This article covers the essential skills of solving quadratic equations – a core topic in DSE Paper 1 Section A(1) and A(2) worth 4–6 marks. You will learn the three main methods for solving quadratic equations: factorization method, completing the square, and the quadratic formula. You will also learn how to choose the most appropriate method for a given problem and how to handle word problems involving quadratic equations. The article includes step-by-step worked examples, DSE exam techniques, practice questions, and a full set of homework with solutions. These are essential skills that appear regularly in DSE papers.
By the end of this article, you should be able to:
Quadratic equations appear every year in DSE Paper 1 Section A(1) and A(2). Questions may ask you to:
These are essential skills that also appear in more advanced topics. Master them early!
Quadratic equation questions often appear in Section A(1) and A(2) as short-answer questions worth 3–4 marks each. They test your algebraic manipulation skills.
The factorization method involves expressing the quadratic expression as a product of two linear factors. This method works best when the quadratic expression can be easily factorized.
Solve: \( x^2 - 5x + 6 = 0 \)
Factorize: \( x^2 - 5x + 6 = (x-2)(x-3) \)
Set each factor to zero:
\( x - 2 = 0 \) → \( x = 2 \)
\( x - 3 = 0 \) → \( x = 3 \)
Answer: \( x = 2 \) or \( x = 3 \)
Factorization: Find two numbers that multiply to \( c \) and add to \( b \) (when \( a = 1 \)).
Completing the square involves rewriting the quadratic expression in the form \( (x + p)^2 + q \). This method is useful for finding the vertex of a parabola and for solving equations when factorization is not straightforward.
Solve: \( x^2 + 6x - 7 = 0 \)
Move constant: \( x^2 + 6x = 7 \)
Add \( \left(\frac{6}{2}\right)^2 = 9 \) to both sides:
\( x^2 + 6x + 9 = 7 + 9 \)
\( (x + 3)^2 = 16 \)
Take square roots: \( x + 3 = \pm 4 \)
\( x = -3 \pm 4 \)
\( x = 1 \) or \( x = -7 \)
Answer: \( x = 1 \) or \( x = -7 \)
The quadratic formula solves any quadratic equation of the form \( ax^2 + bx + c = 0 \):
Solve: \( 2x^2 + 3x - 2 = 0 \)
\( a = 2 \), \( b = 3 \), \( c = -2 \)
\( x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} \)
\( x = \frac{-3 \pm \sqrt{9 + 16}}{4} \)
\( x = \frac{-3 \pm \sqrt{25}}{4} \)
\( x = \frac{-3 \pm 5}{4} \)
\( x = \frac{2}{4} = \frac{1}{2} \) or \( x = \frac{-8}{4} = -2 \)
Answer: \( x = \frac{1}{2} \) or \( x = -2 \)
Quadratic Formula: Sing it to the tune of "Pop Goes the Weasel" – "Negative B, plus or minus square root, B squared minus 4AC, all over 2A!"
| Method | Best Used When | Example |
|---|---|---|
| Factorization | The quadratic can be easily factorized | \( x^2 - 7x + 12 = 0 \) |
| Completing the Square | Finding vertex, or when factorization is difficult | \( x^2 + 4x - 8 = 0 \) |
| Quadratic Formula | Any quadratic (especially when \( a \neq 1 \) or factor is not integer) | \( 3x^2 - 2x - 5 = 0 \) |
In DSE exams, the quadratic formula is the most reliable method. However, if you can factorize quickly, it saves time. Practice all three methods!
The discriminant \( \Delta = b^2 - 4ac \) tells us the nature of the roots.
| Discriminant | Nature of Roots |
|---|---|
| \( \Delta > 0 \) | Two distinct real roots |
| \( \Delta = 0 \) | One repeated root (two equal roots) |
| \( \Delta < 0 \) | No real roots (complex roots) |
For \( 2x^2 - 3x + 1 = 0 \):
\( \Delta = (-3)^2 - 4(2)(1) = 9 - 8 = 1 > 0 \)
Two distinct real roots.
Question: Solve \( x^2 - 8x + 15 = 0 \).
\( x^2 - 8x + 15 = (x-3)(x-5) \)
\( x = 3 \) or \( x = 5 \)
Answer: \( x = 3, 5 \)
Question: Solve \( x^2 - 4x - 12 = 0 \) by completing the square.
\( x^2 - 4x = 12 \)
\( x^2 - 4x + 4 = 12 + 4 \)
\( (x-2)^2 = 16 \)
\( x - 2 = \pm 4 \)
\( x = 6 \) or \( x = -2 \)
Answer: \( x = 6, -2 \)
Question: Solve \( 5x^2 + 6x - 8 = 0 \) using the quadratic formula.
\( a = 5 \), \( b = 6 \), \( c = -8 \)
\( x = \frac{-6 \pm \sqrt{36 - 4(5)(-8)}}{10} \)
\( x = \frac{-6 \pm \sqrt{36 + 160}}{10} = \frac{-6 \pm \sqrt{196}}{10} = \frac{-6 \pm 14}{10} \)
\( x = \frac{8}{10} = \frac{4}{5} \) or \( x = \frac{-20}{10} = -2 \)
Answer: \( x = \frac{4}{5}, -2 \)
Question 1 MC
Solve \( x^2 - 9 = 0 \).
A. \( \pm 3 \) B. \( \pm 9 \) C. \( 3 \) D. \( 9 \)
Question 2 MC
Solve \( 2x^2 - 8 = 0 \).
A. \( \pm 2 \) B. \( \pm 4 \) C. \( \pm \sqrt{2} \) D. \( \pm 2\sqrt{2} \)
Question 3 Short Answer
Solve \( x^2 - 7x + 12 = 0 \) by factorization.
Question 4 Short Answer
Solve \( x^2 + 10x + 24 = 0 \) by completing the square.
Question 5 MC
What is the discriminant of \( x^2 + 4x + 5 = 0 \)?
A. \( -4 \) B. \( -1 \) C. \( 0 \) D. \( 4 \)
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An Exercise on Quadratic Equations | Three Methods (Factorization, Completing Square & Formula)