Quadratic Expressions and Algebraic Techniques: Complete Course Study Guide

Historical Foundations and Context of Algebra

• Arithmetical Principles and Pronumerals:

  • Algebra follows the structural rules of arithmetic, utilising pronumerals (symbols or letters) to represent unknown numerical values.
  • While modern algebraic notation is compact and efficient, for more than $3500$ years mathematical solutions were communicated almost entirely as descriptive, written explanations of sequential steps.

• Chronological Development of Algebraic Thought:

  • Babylonians ($2000\text{ BCE}$): Solved complex arithmetic problems that mirrored algebraic structures using base-60 arithmetic.
  • Egyptians ($1650\text{ BCE}$): Solved linear problems through numerical "guess-and-check" procedures (the method of false position).
  • Euclid of Alexandria ($350\text{ BCE}$): Solved algebraic questions through geometric explanations, visualising products as rectangular and square geometric areas.
  • Chinese Mathematicians ($250\text{ BCE}$): Made significant progress solving systems of equations and finding numerical roots.
  • Diophantus of Alexandria ($275\text{ CE}$): Pioneered geometric algebra while introducing early syncopated symbols to replace written words for powers and unknowns.
  • Brahmagupta ($628\text{ CE}$): Formulated general arithmetic methods to solve linear and quadratic equations, explicitly incorporating negative numbers and irrational roots.
  • Muhammad ibn Musa al-Khwarizmi ($825\text{ CE}$):
    • Authored the foundational treatise Al-jabr, introducing systematic techniques of reducing (restoring subtracted terms to the other side) and balancing (cancelling like terms on opposite sides) to solve linear and quadratic equations.
    • Established algebra as an independent branch of mathematics.
    • Used the Arabic word for "roots" (visualised as hidden tree roots underground) to describe mathematical unknowns, originating modern terms such as "equation root" and "square root".
  • European Scholars ($\approx 1100\text{ CE}$): First encountered algebra via the Latin translation of Al-Khwarizmi's text, translated as Algebrae by Algoritmi, directly providing the English words algebra and algorithm.
  • Leonardo of Pisa (Fibonacci) ($1225\text{ CE}$): Compiled, recorded, and expanded equation-solving techniques across Europe.
  • 16th-Century Italian Mathematicians: Publicly competed in algebraic tournaments to discover general solutions for cubic and quartic equations, attracting large public audiences.
  • René Descartes and John Wallis ($\text{mid-1600s CE}$): Established modern algebraic symbolic notation (such as using $x, y, z$ for unknowns and $a, b, c$ for constants, along with exponential superscripts), creating contemporary symbolic mathematics.

• Modern Applications:

  • Clean Energy and Nuclear Fusion: Developing nuclear fusion power—aiming to create reactions that generate substantially more clean energy than they consume—relies heavily on systems of non-linear and quadratic equations to model plasma dynamics.
  • Curriculum Mapping (NSW Syllabus):
    • Develops mathematical fluency, modeling, and communication ($\text{MAO-WM-01}$).
    • Simplifies algebraic fractions with numerical denominators and expands algebraic expressions ($\text{MA5-ALG-C-01}$).
    • Operates with algebraic fractions, factorises, and simplifies complex expressions ($\text{MA5-ALG-P-02}$).
    • Forms assumed prerequisite mastery for Stage 6 Advanced Mathematics.

Expanding Binomial Products

• Binomial Definition:

  • A binomial is an algebraic expression containing exactly two terms linked by an addition or subtraction sign (e.g. $x + 5$, $2x - 3$, or $x^2 + 3$).
  • A binomial product is the multiplication of two binomial expressions: $(a + b)(c + d)$.

• The Distributive Law:

  • Expanding binomial products requires multiplying every term in the first binomial by every term in the second binomial:         $(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd$

• Geometric Area Model:

  • The expansion can be represented as the total area of a rectangle of length $(a + b)$ and breadth $(c + d)$, divided into four smaller internal rectangular compartments of areas $ac$, $ad$, $bc$, and $bd$:         $\text{Total Area} = ac + ad + bc + bd$

• Worked Examples (Example 1):

  • $(x + 3)(x + 5) = x^2 + 5x + 3x + 15 = x^2 + 8x + 15$
  • $(x - 4)(x + 7) = x^2 + 7x - 4x - 28 = x^2 + 3x - 28$
  • $(2x - 1)(x - 6) = 2x^2 - 12x - x + 6 = 2x^2 - 13x + 6$
  • $(5x - 2)(3x + 7) = 15x^2 + 35x - 6x - 14 = 15x^2 + 29x - 14$

