Master quadratic equations from multiple perspectives: factorise, apply the quadratic formula, complete the square, and interpret the parabola graphically. At Year 4 Advanced, you must prove why methods work, not just apply them.
What You'll Learn
Solve quadratics by factorising and justify using the zero product property
Apply the quadratic formula and interpret the discriminant
Complete the square to find the vertex form y = a(x − h)² + k
Sketch parabolas identifying roots, vertex, axis of symmetry, and y-intercept
Move fluently between factorised form, standard form, and vertex form
Solve real-world problems modelled by quadratic equations
IB Assessment Focus
Criterion A: Apply quadratics to unfamiliar, multi-step problems (projectile motion, area optimisation).
Criterion B: Prove the quadratic formula by completing the square on ax² + bx + c = 0.
Criterion C: Communicate solutions using correct notation; move between algebraic, graphical, and numerical forms.
Criterion D: Justify degree of accuracy and explain whether solutions are reasonable in context.
Key Vocabulary
Term
Definition
Quadratic equation
An equation of the form ax² + bx + c = 0 where a ≠ 0
Roots / Solutions
The values of x where y = 0; the x-intercepts of the parabola
Factorising
Writing a quadratic as a product of two brackets: a(x + p)(x + q)
Zero product property
If ab = 0, then a = 0 or b = 0 (or both)
Quadratic formula
x = (−b ± √(b² − 4ac)) / 2a
Discriminant (Δ)
The expression b² − 4ac; determines the number of real roots
Vertex
The turning point (maximum or minimum) of the parabola
Axis of symmetry
The vertical line x = −b/2a passing through the vertex
Completing the square
Rewriting ax² + bx + c as a(x − h)² + k to reveal the vertex
Parabola
The U-shaped (or inverted U) graph of a quadratic function
Mathematics · Topic 1.1
Solving by Factorising
Factorising is the quickest method when the quadratic has integer roots. It relies on the zero product property: if the product of two factors is zero, at least one factor must be zero.
Method: Factorising x² + bx + c
Write the equation in standard form: ax² + bx + c = 0.
Find two numbers that multiply to c and add to b.
Write the quadratic as (x + p)(x + q) = 0.
Apply the zero product property: x + p = 0 or x + q = 0.
Solve each linear equation.
Example: Solve x² − 7x + 12 = 0
Worked Example — Factorising
Find two numbers: multiply to 12, add to −7 → −3 and −4find factor pair
(x − 3)(x − 4) = 0factorise
x − 3 = 0 → x = 3 or x − 4 = 0 → x = 4zero product property
When the leading coefficient is not 1, use the ac-method (grouping).
Multiply a × c to get the product.
Find two numbers that multiply to ac and add to b.
Split the middle term bx into two terms using those numbers.
Factor by grouping in pairs.
Example: Solve 2x² + 7x + 3 = 0
Worked Example — ac-Method
a × c = 2 × 3 = 6 → two numbers that multiply to 6 and add to 7: 1 and 6ac-method
Rewrite: 2x² + x + 6x + 3 = 0split middle term
Group: x(2x + 1) + 3(2x + 1) = 0factor pairs
(2x + 1)(x + 3) = 0common bracket
x = −12 or x = −3
Special Cases
Difference of two squares
Pattern
a² − b² = (a + b)(a − b)
Example: x² − 25 = (x + 5)(x − 5)
Perfect square trinomial
Pattern
a² ± 2ab + b² = (a ± b)²
Example: x² + 6x + 9 = (x + 3)²
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Common Mistake: Dividing both sides by x when solving x² = 5x. This loses the solution x = 0. Instead, rearrange to x² − 5x = 0, then factor: x(x − 5) = 0, giving x = 0 or x = 5.
Mathematics · Topic 1.1
The Quadratic Formula
The quadratic formula works for every quadratic equation, including those that cannot be factorised. It is derived by completing the square on the general form.
Quadratic Formula
x = −b ± √(b² − 4ac)2a
The Discriminant
The expression under the square root, Δ = b² − 4ac, determines the nature of the roots.
