Quadratics & Equations (AA SL)
A quadratic function produces a parabola, and its behaviour can be read off directly once you know which of its three forms you're looking at. This topic covers moving between general, factorised and vertex form, using the discriminant to tell how many real roots an equation has, and solving quadratic equations by factorising, completing the square, or the quadratic formula.
What the syllabus says
This topic maps onto two points in the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| SL2.6 | The quadratic function \(f(x)=ax^2+bx+c\): its graph, \(y\)-intercept \((0,c)\). Axis of symmetry. The form \(f(x)=a(x-p)(x-q)\), \(x\)-intercepts \((p,0)\) and \((q,0)\). The form \(f(x)=a(x-h)^2+k\), vertex \((h,k)\). |
| SL2.7 | Solution of quadratic equations and inequalities, using factorisation, completing the square (vertex form), and the quadratic formula. The discriminant \(\Delta=b^2-4ac\) and the nature of the roots - two distinct real roots, two equal real roots, or no real roots. |
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a quadratic function?
A quadratic function has the general form \(f(x)=ax^2+bx+c\), where \(a\neq0\). Its graph is a parabola, symmetric about a vertical line through its vertex, opening upward if \(a>0\) and downward if \(a<0\).
e.g. \(f(x)=2x^2+3x-1\) has \(a=2\), \(b=3\), \(c=-1\), and \(y\)-intercept \(f(0)=-1\).
What is the vertex of a parabola?
The vertex is the single turning point of the parabola - its minimum if \(a>0\) or its maximum if \(a<0\). It's read directly from vertex form \(f(x)=a(x-h)^2+k\) as the point \((h,k)\).
e.g. \(f(x)=(x-3)^2+2\) has vertex \((3,2)\), and expanding gives \(x^2-6x+11\).
What is the discriminant?
The discriminant, \(\Delta=b^2-4ac\), is the part of the quadratic formula under the square root. Its sign tells you how many real roots the equation \(ax^2+bx+c=0\) has, without needing to solve it.
e.g. For \(2x^2+3x-5=0\), \(\Delta=3^2-4(2)(-5)=9+40=49>0\), so there are two distinct real roots.
What is the quadratic formula?
The quadratic formula solves any equation \(ax^2+bx+c=0\) directly from its coefficients, giving both roots (when they exist) in one calculation.
e.g. For \(x^2-5x+6=0\): \(x=\dfrac{5\pm\sqrt{25-24}}{2}=\dfrac{5\pm1}{2}\), so \(x=3\) or \(x=2\).
What is completing the square?
Completing the square rewrites \(ax^2+bx+c\) in vertex form by forcing a perfect square out of the \(x^2\) and \(x\) terms. It's the method that turns general form into vertex form directly.
e.g. \(x^2+6x+5 = (x+3)^2-9+5 = (x+3)^2-4\), so the vertex is \((-3,-4)\).
Key formulas
Two formulas and one useful line drive every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The quadratic formula and discriminant are on the official formula booklet; the axis of symmetry shortcut is assumed prior knowledge, derived directly from the formula.
| Formula | Used for | Booklet? |
|---|---|---|
| \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) | Quadratic formula | ✓ Yes |
| \(\Delta=b^2-4ac\) | Discriminant | ✓ Yes |
| \(f(x)=a(x-h)^2+k\), vertex \((h,k)\) | Vertex form | Not in booklet - prior knowledge |
| \(x=-\dfrac{b}{2a}\) | Axis of symmetry | Not in booklet - prior knowledge |
The discriminant and the roots
Before solving anything, the discriminant's sign tells you exactly what kind of roots to expect.
| Discriminant | Nature of roots | Graph |
|---|---|---|
| \(\Delta>0\) | Two distinct real roots | Crosses the \(x\)-axis twice |
| \(\Delta=0\) | Two equal real roots (one repeated root) | Touches the \(x\)-axis at the vertex |
| \(\Delta<0\) | No real roots | Never meets the \(x\)-axis |
Three forms of a quadratic
The same parabola can be written three different ways - which one you use depends on what the question is asking for.
General form
\[f(x)=ax^2+bx+c\]
Reads off the \(y\)-intercept \((0,c)\) directly.
Factorised form
\[f(x)=a(x-p)(x-q)\]
Reads off the \(x\)-intercepts \((p,0)\) and \((q,0)\) directly.
Vertex form
\[f(x)=a(x-h)^2+k\]
Reads off the vertex \((h,k)\) directly.
Solving quadratic equations
Three methods, each suited to a different situation.
Factorising
Fastest when the coefficients are small integers - write as \(a(x-p)(x-q)=0\) and read off \(x=p\), \(x=q\).
Completing the square
Best when you also need the vertex - rearrange to \(a(x-h)^2+k=0\) and isolate \(x\) with a square root.
