Grade 10 core mathematics Numbers and Algebra – Quadratic Expressions and Equations Notes

Specific learning outcomes

• (a) Identify and outline the sub-sub-strands:
  • Quadratic expressions
  • Quadratic identities
  • Factorisation of quadratic expressions
  • Formation of quadratic equations
  • Solution of quadratic equations by factorisation
  • Application of quadratic equations to real-life situations
• (b) Form quadratic expressions from different situations, derive quadratic identities from the concept of area, and apply identities in numerical cases.
• (c) Factorise quadratic expressions from different situations, form quadratic equations, and solve them by factorisation.
• (d) Apply quadratic equations to real-life situations.
• (e) Explore uses of quadratic equations in real life.

1. What is a quadratic expression?

A quadratic expression in one variable x has the form ax² + bx + c where a, b, c are numbers and a ≠ 0. Example: 2x² − 3x + 1.

Quadratic equation

A quadratic equation sets a quadratic expression to zero: ax² + bx + c = 0. The solutions (roots) are values of x that satisfy this equation.

2. Quadratic identities (derived using area model)

We can use areas of squares and rectangles to derive common identities.

3. Forming quadratic expressions from situations (examples)

1. Area problem:
   A square has side (x + 3). Its area = (x + 3)² = x² + 6x + 9 (quadratic expression).
2. Rectangle dimensions:
   Length = x + 2, Breadth = 2x − 1. Area = (x + 2)(2x − 1) = 2x² + 3x − 2.
3. Motion (vertical throw, simplified): height h(t) = −5t² + 20t + 15 (a quadratic expression in time t).

4. Factorisation of quadratic expressions

Goal: write ax² + bx + c as a product of two linear factors if possible:

• Common patterns:
  • Perfect square: a² + 2ab + b² = (a + b)²
  • Difference of squares: a² − b² = (a − b)(a + b)
• General factoring (when a = 1): x² + bx + c → find m, n with m + n = b and mn = c, then (x + m)(x + n).
• When a ≠ 1: use split-the-middle term or grouping or the AC method.

5. Forming and solving quadratic equations by factorisation

Steps to solve ax² + bx + c = 0 by factorisation:

1. Bring the equation to standard form ax² + bx + c = 0.
2. Factor the left-hand side into two linear factors, if possible.
3. Use the zero-product property: if (p)(q) = 0 then p = 0 or q = 0.
4. Solve the resulting linear equations.

6. Applications to real-life situations (simple, Kenyan context)

Quadratic equations model many real-life situations. Here are accessible examples for students age 15:

1. Area/enclosure problem:
   A rectangular maize plot has length (x + 5) m and width (x − 1) m. If area is 84 m², form and solve the quadratic.
   Equation: (x + 5)(x − 1) = 84 → x² + 4x − 5 − 84 = 0 ⇒ x² + 4x − 89 = 0 (solve numerically — this does not factor nicely; use quadratic formula or approximate).
   (Choice of numbers in exercises may lead to factorable quadratics.)
2. Simple projectile (vertical throw) — school sports:
   A ball is thrown up and its height after t seconds is h = −5t² + 10t + 1 (metres). To find times when h = 0, form −5t² + 10t + 1 = 0 and solve (factor or use formula).
3. Profit or revenue approximation:
   If profit P(x) = −2x² + 40x − 120, where x is number of units produced (small scale), set P(x) = 0 to find break-even levels.

7. Guided practice (short exercises)

1. Expand: (x + 4)² and (2x − 3)². (Show working.)
2. Factorise: x² + 7x + 10; x² − 6x + 9; 3x² + 11x + 6.
3. Form a quadratic expression: The area of a rectangle is 3x² + x; if one side is x, what is the other side?
4. Form and solve: A square field has side (x + 2). Its area is 81 m². Find x.
5. Application: A toy car is launched; height h(t) = −5t² + 15t + 2. Find when h = 0 (times when it hits the ground).

8. Answers and notes to guided practice

1. (x + 4)² = x² + 8x + 16. (2x − 3)² = 4x² − 12x + 9.
2. x² + 7x + 10 = (x + 5)(x + 2); x² − 6x + 9 = (x − 3)²; 3x² + 11x + 6 = (3x + 2)(x + 3).
3. The other side = 3x + 1 (since area = side × other side → x × ? = 3x² + x → ? = 3x + 1).
4. Square: (x + 2)² = 81 → x + 2 = ±9. Since side length is positive: x + 2 = 9 ⇒ x = 7 (x + 2 = −9 gives negative side, discard).
5. Solve −5t² + 15t + 2 = 0. Multiply by −1: 5t² − 15t − 2 = 0. This does not factor neatly; use quadratic formula: t = [15 ± sqrt(225 + 40)]/(10) = [15 ± sqrt(265)]/10. (Keep exact or approximate values.)

9. Suggested learning experiences for the classroom (Kenya, age 15)

• Start with area models: let students draw squares and rectangles with sides (a + b) to derive identities visually.
• Use local contexts: maize/vegetable garden plots, school sports (throwing or jumping heights), small business profit problems.
• Pair work: give factorable quadratics for factoring races; encourage different methods (trial pairs, grouping, AC method).
• Problem-solving: set tasks to form quadratic equations from word problems, then solve by factorisation; discuss reasonableness of answers (reject negative lengths where inappropriate).
• Extension: explore how quadratics model projectile motion and simple cost/revenue curves; use spreadsheets to tabulate values and plot parabolas.

10. Exploration and extension

Invite learners to:

• Find real-life examples in their community where areas or profits lead to quadratic expressions.
• Experiment with completing the square to rewrite ax² + bx + c in vertex form to understand maximum/minimum values (e.g., highest point of a thrown ball).
• Use a graphing calculator or mobile app to plot parabolas and link algebraic factors to x-intercepts (roots).