Factorising — first choice when integer/simple roots exist; fastest method
Quadratic formula — universal method; use when factorising is not obvious
Completing the square — use when you need vertex form or to prove the formula
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Critical Rule: Always check the discriminant FIRST. If Δ < 0, state "no real solutions" immediately — do not attempt to take the square root of a negative number at this level.
Mathematics · Topic 1.1
Graphing Parabolas
The graph of y = ax² + bx + c is a parabola. Understanding its features lets you move between algebraic and graphical representations.
Parabola y = a(x−p)(x−q) — vertex, roots, y-intercept, and axis of symmetry
Key Features of a Parabola
Feature
How to find it
Direction
a > 0: opens upward (minimum); a < 0: opens downward (maximum)
y-intercept
Set x = 0: y-intercept = c
x-intercepts (roots)
Set y = 0 and solve ax² + bx + c = 0
Axis of symmetry
x = −b / 2a
Vertex
(−b/2a, f(−b/2a)) — substitute the axis of symmetry value back into the equation
Example: Sketch y = x² − 4x − 5
Worked Example — Sketching a Parabola
Direction: a = 1 > 0 → opens upward (minimum)sign of a
y-intercept: c = −5 → (0, −5)set x = 0
x-intercepts: (x − 5)(x + 1) = 0 → x = 5 or x = −1factorise & solve
Axis of symmetry: x = −(−4)2 × 1 = 42 = 2x = −b/2a
Vertex: y = 4 − 8 − 5 = −9 → vertex (2, −9)substitute x = 2
Roots: x = −1 and x = 5 Vertex: (2, −9)
Three Forms of a Quadratic
Form
Equation
Reveals
Standard form
y = ax² + bx + c
y-intercept (c); use formula for roots
Factorised form
y = a(x − p)(x − q)
x-intercepts directly (p and q)
Vertex form
y = a(x − h)² + k
Vertex directly at (h, k)
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Criterion C Tip: In IB assessments, show the same quadratic in multiple representations — algebraic, graphical, and numerical (table of values). Moving fluently between representations is a mark of higher achievement.
Mathematics · Topic 1.1
Completing the Square
Completing the square transforms a quadratic from standard form to vertex form. It also provides the foundation for proving the quadratic formula.
Method (when a = 1)
Start with x² + bx + c.
Take half the coefficient of x: b/2.
Write (x + b/2)² and subtract (b/2)² to compensate: x² + bx + c = (x + b/2)² − (b/2)² + c.
Derive the quadratic formula by completing the square on ax² + bx + c = 0.
Proof — Derivation of the Quadratic Formula (Criterion B)
x² + bax + ca = 0divide both sides by a
x² + bax = −camove constant to RHS
Half the x-coefficient: b2a Square it: b²4a²prepare to complete square
(x + b2a)² = b²4a² − ca = b² − 4ac4a²add to both sides & combine
x + b2a = ±√(b² − 4ac)2asquare root both sides
x = −b ± √(b² − 4ac)2a ∎
This proof is a key Criterion B target at Year 4 — you must be able to reproduce it.
Mathematics · Topic 1.1
Worked Examples
Multi-step solutions showing the reasoning expected at Year 4 Advanced.
EXAMPLE 1Solve x² − 5x + 6 = 0 by factorising. Justify every step.
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Full Solution
I need two numbers that multiply to +6 and add to −5. Those are −2 and −3 (since (−2)(−3) = 6 and −2 + (−3) = −5).
x² − 5x + 6 = (x − 2)(x − 3) = 0.
Applying the zero product property: if the product of two factors equals zero, at least one factor must be zero.
x − 2 = 0 → x = 2, or x − 3 = 0 → x = 3.
EXAMPLE 3Find the vertex of y = x² − 8x + 12 by completing the square.
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Full Solution
Half of −8 is −4. Square it: (−4)² = 16.
x² − 8x + 12 = (x − 4)² − 16 + 12 = (x − 4)² − 4
Vertex form: y = (x − 4)² − 4. The vertex is at (4, −4).
Since a = 1 > 0, this is a minimum point.