• Real-World Modeling Applications:

  • Business Profit Modeling: Analysts expand binomial products to determine corporate profitability when both sales volume and profit margin depend on the selling price $c$:         $P = 50(20 - c)(c - 4) = 50(20c - 80 - c^2 + 4c) = -50c^2 + 1200c - 4000$
  • Architectural Extensions: A room measuring $4\,\text{m} \times 5\,\text{m}$ expanded by $x\,\text{m}$ on each side has:         $\text{Area} = (x + 4)(x + 5) = x^2 + 9x + 20$         For an increase of $x = 3\,\text{m}$, $\text{Area} = 3^2 + 9(3) + 20 = 56\,\text{m}^2$, which represents an area increase of $56 - 20 = 36\,\text{m}^2$.
  • Picture Frame Dimensions: A rectangular frame of uniform border width $5\,\text{cm}$ has outer breadth $x\,\text{cm}$ and length $2x\,\text{cm}$:         $\text{Total Area} = 2x \times x = 2x^2\,\text{cm}^2$$\text{Inner Picture Area} = (2x - 10)(x - 10) = 2x^2 - 30x + 100\,\text{cm}^2$
  • Swimming Pool Path: A swimming pool surrounded by a concrete border of uniform width $x\,\text{m}$ within an outer boundary of $15\,\text{m} \times 10\,\text{m}$:         $\text{Pool Area} = (15 - 2x)(10 - 2x) = 150 - 50x + 4x^2$         For $x = 2\,\text{m}$, $\text{Area} = (15 - 4)(10 - 4) = 11 \times 6 = 66\,\text{m}^2$.

• Trinomial Expansions and Triple Products (Enrichment):

  • Expanding a binomial and a trinomial:         $(a + b)(c + d + e) = ac + ad + ae + bc + bd + be$
  • Example: $(x + 1)(x^2 + x + 1) = x^3 + x^2 + x + x^2 + x + 1 = x^3 + 2x^2 + 2x + 1$
  • Example: $(x - 2)(x^2 - x + 3) = x^3 - x^2 + 3x - 2x^2 + 2x - 6 = x^3 - 3x^2 + 5x - 6$
  • Product of three binomials:         $(x + 1)(x + 2)(x + 3) = (x^2 + 3x + 2)(x + 3) = x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 = x^3 + 6x^2 + 11x + 6$

Special Binomial Products: Perfect Squares and Difference of Two Squares

• Perfect Squares:

  • A perfect square is the product of an algebraic expression multiplied by itself: $3^2 = 9$, $a^2$, $(2y)^2 = 4y^2$, $(x - 1)^2$, and $(3 - 2y)^2$.
  • Expansion Identities:$(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$$(a - b)^2 = (a - b)(a - b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2$
  • Rule: The square of a binomial equals the square of the first term, plus (or minus) twice the product of the two terms, plus the square of the second term.
  • Worked Examples (Example 2):
    • $(x - 4)^2 = x^2 - 2(x)(4) + 4^2 = x^2 - 8x + 16$
    • $(2x + 3)^2 = (2x)^2 + 2(2x)(3) + 3^2 = 4x^2 + 12x + 9$

• Difference of Two Squares (DOTS):

  • Multiplying the sum of two terms by their difference results in the difference of their squares:         $(a + b)(a - b) = a^2 - ab + ba - b^2 = a^2 - b^2$
  • Because $ab = ba$, the two middle cross-terms $-ab + ba$ sum to zero.
  • Worked Examples (Example 3):
    • $(x + 2)(x - 2) = x^2 - 2^2 = x^2 - 4$
    • $(3x - 2y)(3x + 2y) = (3x)^2 - (2y)^2 = 9x^2 - 4y^2$

• Geometric Applications:

  • Fabricated Sheet Metal Tray: A square tin sheet measuring $20\,\text{cm} \times 20\,\text{cm}$ has square corners of side length $x\,\text{cm}$ removed to fold into an open tray:
    • Side length of tray base: $(20 - 2x)\,\text{cm}$
    • Base area: $(20 - 2x)^2 = 400 - 80x + 4x^2\,\text{cm}^2$
    • For $x = 3\,\text{cm}$:             $\text{Base Area} = (20 - 6)^2 = 14^2 = 196\,\text{cm}^2$$\text{Volume} = \text{Base Area} \times \text{Height} = 196 \times 3 = 588\,\text{cm}^3$

