: equations like x⁴ − 13x² + 36 = 0 become quadratics under a substitution like u = x². Solve for u, then back-substitute to find x.

Picking the right method

Three methods, each with its sweet spot:

Factorising

brackets = 0

fastest when factors are integer-friendly

Quadratic formula

x = −b ± √D2a

always works — your default safety net

Completing the square

±√ both sides

cleanest exact answers in surd form

Quick decision rule: try factorising for 10 seconds. If you don’t see the factors fast, switch to the formula. Don’t burn 5 minutes hunting for a factorisation that doesn’t exist.

Method 1 — Factorising

Once the equation is in the form a(x − p)(x − q) = 0, the answers fall out instantly: a product of factors equals zero only if at least one factor is zero. So either x − p = 0 or x − q = 0, giving x = p or x = q.

Watch the rearrangement. If the equation is given as 2x² = 7x − 3, move everything to one side first: 2x² − 7x + 3 = 0. Don’t try to factorise something that isn’t equal to zero — the zero-product rule won’t apply.

Method 2 — The quadratic formula

The big one. It always works, regardless of whether the quadratic factors:

Quadratic formula x = −b ± √(b² − 4ac)2a ✓ in the formula booklet

🧭 Recipe — using the quadratic formula

Rearrange to ax² + bx + c = 0.
Identify a, b, c — including signs.
Compute the discriminant D = b² − 4ac first, separately. Catches errors early.
Plug in: x = (−b ± √D) / (2a).
Simplify the surd or convert to decimals depending on what’s asked.

Compute the discriminant on its own first — it’s where most sign errors live. If D < 0 there are no real solutions and you can stop. If D ≥ 0 you’ve already done the hardest bit.

Method 3 — Completing the square

Useful when the quadratic doesn’t factor and the question asks for an exact answer in surd form. The route:

🧭 Recipe — solving by completing the square

Complete the square: rewrite as a(x − h)² + k = 0.
Isolate the squared term: (x − h)² = −k/a.
Take ±√ both sides: x − h = ±√(−k/a).
Solve for x: x = h ± √(−k/a).

If −k/a turns out negative, no real solutions exist (you can’t square-root a negative real number).

Hidden quadratics — equations in disguise

Some equations look fourth-degree or worse but are quadratics in disguise. The classic example:

Quartic disguised as quadratic x⁴ − 13x² + 36 = 0 sub u = x² ⟹ u² − 13u + 36 = 0

Solve the quadratic in u, then back-substitute to find the original variable. Each value of u can give two values of x (one positive, one negative). Other patterns to spot:

Even-power quartic

ax⁴ + bx² + c = 0

substitute u = x²

Exponential disguise

a·e^2x + b·e^x + c = 0

substitute u = e^x

Worked examples

WE 1

Solve by factorising — monic

Solve x² + 7x − 18 = 0.

Step 1: Find p, q with sum 7 and product −18 9 + (−2) = 7 ✓ and 9 × (−2) = −18 ✓ Step 2: Factorise (x + 9)(x − 2) = 0 Step 3: Set each bracket to zero x + 9 = 0 → x = −9 x − 2 = 0 → x = 2 x = −9 or x = 2 check: at x = 2, 4 + 14 − 18 = 0 ✓

WE 2

Solve by factorising — non-monic

Solve 3x² − 14x + 8 = 0.

Step 1: Find m, n with sum −14 and product ac = 24 −12 + (−2) = −14 ✓ and (−12)(−2) = 24 ✓ Step 2: Split the middle and factor by grouping 3x² − 12x − 2x + 8 = 0 3x(x − 4) − 2(x − 4) = 0 (3x − 2)(x − 4) = 0 Step 3: Solve each factor 3x − 2 = 0 → x = 2/3 x − 4 = 0 → x = 4 x = 2/3 or x = 4 check at x = 4: 48 − 56 + 8 = 0 ✓

WE 3

Solve using the quadratic formula

Solve 2x² + 5x − 4 = 0, giving exact answers in surd form.

