Quadratic Equations
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Master the core principles of quadratic equations \(ax^2 + bx + c = 0\) (where \(a \neq 0\)) by exploring geometric area models, the zero-product property, and the discriminant.
Foundational Concept: A linear equation represents a constant rate of change (a straight line), while a quadratic equation introduces a squared term \(x^2\), representing area, acceleration under gravity, or curved trajectories (parabolas).
(a) Concrete Scenario: The Shamba Expansion
Farmer Mwangi in Kiambu has a square plot of land of side length \(x\) metres. He expands one side by \(3\text{ m}\) and the other side by \(2\text{ m}\). The total area of his new rectangular plot is \(30\text{ m}^2\).
The area is given by the algebraic expression:
\[ (x + 3)(x + 2) = 30 \implies x^2 + 5x + 6 = 30 \implies x^2 + 5x - 24 = 0 \](b) Geometric Insight: Why Quadratic Equations Have Up to Two Roots
The graph of \(y = ax^2 + bx + c\) is a symmetric curve called a parabola:
- When the parabola crosses the \(x\)-axis in two places, the equation \(ax^2 + bx + c = 0\) has two distinct real roots (\(\Delta > 0\)).
- When the parabola just touches the \(x\)-axis at its turning point (vertex), it has one repeated real root (\(\Delta = 0\)).
- When the parabola floats entirely above or below the \(x\)-axis, it has no real roots (\(\Delta < 0\)).
(c) The Zero-Product Principle
If \(A \times B = 0\), then either \(A = 0\), \(B = 0\), or both are \(0\). We factorise quadratic expressions to transform a complex second-degree problem into two simple linear equations.
Interactive Visualiser: Parabola & Discriminant Dial
Adjust the slider for \(c\) to see how shifting the parabola changes the number of real roots (intersections with the \(x\)-axis).
Key Formulas
- \(\Delta > 0\): Two distinct real roots
- \(\Delta = 0\): Two equal real roots (perfect square trinomial)
- \(\Delta < 0\): No real roots
Summary of Solution Methods
| Method | Best Used When |
|---|---|
| Factorisation | Roots are rational; factors of \(ac\) summing to \(b\) are easily spotted. |
| Completing the Square | Coefficient \(a = 1\) and \(b\) is an even integer, or finding the vertex. |
| Quadratic Formula | General cases involving decimals, surds, or difficult factorisations. |
Worked Examples
Solve the quadratic equation: \[ x^2 - 7x + 10 = 0 \]
Step-by-step Solution:- Identify product \(P = 10\) and sum \(S = -7\).
- Find two integers whose product is \(+10\) and sum is \(-7\): they are \(-2\) and \(-5\).
- Express as factors: \((x - 2)(x - 5) = 0\).
- Apply zero-product rule: \[ x - 2 = 0 \implies x = 2 \] \[ x - 5 = 0 \implies x = 5 \]
- Answer: \(x = 2\) or \(x = 5\).
Solve the equation: \[ 2x^2 + 5x - 3 = 0 \]
Step-by-step Solution:- Identify coefficients: \(a = 2\), \(b = 5\), \(c = -3\).
- Calculate the discriminant \(\Delta\): \[ \Delta = b^2 - 4ac = (5)^2 - 4(2)(-3) = 25 + 24 = 49 \]
- Substitute into the formula \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\): \[ x = \frac{-5 \pm \sqrt{49}}{2(2)} = \frac{-5 \pm 7}{4} \]
- Separate the two cases: \[ x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \] \[ x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3 \]
- Answer: \(x = \frac{1}{2}\) or \(x = -3\).
A rectangular plot in Naivasha has an area of \(24\text{ m}^2\). The length is \(4\text{ m}\) longer than the width. Find the dimensions of the plot by completing the square.
Step-by-step Solution:- Let width \(= w\). Length \(= w + 4\). Area equation: \[ w(w + 4) = 24 \implies w^2 + 4w = 24 \]
- Complete the square: take half the coefficient of \(w\) (\(\frac{4}{2} = 2\)) and square it (\(2^2 = 4\)). Add \(4\) to both sides: \[ w^2 + 4w + 4 = 24 + 4 \]
- Factor the left side as a perfect square: \[ (w + 2)^2 = 28 \]
- Take square roots on both sides: \[ w + 2 = \pm\sqrt{28} = \pm 2\sqrt{7} \implies w = -2 \pm 2\sqrt{7} \]
- Since physical width must be positive: \[ w = -2 + 2\sqrt{7} \approx -2 + 5.292 = 3.292\text{ m} \]
- Length \(= w + 4 = 2 + 2\sqrt{7} \approx 7.292\text{ m}\).
- Answer: \(\text{Width} = 3.29\text{ m}\), \(\text{Length} = 7.29\text{ m}\) (to 2 d.p.).
Common Mistakes
Incorrect: \(x(x - 3) = 10 \implies x = 10\) or \(x - 3 = 10\).
Why it happens: Learners generalize \(A \cdot B = 0\) to any number \(k\).
Correction: The product property only holds for zero! Expand and rearrange to standard form first: \(x^2 - 3x - 10 = 0 \implies (x - 5)(x + 2) = 0\), giving \(x = 5\) or \(x = -2\).
Incorrect: In \(x^2 - 3x - 4 = 0\), evaluating \(b^2 - 4ac\) as \((-3)^2 - 4(1)(4) = 9 - 16 = -7\).
Why it happens: Dropping the negative sign of \(c = -4\).
Correction: Write brackets carefully: \((-3)^2 - 4(1)(-4) = 9 - (-16) = 9 + 16 = 25\).
Incorrect: \((x - 3)^2 = 16 \implies x - 3 = 4 \implies x = 7\) (losing the second root).
Correction: Every positive number has two square roots: \(x - 3 = \pm 4\), so \(x = 3 + 4 = 7\) or \(x = 3 - 4 = -1\).
Incorrect: Adding \(\left(\frac{6}{2}\right)^2\) directly to \(2x^2 + 6x = 8\).
Correction: Always divide the entire equation by \(a\) first: \(x^2 + 3x = 4\), then add \(\left(\frac{3}{2}\right)^2\).
Real World
1. Sports: Football Trajectory at Kasarani
When Harambee Stars striker shoots a ball, its height \(h\) in metres after \(t\) seconds is modeled by \(h = -5t^2 + 15t\). Solving \(-5t^2 + 15t = 0\) reveals \(t = 0\) (kickoff) and \(t = 3\text{ s}\) (landing back on the pitch).
2. Business: Kiosk Revenue Maximisation
An M-Pesa agent & electronics kiosk in Nairobi sells flash drives. If price is lowered by \(x\) shillings, sales increase, yielding total profit \(P(x) = -2x^2 + 80x + 1200\). Finding the vertex quadratic roots determines the break-even and peak profit pricing.
3. Engineering: Suspension Bridges & Solar Dishes
Cables on suspension bridges across rivers hang in parabolic curves \(y = ax^2\). Parabolic solar concentrators in northern Kenya focus all incoming sunlight directly to a single focal point receiver at the vertex focus.
Practice