• Four-Court Sports Complex Layout: Four courts of dimensions $a \times b$ surround a central square storage zone:
  • Overall perimeter side length: $a + b$
  • Total enclosure area: $(a + b)^2 = a^2 + 2ab + b^2$
  • Inner storage zone side length: $a - b$
  • Inner storage zone area: $(a - b)^2 = a^2 - 2ab + b^2$
  • Combined area of the four courts:         $(a + b)^2 - (a - b)^2 = (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) = 4ab$
  • Individual court area: $ab$, confirming internal consistency.

Factorisation Techniques: Common Factors and Grouping in Pairs

• Principles of Factorisation:

  • Factorisation is the exact inverse process of expansion, transforming an expanded sum or difference into a product of factors.
  • First Rule of Factorisation: Always extract the highest common factor (HCF) before applying further techniques.

• Categories of Highest Common Factors:

  • Numerical constant: $40 - 16b = 8(5 - 2b)$
  • Pronumeral/Variable: $x^2 + 5x = x(x + 5)$
  • Product of constants and pronumerals: $2x^2 + 10x = 2x(x + 5)$
  • Negative common factor: $-8x^2 - 12x = -4x(2x + 3)$
  • Binomial factor:         $3(x + y) + x(x + y) = (x + y)(3 + x)$$(7 - 2x) - x(7 - 2x) = 1(7 - 2x) - x(7 - 2x) = (7 - 2x)(1 - x)$

• Reversing Signs Property:

  • Recognise the negative relationship: $(a - b) = -(b - a)$.
  • Application: $x(x - 2) - 5(2 - x) = x(x - 2) + 5(x - 2) = (x - 2)(x + 5)$.

• Factorising by Grouping in Pairs:

  • Employed for expressions containing four distinct terms.
  • Methodology:
    1. Group the four terms into two pairs sharing common factors.
    2. Factor out the HCF from each pair.
    3. Factor out the resulting common binomial factor.
    4. If the initial pairs share no common factor, rearrange the terms.
  • Worked Examples (Examples 9 & 10):
    • $x^2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3)$
    • $x^2 + 3x - 5x - 15 = x(x + 3) - 5(x + 3) = (x + 3)(x - 5)$
    • Rearranging terms for $2x^2 - 9 - 18x + x$:             $2x^2 + x - 18x - 9 = x(2x + 1) - 9(2x + 1) = (2x + 1)(x - 9)$             Alternative arrangement:             $2x^2 - 18x + x - 9 = 2x(x - 9) + 1(x - 9) = (x - 9)(2x + 1)$

• Extended Grouping (Enrichment):

  • Expressions with six terms can be grouped in pairs or triples:         $a^2 - 3ac - 2ab + 6bc + 3abc - 9bc^2 = a(a - 3c) - 2b(a - 3c) + 3bc(a - 3c) = (a - 3c)(a - 2b + 3bc)$

Factorising the Difference of Two Squares

• Fundamental Identity:$a^2 - b^2 = (a + b)(a - b)$

• Execution Rule:

  • Always extract any numerical or algebraic common factor prior to applying the difference of two squares identity.

• Worked Examples (Examples 7 & 8):

  • Standard forms:
    • $x^2 - 4 = x^2 - 2^2 = (x + 2)(x - 2)$
    • $9a^2 - 25 = (3a)^2 - 5^2 = (3a + 5)(3a - 5)$
    • $81x^2 - y^2 = (9x)^2 - y^2 = (9x + y)(9x - y)$
  • Common factor first:
    • $2b^2 - 32 = 2(b^2 - 16) = 2(b^2 - 4^2) = 2(b + 4)(b - 4)$
  • Compound binomials:
    • $(x + 1)^2 - 4 = (x + 1)^2 - 2^2 = ((x + 1) + 2)((x + 1) - 2) = (x + 3)(x - 1)$
  • Reordered negative terms:
    • $-4 + 9x^2 = 9x^2 - 4 = (3x + 2)(3x - 2)$

• Common Error Warning:

