Quadratic Expressions
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Core Objective: Master the geometric and algebraic foundations of expanding and factorising quadratic expressions of the form \(ax^2 + bx + c\).
Definition: A quadratic expression is a polynomial of degree 2, standardly written as \(ax^2 + bx + c\) where \(a, b, c \in \mathbb{R}\) and \(a \neq 0\). The term \(ax^2\) is the quadratic term, \(bx\) is the linear term, and \(c\) is the constant term.
1. Concrete Scenario: Mzee Juma's Maize Shamba
Imagine Mzee Juma in Kitale is expanding his rectangular vegetable plot. Originally, the plot was a square with side length \(x\) metres (area \(= x^2\) m\(^2\)). He extends the length by \(3\) metres and the width by \(2\) metres. The new total length is \((x + 3)\) metres and the new width is \((x + 2)\) metres.
The total area of the shamba is the product of its new dimensions: \(\text{Area} = (x + 3)(x + 2)\).
2. The Geometric Area Model
When we partition the expanded shamba into four distinct sub-plots, we observe:
- A square plot: \(x \times x = x^2\)
- A top rectangular strip: \(x \times 2 = 2x\)
- A side rectangular strip: \(3 \times x = 3x\)
- A corner rectangular patch: \(3 \times 2 = 6\)
Summing these individual areas gives: \[\text{Total Area} = x^2 + 2x + 3x + 6 = x^2 + 5x + 6\]
Two Sides of the Same Coin:
- Expansion: \((x + 3)(x + 2) \longrightarrow x^2 + 5x + 6\) (Multiplying dimensions to find total area)
- Factorisation: \(x^2 + 5x + 6 \longrightarrow (x + 3)(x + 2)\) (Decomposing total area back into side lengths)
Key Formulas
1. Standard Form of Quadratic Expression:
\[ax^2 + bx + c \quad (a \neq 0)\]Where \(a\) is the quadratic coefficient, \(b\) is the linear coefficient, and \(c\) is the constant term.
2. Binomial Expansion Identities:
\[(x + p)(x + q) = x^2 + (p + q)x + pq\] \[(a + b)^2 = a^2 + 2ab + b^2 \quad \text{(Perfect Square Trinomial)}\] \[(a - b)^2 = a^2 - 2ab + b^2 \quad \text{(Perfect Square Trinomial)}\] \[(a + b)(a - b) = a^2 - b^2 \quad \text{(Difference of Two Squares)}\]3. Factorisation by Grouping (General Trinomial \(ax^2 + bx + c\)):
To factorise \(ax^2 + bx + c\):
- Find two integers \(u\) and \(v\) such that \(u \cdot v = a \cdot c\) (Product) and \(u + v = b\) (Sum).
- Split the middle term: \(ax^2 + ux + vx + c\).
- Factor by grouping: \(x(ax + u) + \dots = (mx + r)(sx + t)\).
4. Discriminant and Nature of Roots:
\[\Delta = b^2 - 4ac\]- If \(\Delta > 0\) and a perfect square: Expression factorises into rational linear factors.
- If \(\Delta = 0\): Expression is a perfect square trinomial \((px + q)^2\).
- If \(\Delta < 0\): Expression cannot be factorised over real numbers.
Worked Examples
Example 1 (Easy — Expansion of Binomials)
Problem: Expand and simplify the expression \((2x + 5)(x - 3)\).
Step-by-step Solution:
- Apply the distributive law (multiply each term in the first bracket by each term in the second):
\[2x(x - 3) + 5(x - 3)\] - Expand each term:
\[= 2x^2 - 6x + 5x - 15\] - Collect and combine like terms (\(-6x + 5x = -x\)):
\[= 2x^2 - x - 15\]
Final Answer: \(2x^2 - x - 15\)
Example 2 (Medium — Factorising Monic Quadratic \(a = 1\))
Problem: Factorise completely: \(x^2 - 9x + 20\).
