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Quadratic Equations

Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.

Grade 10 Pathway: N/A

First Principles

Objective: Solve quadratic equations of the form \(ax^2 + bx + c = 0\) by factorising into linear binomials and applying the zero-product property.

Interactive Quadratic Factoriser

Select a quadratic equation to see how it splits into linear factors and reveals roots!

(a) Concrete Scenario: A farmer in Trans Nzoia wants to enclose a rectangular plot whose length is \(3\text{ m}\) longer than its width \(w\). If the total plot area is \(40\text{ m}^2\), the relationship is given by \(w(w + 3) = 40\), which rearranges to \(w^2 + 3w - 40 = 0\). Solving for \(w\) requires factorising the quadratic.

(b) Geometric Insight (Algebra Tiles): Expanding \((x + p)(x + q)\) gives an area composed of one large square \(x^2\), two rectangular side strips \(px\) and \(qx\), and a corner block \(pq\): \[x^2 + (p + q)x + pq = (x + p)(x + q)\] Factorisation reverses this process: we find two numbers \(p\) and \(q\) whose sum is the coefficient of \(x\) (\(b\)) and whose product is the constant term (\(c\)).

(c) The Zero-Product Property: \[\text{If } A \times B = 0, \text{ then either } A = 0 \text{ or } B = 0.\] Once a quadratic is written as \((x - p)(x - q) = 0\), we set each factor equal to zero to obtain the two roots: \(x = p\) and \(x = q\).

Critical Warning: ALWAYS ensure your quadratic equation is set equal to zero (\(ax^2 + bx + c = 0\)) BEFORE attempting to factorise!

Key Formulas

\(ax^2 + bx + c = 0\) — Standard general form of a quadratic equation (where \(a \neq 0\)).
\(x^2 + (p + q)x + pq = (x + p)(x + q) = 0\) — Monic quadratic factorisation rule (where \(p + q = b\) and \(pq = c\)).
\(a \cdot c \text{ Method}: ax^2 + b_1 x + b_2 x + c = 0\) — Split middle term \(b = b_1 + b_2\) such that \(b_1 \cdot b_2 = a \cdot c\).
\(A \cdot B = 0 \implies A = 0 \text{ or } B = 0\) — Zero-Product Property for finding roots.

Worked Examples

Example 1 (Easy - Monic Quadratic): Solve \(x^2 - 5x + 6 = 0\) by factorisation.
  1. Identify target product and sum: \(\text{Product} = 6\), \(\text{Sum} = -5\).
  2. Find pair: \(-2\) and \(-3\) (since \((-2) \times (-3) = 6\) and \((-2) + (-3) = -5\)).
  3. Write in factored form: \((x - 2)(x - 3) = 0\).
  4. Apply Zero-Product Property: \(x - 2 = 0 \implies x = 2\) or \(x - 3 = 0 \implies x = 3\).
  5. Answer: \(x = 2\) or \(x = 3\).
Example 2 (Medium - Non-Monic Quadratic): Solve \(2x^2 + 5x - 3 = 0\) by factorisation.
  1. Multiply \(a \times c = 2 \times (-3) = -6\).
  2. Find two numbers that multiply to \(-6\) and add to \(+5\): \(+6\) and \(-1\).
  3. Split middle term: \(2x^2 + 6x - 1x - 3 = 0\).
  4. Factor by grouping: \(2x(x + 3) - 1(x + 3) = 0 \implies (2x - 1)(x + 3) = 0\).
  5. Solve linear equations: \(2x - 1 = 0 \implies x = \frac{1}{2}\) ; \(x + 3 = 0 \implies x = -3\).
  6. Answer: \(x = \frac{1}{2}\) or \(x = -3\).
Example 3 (Hard - Geometry Word Problem): A rectangular table top in a workshop has an area of \(12\text{ m}^2\). The length is \(1\text{ m}\) longer than the width \(w\). Find the width of the table.
  1. Set up equation: \(w(w + 1) = 12 \implies w^2 + w = 12\).
  2. Rearrange to standard form: \(w^2 + w - 12 = 0\).
  3. Factorise: Find numbers multiplying to \(-12\) and adding to \(+1\): \(+4\) and \(-3\).
    \((w + 4)(w - 3) = 0\).
  4. Solve: \(w + 4 = 0 \implies w = -4\) (reject, length cannot be negative) or \(w - 3 = 0 \implies w = 3\).
  5. Answer: Width = \(3\text{ m}\).

Common Mistakes

Mistake Attempting to factorise before moving all terms to one side (e.g., trying to factorise \(x^2 + 5x = 6\) as \(x(x + 5) = 6\) and setting \(x = 6\) or \(x + 5 = 6\)).
Correction The Zero-Product Property ONLY works when the equation equals ZERO! First rearrange to \(x^2 + 5x - 6 = 0\), then factorise to \((x + 6)(x - 1) = 0\).
Mistake Forgetting the negative sign when extracting roots from binomial factors (e.g., concluding that \((x + 5) = 0 \implies x = 5\)).
Correction \(x + 5 = 0 \implies x = -5\). The root has the opposite sign of the constant in the linear factor!

Real World

Agricultural Land Division: Kenyan farmers use quadratic factorisation to calculate field dimensions when plot area constraints and side-length differences are specified.
Architecture & Construction: Carpenters and builders in Nairobi determine exact doorway and tabletop dimensions from specified total surface areas.
Physics of Projectiles: Determining the exact flight duration of a launched ball before it touches the ground (height \(h(t) = -5t^2 + 20t = 0 \implies -5t(t - 4) = 0\)).

Practice

Solve the quadratic equation x^2 + 7x + 10 = 0 by factorisation. What is the larger (more positive) root of x? (Type only the number, e.g., -2)
Review the concepts above.
A pharmacist designs a rectangular storage box for medicines. The length is 1 cm longer than the width w, and the box must have an area of 12 cm². What is the width w in centimetres? (Type only the number, e.g., 3)
Review the concepts above.
A carpenter builds a rectangular tabletop whose length is 3 m longer than its width w. The area is 10 m². Find the width w in metres. (Type only the number, e.g., 2)
Review the concepts above.
A teacher in Nairobi asks students to factorise x^2 + 10x + 25 = 0 and solve for x. What is the value of x? (Type only the number, e.g., -5)
Review the concepts above.
Solve the equation x^2 - 10x + 9 = 0 by factorisation. What is the smaller root for x? (Type only the number, e.g., 1)
Review the concepts above.
A SACCO manager determines a performance bonus B in shillings satisfying B^2 - 50B + 600 = 0. Taking the smaller of the two possible positive solutions, what is the bonus amount in shillings? (Type only the number, e.g., 20)
Review the concepts above.