Quadratic formula
Always works, including irrational roots - substitute \(a\), \(b\), \(c\) directly and simplify.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
The function \(f\) is defined by \(f(x)=x^2-x-6\), for \(x\in\mathbb{R}\).
(a) Factorise \(f(x)\).
(b)(i) Hence write down the \(x\)-intercept of the graph of \(f\) with \(x=3\).
(b)(ii) Hence write down the \(x\)-intercept of the graph of \(f\) with \(x=-2\).
(c) Write down the \(y\)-intercept of the graph of \(f\).
Worked solution
(a) \(f(x)=(x-3)(x+2).\) A1
(b)(i) \(x=3.\) A1
(b)(ii) \(x=-2.\) A1
(c) \(f(0)=-6,\) so the \(y\)-intercept is \(-6.\) A1
Let \(f(x) = x^2 - 2x - 8\).
(a)(i) Find the \(x\)-intercept with \(x=4\).
(a)(ii) Find the \(x\)-intercept with \(x=-2\).
(b) Find the coordinates of the vertex.
(c) State the range of \(f\).
Worked solution
(a)(i) Set \(f(x)=0\): \(x^2-2x-8=(x-4)(x+2)=0.\) M1
\(\Rightarrow x=4,\) intercept \((4,0).\) A1
(a)(ii) Or \(x=-2,\) intercept \((-2,0).\) A1
(b) Axis of symmetry is the midpoint of the roots: \(x=\dfrac{4+(-2)}{2}=1.\) M1
Then \(f(1)=1-2-8=-9,\) so the vertex is \((1,-9).\) A1
(c) An upward parabola has range bounded below by its vertex, minimum \(-9.\) R1
So the range is \(f(x)\ge-9.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Reading the wrong sign off factorised form. If \(f(x)=(x-p)(x-q)\), the roots are \(x=p\) and \(x=q\) - not \(-p\) and \(-q\). Check by expanding if you're unsure.
- Forgetting the \(\pm\) when isolating a squared term. Both completing the square and the quadratic formula involve a square root, which always gives two possible values, not one.
- Not rearranging to \(=0\) before reading off \(a\), \(b\), \(c\). If the equation is given as \(x^2+3x=5\), you must rewrite it as \(x^2+3x-5=0\) before substituting into the formula or discriminant.
- Misreading the sign of \(h\) in vertex form. \(f(x)=a(x-h)^2+k\) has vertex \((h,k)\) - for \(f(x)=(x+3)^2-4\), that means \(h=-3\), not \(h=3\).
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Faster and safer than algebra for quadratics with irrational or messy roots - and it finds every solution, not just one.
- Graph \(f(x)\) first so you can see how many solutions exist and roughly where they are.
- Rearrange so everything is on one side: \(f(x) = 0\) - or graph both sides as separate functions and find intersections.
- MATH → Solver: enter the expression, type a starting guess close to one root, press ALPHA + ENTER. Move the guess to near a different root and repeat for each solution.TI-84
- Type nSolve(f(x)=0, x, guess) - include a guess or interval e.g. nSolve(f(x)=0, x, 2) or nSolve(f(x)=0, x, {1,5}) to target a specific root.Nspire
- Run-Matrix → SolveN(f(x), x) returns all real roots at once; or use the Equation app for a visual approach.Casio
- Always verify each solution by substituting back into the original equation.
Tip: The solver finds ONE root near your starting guess - change the guess to find others. The graph shows you how many to expect.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Quadratics & equations questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between the roots of a quadratic and its vertex?
The roots are the \(x\)-intercepts, where the graph crosses the \(x\)-axis and \(f(x)=0\). The vertex is the single turning point of the parabola - its minimum or maximum. A quadratic can have two roots, one root, or none, but it always has exactly one vertex.
When does a quadratic have no real roots?
When the discriminant \(b^2-4ac\) is negative. Graphically this means the parabola never crosses the \(x\)-axis - it stays entirely above it (if \(a>0\)) or entirely below it (if \(a<0\)).
Which method should I use to solve a quadratic - factorising, completing the square, or the formula?
Try factorising first if the numbers look nice - it's fastest. Completing the square is best when you also need the vertex. The quadratic formula always works, including for irrational or complex roots, so use it when factorising isn't obvious.
Can my GDC solve a quadratic equation for me?
Yes, on Paper 2 your GDC can solve an equation numerically or find where a graph crosses the \(x\)-axis, which handles quadratics with messy or irrational roots. On Paper 1 you'll need factorising, completing the square or the formula by hand. See the GDC guide for model-specific instructions.
Sub-topics
Quadratics & Equations broken down into its individual skills, each with its own focused page.
Related topics
More Functions topics from the same AA SL syllabus unit, in case you want to keep going.