Quadratics (AA HL)
Quadratics show up everywhere in AA HL, from a stand-alone "solve this equation" question to a hidden step buried inside a much longer calculus or complex-numbers problem. This page pulls together the two most exam-relevant tools - the vertex form and the discriminant - along with worked examples that go beyond routine factorising, and the mistakes that quietly cost marks. It's part of the broader Quadratics & Polynomials topic.
24 questions on this sub-topic.
Vertex form and the discriminant
Covered under IB syllabus references SL2.6 (the quadratic function, its graph and equivalent forms) and SL2.7 (solving quadratic equations and inequalities, and the discriminant), content common to both AA SL and AA HL.
Vertex form
\(f(x)=a(x-h)^2+k\)
Reached by completing the square. The vertex sits at \((h,k)\) - this is the fastest way to read off a maximum or minimum value, or the range of \(f\), without any calculus.
The discriminant
\(\Delta=b^2-4ac\)
For \(ax^2+bx+c=0\): \(\Delta>0\) gives two distinct real roots, \(\Delta=0\) gives one repeated real root, \(\Delta<0\) gives no real roots. It's also the standard tool for "for what values of \(k\)..." questions about a family of quadratics.
Need the full syllabus wording and formula-booklet reference table? See Quadratics & Polynomials.
Worked examples
\(f(x)=2x^2 - 8x + 5\).
(a) Express \(f\) in the form \(a(x-h)^2+k\).
(b) State the range.
Worked solution
(a) Complete the square: \(2(x^2 - 4x) + 5 = 2(x-2)^2 - 8 + 5\) M1
\(= 2(x-2)^2 - 3.\) A1
(b) Range: the minimum value is \(-3\) A1
(since \(2(x-2)^2 \ge 0\)) M1
so the range is \(f(x) \ge -3.\) A1
Find the set of values of \(m\) for which \(x^2 + mx + (m+3) > 0\) for all real \(x\).
Worked solution
an upward parabola is positive for all \(x\) iff \(\Delta < 0.\) M1 \(m^2 - 4(m+3) < 0.\) A1
\(m^2 - 4m - 12 < 0 \Rightarrow (m-6)(m+2) < 0.\) M1 A1 So \(-2 < m < 6.\) A1
Find the value of \(k\) for which \(x^2+kx+9=0\) has equal roots, given \(k>0.\)
Worked solution
Equal roots \(\Rightarrow \Delta = k^2 - 36 = 0\) M1
\(\Rightarrow k^2 = 36.\) A1
Since \(k > 0,\ k = 6.\) A1
Common mistakes
- Getting the sign of the vertex wrong. In \(a(x-h)^2+k\) the vertex is at \((h,k)\) - if you complete the square to get \(2(x-2)^2-3\), the vertex is \((2,-3)\), not \((-2,-3)\).
- Forgetting the "always positive/negative" condition needs the leading coefficient too. \(\Delta<0\) alone only guarantees no real roots - you also need \(a>0\) (or \(a<0\)) to conclude the quadratic is always positive (or always negative).
- Treating \(\Delta \ge 0\) and \(\Delta > 0\) as interchangeable. "Real roots" allows the repeated case \(\Delta = 0\); "two distinct real roots" strictly needs \(\Delta > 0\) - reread the question to see which one is being asked for.
Ready to practise properly?
23 quadratics questions, marked instantly like the real exam. GDC guidance for graphing and solving is on the full topic page.
Quick answers
What does the discriminant tell you about a quadratic equation?
The discriminant is \(\Delta=b^2-4ac\) for \(ax^2+bx+c=0\). If \(\Delta>0\) there are two distinct real roots; if \(\Delta=0\) there is one repeated real root; if \(\Delta<0\) there are no real roots.
How do you write a quadratic in vertex form?
Complete the square on \(f(x)=ax^2+bx+c\) to get \(f(x)=a(x-h)^2+k\), where \((h,k)\) is the vertex of the parabola - the maximum or minimum point of the graph.