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·e2x + b·ex + c = 0
substitute u = ex
Worked examples
WE 1Solve 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 2Solve 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 3Solve 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 4Solve 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 5Rearrange 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 6Hidden 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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