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📚 MATHSKILLER TEXTBOOK SERIES

Quadratic Equations

in One Unknown

📖 Detailed Explanations

📝 DSE Past Papers

⚡ Quick Tips

✏️ Practice Questions

For HKDSE Mathematics (Compulsory Part)

📋 Table of Contents

1.1 Definition and Standard FormP.3

1.2 Four Methods of SolvingP.4

• FactorizationP.4

• Quadratic FormulaP.5

• Completing the SquareP.6

• Graphical MethodP.7

1.3 The DiscriminantP.8

1.4 Vieta's FormulasP.10

1.5 Forming EquationsP.12

1.6 DSE Past Paper QuestionsP.14

1.7 Practice QuestionsP.20

1.8 AnswersP.24

CHAPTER 1

Quadratic Equations in One Unknown

🎯 Learning Objectives

Master the four methods of solving quadratic equations
Understand and apply the discriminant to determine the nature of roots
Use Vieta's formulas to find sum and product of roots
Form quadratic equations from given conditions
Solve word problems involving quadratic equations

1.1

Definition and Standard Form

📖 Definition

A quadratic equation is a polynomial equation of degree 2 in one variable.

Standard Form (General Form)

$ax^2 + bx + c = 0$

where $a \neq 0$, and $a$, $b$, $c$ are constants

Coefficient	Name	Example: $2x^2 - 5x + 3 = 0$
$a$	Coefficient of $x^2$	$a = 2$
$b$	Coefficient of $x$	$b = -5$ (note the negative!)
$c$	Constant term	$c = 3$

⚡ Quick Tip: Identifying Coefficients

Given $3x^2 = 5x - 2$, first convert to standard form!

$3x^2 - 5x + 2 = 0$

Here: $a = 3$, $b = -5$, $c = 2$

⚠️ Be careful with the sign of $b$!

1.2

Four Methods of Solving

Method 1: Factorization

📖 When to Use

When the equation can be factored into two linear factors.

Principle: If $AB = 0$, then $A = 0$ or $B = 0$

Example 1

Solve: $x^2 - 5x + 6 = 0$

📝 Solution

Step 1: Find two numbers that add to -5 and multiply to 6

Answer: -2 and -3

Step 2: Factorize

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

Step 3: Solve

$x - 2 = 0$ or $x - 3 = 0$

$\therefore x = 2$ or $x = 3$

Method 2: Quadratic Formula

The Quadratic Formula

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Works for ALL quadratic equations

Example 2

Solve: $2x^2 - 5x - 3 = 0$

📝 Solution

Identify: $a = 2$, $b = -5$, $c = -3$

Substitute into formula:

$x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$

$= \dfrac{5 \pm \sqrt{25 + 24}}{4} = \dfrac{5 \pm 7}{4}$

$\therefore x = 3$ or $x = -\dfrac{1}{2}$

1.3

The Discriminant

Discriminant

$\Delta = b^2 - 4ac$

Discriminant Value	Nature of Roots	Graph Intersection
$\Delta > 0$	Two distinct real roots	Crosses x-axis at 2 points
$\Delta = 0$	Two equal real roots (repeated)	Touches x-axis at 1 point
$\Delta < 0$	No real roots	Does not cross x-axis

⚡ DSE Trap Warning!

"Has real roots" ≠ "Has distinct real roots"

"Has real roots" → $\Delta \geq 0$ (includes repeated!)
"Has distinct real roots" → $\Delta > 0$
"Has no real roots" → $\Delta < 0$

1.4

Vieta's Formulas

📖 Vieta's Formulas

If $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$, then:

Sum of roots: $\alpha + \beta = -\dfrac{b}{a}$

Product of roots: $\alpha\beta = \dfrac{c}{a}$

🧠 Memory Tip

Sum has a negative sign: $-\dfrac{b}{a}$
Product has no negative: $\dfrac{c}{a}$

⚡ Useful Derived Formulas

Expression	Formula
$\alpha^2 + \beta^2$	$= (\alpha + \beta)^2 - 2\alpha\beta$
$(\alpha - \beta)^2$	$= (\alpha + \beta)^2 - 4\alpha\beta$
$\dfrac{1}{\alpha} + \dfrac{1}{\beta}$	$= \dfrac{\alpha + \beta}{\alpha\beta}$

1.6

DSE Past Paper Questions

⚡ Quadratic Equations - Quick Solve Strategies

Type	Quick Strategy	Frequency
Factorization	Find two numbers: sum = -b, product = c	85%
Discriminant Δ = 0	b² = 4ac, expand and solve for k	70%
Root substitution	If β is root, then aβ² + bβ + c = 0	60%
Vieta's Formulas	α+β = -b/a, αβ = c/a	55%

📋 Paper 1 Questions

DSE 2015

Q9. The equation $x^2 - (k+1)x + k = 0$ has two equal real roots. (a) Find $k$. (b) Find the root.

📝 Solution

(a) Equal roots → $\Delta = 0$

$(k+1)^2 - 4k = 0$

$k^2 - 2k + 1 = 0$

$(k-1)^2 = 0$, so $k = 1$

(b) When $k = 1$: $x^2 - 2x + 1 = 0$

$(x-1)^2 = 0$, so $x = 1$

DSE 2022

Q9. If $\alpha$ and $\beta$ are roots of $3x^2 - 8x + 2 = 0$, find (a) $\alpha + \beta$ and $\alpha\beta$ (b) $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$

📝 Solution

(a) $\alpha + \beta = \dfrac{8}{3}$, $\alpha\beta = \dfrac{2}{3}$

(b) $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{8/3}{2/3} = 4$

📋 Paper 2 Questions (Multiple Choice)

DSE 2014 Q24

Solve $(x-k)^2=4k^2$

A. $x=-k$
B. $x=3k$
C. $x=-k$ or $x=3k$
D. $x=k$ or $x=-3k$

📝 Quick Solution

$(x-k)^2 = 4k^2$

$x - k = \pm 2k$

$x = k + 2k = 3k$ or $x = k - 2k = -k$

Answer: C

DSE 2016 Q26

If $x^2+ax+a=1$ has equal roots, find $a$.

A. $-1$ B. $2$ C. $0$ or $-4$ D. $-2$ or $2$

📝 Quick Solution

Standard form: $x^2 + ax + (a-1) = 0$

$\Delta = a^2 - 4(a-1) = 0$

$(a-2)^2 = 0$, so $a = 2$

Answer: B

1.7

Practice Questions

📝 Self Test

1. Solve: $x^2 - 7x + 12 = 0$

2. Solve: $2x^2 + 5x - 3 = 0$

3. If $x^2 - 4x + k = 0$ has equal roots, find $k$.

4. If $\alpha$, $\beta$ are roots of $x^2 - 5x + 3 = 0$, find $\alpha^2 + \beta^2$.

5. If one root of $x^2 + px + 12 = 0$ is 3, find $p$ and the other root.

6. The equation $kx^2 - 6x + 3 = 0$ has real roots. Find the range of $k$.

7. If $\alpha$, $\beta$ are roots of $2x^2 - 3x - 5 = 0$, find $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$.

8. Form a quadratic equation with roots $2 + \sqrt{3}$ and $2 - \sqrt{3}$.

📋 Answers

1. $x=3$ or $4$

2. $x=\frac{1}{2}$ or $-3$

3. $k=4$

4. $19$

5. $p=-7$, root $4$

6. $k \leq 3$, $k \neq 0$

7. $-\frac{3}{5}$

8. $x^2-4x+1=0$

📘 MathsKiller Textbook Series

Chapter 1: Quadratic Equations