Step 1: Identify a, b, c a = 2, b = 5, c = −4 Step 2: Compute the discriminant separately D = b² − 4ac = 25 − 4(2)(−4) = 25 + 32 = 57 Step 3: Plug into the formula x = (−5 ± √57) / (2 × 2) x = (−5 ± √57) / 4 x = (−5 + √57)/4 or x = (−5 − √57)/4 since √57 doesn’t simplify and D > 0 there are two distinct irrational roots

WE 4

Solve by completing the square

Solve x² + 6x − 1 = 0 by completing the square. Give exact answers.

Step 1: Complete the square half of 6 is 3 → (x + 3)² − 9 x² + 6x − 1 = (x + 3)² − 9 − 1 = (x + 3)² − 10 Step 2: Set to zero and isolate the square (x + 3)² − 10 = 0 (x + 3)² = 10 Step 3: Take ±√ both sides x + 3 = ±√10 x = −3 ± √10 x = −3 + √10 or x = −3 − √10 don’t forget the ± — taking just the positive root halves your marks

WE 5

Rearrange before solving

Solve 6x² = 11x + 10.

Step 1: Rearrange to standard form 6x² − 11x − 10 = 0 Step 2: Try factorising — find m, n with sum −11, product −60 −15 + 4 = −11 ✓ and (−15)(4) = −60 ✓ Step 3: Split the middle and group 6x² − 15x + 4x − 10 = 0 3x(2x − 5) + 2(2x − 5) = 0 (3x + 2)(2x − 5) = 0 Step 4: Solve each factor 3x + 2 = 0 → x = −2/3 2x − 5 = 0 → x = 5/2 x = −2/3 or x = 5/2 always rearrange to “= 0” first — the zero-product rule needs zero on the right

WE 6

Hidden quadratic — quartic equation

Solve x⁴ − 13x² + 36 = 0.

Step 1: Substitute u = x² u² − 13u + 36 = 0 Step 2: Factorise — find p, q with sum −13, product 36 −4 + (−9) = −13 ✓ and (−4)(−9) = 36 ✓ (u − 4)(u − 9) = 0 u = 4 or u = 9 Step 3: Back-substitute u = x² x² = 4 → x = ±2 x² = 9 → x = ±3 x = −3, −2, 2, or 3 (four solutions) a quartic can have up to four roots — each positive value of u gives a ± pair of x-values

💡 Top tips

Always rearrange to “= 0” first, no matter what method you’re using. Zero on the right is non-negotiable.
Try factorising first if the numbers look clean. If integer factors don’t appear in 10–15 seconds, switch to the formula.
Compute the discriminant separately when using the formula — easier to spot sign errors.
Use completing the square when the question says “exact” or asks for surd form — the formula works too, but CTS often gives a cleaner expression.
Don’t drop the ± when square-rooting. A single root is half the answer.
For hidden quadratics, always back-substitute — the question asks for the original variable, not the substitution variable.
Use your GDC on Paper 2. Polynomial root finder gives all real roots in seconds. Always sketch the graph as part of your working — examiners reward the visual.

⚠ Common mistakes

Trying to factorise an equation that isn’t = 0. 2x² + x = 6 cannot be factored as it stands — rearrange first.
Sign errors in the formula. The minus on −b at the front, and the −4ac inside the square root, are the two trouble spots.
Forgetting that c can be negative. If c = −5, then −4ac becomes −4a(−5) = +20a — the negatives cancel.
Rounding too early when using the formula. Keep exact values until the very last step, especially in multi-mark questions.
Reporting only one solution after taking ±√. There should be two unless the discriminant is exactly zero.
Forgetting to back-substitute on hidden quadratics. “u = 4 or 9″ isn’t the answer — those are u-values. You need x-values.
Discarding negative x-values without checking. When x² = 9, both x = 3 and x = −3 are real solutions — keep both unless the context (like a length) forces positive.

Solving quadratics is one of those skills that becomes muscle memory — by the end of the IB course you’ll do it without thinking. The next note shifts the goalposts: instead of “where is the quadratic equal to zero?” we ask “where is the quadratic greater than or less than zero?”. That’s quadratic inequalities, and the technique builds directly on what you’ve just learned.

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