Check: The roots are x² − 8x + 12 = (x − 2)(x − 6) = 0, giving x = 2 and x = 6. The axis of symmetry x = (2 + 6)/2 = 4 confirms the vertex x-coordinate. ✓
EXAMPLE 4Determine the number of real roots of 5x² − 4x + 1 = 0 without solving.
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Full Solution
Δ = b² − 4ac = (−4)² − 4(5)(1) = 16 − 20 = −4.
Since Δ < 0, the equation has no real roots.
Graphically, the parabola y = 5x² − 4x + 1 opens upward (a = 5 > 0) and sits entirely above the x-axis — it never crosses it.
EXAMPLE 5A ball is thrown upward. Its height is h = −5t² + 20t + 1 metres after t seconds. Find the maximum height and when the ball hits the ground.
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Full Solution
Maximum height (complete the square):
h = −5(t² − 4t) + 1 = −5[(t − 2)² − 4] + 1 = −5(t − 2)² + 20 + 1 = −5(t − 2)² + 21.
Maximum height: 21 metres at t = 2 seconds (since a = −5 < 0, vertex is a maximum).
When it hits the ground (h = 0):
−5t² + 20t + 1 = 0 → 5t² − 20t − 1 = 0
Δ = 400 + 20 = 420.
t = (20 ± √420) / 10 = (20 ± 20.49) / 10.
t = 4.049 or t = −0.049 (rejected, t ≥ 0).
The ball hits the ground at t ≈ 4.05 seconds.
Justification of accuracy: I rounded to 2 d.p. because time measurements in this context are practical (to the nearest hundredth of a second).
EXAMPLE 6Solve 6x² + x − 2 = 0 using the ac-method.
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Full Solution
a × c = 6 × (−2) = −12. Find two numbers that multiply to −12 and add to 1: 4 and −3.
EXAMPLE 7Find the value(s) of k for which kx² + 6x + k = 0 has exactly one repeated root.
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Full Solution
For exactly one repeated root, the discriminant must equal zero:
Δ = b² − 4ac = 36 − 4(k)(k) = 36 − 4k² = 0.
4k² = 36 → k² = 9 → k = ±3.
Check k ≠ 0 (otherwise it's not quadratic). Both k = 3 and k = −3 are valid. k = 3 or k = −3.
Mathematics · Topic 1.1
Practice Q&A
Attempt each question before revealing the model answer. Focus on justifying each step.
SOLVESolve x² + 2x − 15 = 0 by factorising.
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Model Answer
Two numbers that multiply to −15 and add to 2: 5 and −3.
(x + 5)(x − 3) = 0 → x = −5 or x = 3.
Verify: (−5)² + 2(−5) − 15 = 25 − 10 − 15 = 0 ✓
APPLYSolve 3x² − 5x − 2 = 0 using the quadratic formula.
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Model Answer
a = 3, b = −5, c = −2. Δ = 25 + 24 = 49 > 0.
x = (5 ± 7)/6 → x = 12/6 = 2 or x = −2/6 = −1/3.
CONVERTWrite y = x² − 10x + 21 in vertex form and state the vertex.
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Model Answer
Half of −10 is −5; (−5)² = 25.
y = (x − 5)² − 25 + 21 = (x − 5)² − 4.
Vertex: (5, −4). Minimum value is −4.
ANALYSEWithout solving, determine the nature of the roots of 4x² + 4x + 1 = 0.
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Model Answer
Δ = 4² − 4(4)(1) = 16 − 16 = 0.
Since Δ = 0, the equation has one repeated root. The parabola touches the x-axis at exactly one point.
(In fact: 4x² + 4x + 1 = (2x + 1)² = 0, so x = −1/2 is the repeated root.)
SOLVESolve 4x² − 49 = 0.
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Model Answer
This is a difference of two squares: (2x)² − 7² = (2x + 7)(2x − 7) = 0.
2x + 7 = 0 → x = −7/2 = −3.5
2x − 7 = 0 → x = 7/2 = 3.5 x = ±3.5
SKETCHSketch y = −x² + 4x + 5. State the vertex, axis of symmetry, and x-intercepts.