Step-by-step Solution:
- Identify target sum and target product:
\(\text{Sum } (b) = -9\), \(\text{Product } (c) = +20\). - List factor pairs of \(20\) whose signs yield a positive product and negative sum (both factors must be negative):
\((-1) \times (-20) \implies \text{sum } = -21\)
\((-2) \times (-10) \implies \text{sum } = -12\)
\((-4) \times (-5) \implies \text{sum } = -9\) (Correct pair!) - Write down the linear factors:
\[(x - 4)(x - 5)\]
Final Answer: \((x - 4)(x - 5)\)
Example 3 (Hard — Factorising Non-Monic Quadratic \(a > 1\) with Grouping)
Problem: Factorise completely: \(6x^2 + 7x - 5\).
Step-by-step Solution:
- Find the product \(a \cdot c\):
\(a \cdot c = 6 \times (-5) = -30\). Middle term sum \(b = +7\). - Find two integers that multiply to \(-30\) and add to \(+7\):
Testing factor pairs of \(-30\): \(10 \times (-3) = -30\) and \(10 + (-3) = 7\). - Split the middle linear term \(7x\) into \(+10x - 3x\):
\[6x^2 + 10x - 3x - 5\] - Factor by grouping in pairs:
\[2x(3x + 5) - 1(3x + 5)\] - Extract the common binomial factor \((3x + 5)\):
\[(3x + 5)(2x - 1)\]
Final Answer: \((3x + 5)(2x - 1)\)
Common Mistakes
1. The "Freshman's Dream" Fallacy:
Mistake: Expanding \((a + b)^2\) as \(a^2 + b^2\) or \((x - 3)^2\) as \(x^2 - 9\).
Why it happens: Students falsely distribute the exponent across addition/subtraction, forgetting that squaring a binomial means multiplying the entire expression by itself: \((a+b)(a+b)\).
Correction: Always remember the middle cross-term: \[ (a \pm b)^2 = a^2 \pm 2ab + b^2 \] For instance, \((x - 3)^2 = x^2 - 2(x)(3) + 3^2 = x^2 - 6x + 9\).
2. Ignoring Signs When Identifying Factor Pairs:
Mistake: Factorising \(x^2 - 5x + 6\) as \((x + 2)(x + 3)\) because \(2 \times 3 = 6\).
Why it happens: Looking only at absolute values of the product without ensuring the sum equals \(-5\).
Correction: If the constant term is positive (\(+6\)) and the middle term is negative (\(-5\)), both numbers must be negative: \((-2) \times (-3) = +6\) and \((-2) + (-3) = -5\). Thus, \(x^2 - 5x + 6 = (x - 2)(x - 3)\).
3. Incorrect Grouping Sign Changes:
Mistake: Grouping \(2x^2 - 6x - x + 3\) as \(2x(x - 3) - (x + 3)\).
Why it happens: Forgetting that factoring out \(-1\) changes the sign inside the brackets: \(- (x - 3) = -x + 3\).
Correction: Check by expanding back immediately. The correct grouping is \(2x(x - 3) - 1(x - 3) = (2x - 1)(x - 3)\).
Real World
1. Agribusiness Land Demarcation (Kenya)
A tea cooperative in Kericho plans to build a storage warehouse surrounded by a uniform security perimeter path of width \(x\) metres. If the warehouse is \(20\) m by \(10\) m, the total fenced ground area is \((20 + 2x)(10 + 2x) = 4x^2 + 60x + 200\). By factorising or solving this quadratic expression for a specified budget area, engineers determine the exact allowable perimeter clearance \(x\).
2. Revenue Maximisation in Matatu Transport
A Sacco operating matatus along the Thika Superhighway notices that if they charge Ksh 100 per seat, all 50 seats are filled. For every Ksh 10 fare hike (\(x\)), they lose 2 passengers. The total revenue \(R(x)\) is given by: \[ R(x) = (100 + 10x)(50 - 2x) = -20x^2 + 300x + 5000 \] Factorising and finding the vertex of this quadratic expression allows the Sacco to determine the optimal fare to maximise revenue.
3. Ballistics and Sports Science
During a football match at Nyayo National Stadium, a striker kicks the ball. Its height \(h\) in metres after \(t\) seconds is modeled by \(h = -5t^2 + 20t\). Factorising into \(h = -5t(t - 4)\) reveals immediately that the ball is at ground level (\(h = 0\)) at \(t = 0\) seconds (kick-off) and lands back on the turf at \(t = 4\) seconds.
Practice