Quadratic equations are a core topic in DSE Mathematics, appearing in both Paper 1 Section A(1) and A(2). This article covers the three essential solution methods: factorization, completing the square, and the quadratic formula. You will learn how to choose the most appropriate method for each equation, step-by-step techniques, and how to avoid common mistakes. Worked examples, DSE exam tips, practice questions, and homework are all included to build your confidence and speed.
A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). In DSE Paper 1 Section A(1) and A(2), you are often asked to solve such equations. There are three main methods:
This article will explain each method with clear steps, worked examples, and exam strategies.
If the quadratic expression \(ax^2 + bx + c\) can be factored into \((px + q)(rx + s)\), then the equation \(ax^2 + bx + c = 0\) is solved by setting each factor to zero.
Factorization is the fastest method for Paper 2 MC. Practise recognising factorable quadratics at a glance. For Paper 1, show the factorisation step clearly to gain method marks.
Solve \(2x^2 - 5x - 3 = 0\).
Solution:
Factorise: \((2x + 1)(x - 3) = 0\).
Set each factor to zero:
\[
2x + 1 = 0 \Rightarrow x = -\frac{1}{2}, \quad x - 3 = 0 \Rightarrow x = 3.
\]
Answer: \(x = -\frac{1}{2}\) or \(x = 3\).
This method rewrites the quadratic in the form \((x + p)^2 = q\), then solves by taking square roots. It is especially useful when the quadratic cannot be factored easily or when you need the vertex.
Remember to add the same value to both sides. If the coefficient of \(x^2\) is not 1, divide first – otherwise the square will not be perfect.
Solve \(x^2 - 6x - 7 = 0\) by completing the square.
Solution:
Move constant: \(x^2 - 6x = 7\).
Add \(\left(\frac{-6}{2}\right)^2 = 9\) to both sides:
\[
x^2 - 6x + 9 = 7 + 9 \Rightarrow (x - 3)^2 = 16.
\]
Take square roots: \(x - 3 = \pm 4\).
So \(x = 7\) or \(x = -1\).
Answer: \(x = 7\) or \(x = -1\).
The quadratic formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) gives the solutions of \(ax^2 + bx + c = 0\). It works for all quadratic equations.
The quadratic formula is a safety net – use it when factorisation is not obvious or completing the square is too time-consuming. In Paper 1, you can use it directly.
Solve \(2x^2 + 3x - 2 = 0\) using the quadratic formula.
Solution:
Here \(a = 2\), \(b = 3\), \(c = -2\).
Discriminant: \(\Delta = 3^2 - 4(2)(-2) = 9 + 16 = 25\).
\[
x = \frac{-3 \pm \sqrt{25}}{2(2)} = \frac{-3 \pm 5}{4}.
\]
So \(x = \frac{2}{4} = \frac{1}{2}\) or \(x = \frac{-8}{4} = -2\).
Answer: \(x = \frac{1}{2}\) or \(x = -2\).
| Situation | Recommended Method | Reason |
|---|---|---|
| Expression factorises easily | Factorization | Fastest, no square roots or fractions. |
| \(a=1\) and \(b\) is even | Completing the square | Gives vertex form and is systematic. |
| Numbers are large or factorisation not obvious | Quadratic formula | Guaranteed to work, minimal thinking. |
Solve each equation using the specified method.
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An Exercise on Quadratic Equations – 3 Solution Methods: Factorization + Completing Square + Formula