  • Factorising $16x^2 - 4$ directly as $(4x + 2)(4x - 2)$ is incomplete because common factors remain inside each bracket: $(4x + 2) = 2(2x + 1)$ and $(4x - 2) = 2(2x - 1)$, producing $2 \times 2(2x + 1)(2x - 1) = 4(2x + 1)(2x - 1)$. Taking the HCF out first eliminates this error:         $16x^2 - 4 = 4(4x^2 - 1) = 4(2x + 1)(2x - 1)$

• Engineering and Physical Applications:

  • Turbine Annular Surface Area: Power station turbine discs have ring cross-sections with outer radius $R$ and inner radius $r$:         $A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi(R + r)(R - r)$
  • Free-Falling Projectile Height: An object dropped from a building has height $h = 36 - 4t^2$:         $h = 4(9 - t^2) = 4(3 - t)(3 + t)$         Initial height ($t = 0$): $36\,\text{m}$. Ground impact ($h = 0$): $t = 3\,\text{s}$.

Factorising Monic Quadratic Trinomials

• Monic Quadratic Trinomial Definition:

  • A monic quadratic trinomial is a polynomial of degree $2$ with a leading coefficient of $1$, taking the general form:         $x^2 + bx + c$

• Factorisation Principle:

  • To factorise $x^2 + bx + c$ into $(x + p)(x + q)$, identify two numbers $p$ and $q$ such that:         $p \times q = c \quad \text{and} \quad p + q = b$

• Sign Analysis Framework:

  • When $c > 0$ (Constant is positive):
    • Both factors must share the same sign.
    • If $b > 0$, both factors are positive: $x^2 + 5x + 6 = (x + 2)(x + 3)$.
    • If $b < 0$, both factors are negative: $x^2 - 5x + 6 = (x - 2)(x - 3)$.
  • When $c < 0$ (Constant is negative):
    • The two factors must have opposite signs.
    • The factor with the larger magnitude takes the sign of $b$:             $x^2 + 2x - 8 = (x + 4)(x - 2)$$x^2 - 7x - 18 = (x - 9)(x + 2)$

• Worked Examples (Examples 11 & 12):

  • $x^2 + 7x + 10 = (x + 5)(x + 2)$
  • $x^2 + 2x - 8 = (x + 4)(x - 2)$
  • $x^2 - 7x + 10 = (x - 2)(x - 5)$
  • Common factor removal:         $2x^2 - 2x - 12 = 2(x^2 - x - 6) = 2(x - 3)(x + 2)$

• Completing the Square (Enrichment):

  • Transforms non-factorisable or general trinomials into the form $(x + p)^2 + q$ by adding and subtracting $\left(\frac{b}{2}\right)^2$:         $x^2 - 6x - 8 = x^2 - 6x + \left(-\frac{6}{2}\right)^2 - \left(-\frac{6}{2}\right)^2 - 8 = (x - 3)^2 - 9 - 8 = (x - 3)^2 - 17$

Factorising Non-Monic Quadratic Trinomials

• Definition:

  • A non-monic quadratic trinomial takes the form $ax^2 + bx + c$, where the leading coefficient $a \neq 1$ and $a$ cannot be factored out as a common factor.

• Method 1: Grouping in Pairs (The $ac$ Method):

  1. Compute the product of the leading coefficient and the constant: $a \times c$.
  2. Find two numbers that multiply to give $ac$ and add to give $b$.
  3. Split the middle linear term $bx$ into these two parts.
  4. Factorise by grouping in pairs.
  • Worked Examples (Examples 13 & 14):
    • Factorise $2x^2 + 7x + 3$:             $ac = 2 \times 3 = 6$. Factors of $6$ that sum to $7$ are $1$ and $6$.             $2x^2 + 7x + 3 = 2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$
    • Factorise $4x^2 + 5x - 6$:             $ac = 4 \times (-6) = -24$. Factors of $-24$ summing to $5$ are $8$ and $-3$.             $4x^2 + 5x - 6 = 4x^2 + 8x - 3x - 6 = 4x(x + 2) - 3(x + 2) = (x + 2)(4x - 3)$
    • Factorise $6x^2 - 17x + 12$:             $ac = 6 \times 12 = 72$. Factors of $72$ summing to $-17$ are $-9$ and $-8$.             $6x^2 - 17x + 12 = 6x^2 - 9x - 8x + 12 = 3x(2x - 3) - 4(2x - 3) = (2x - 3)(3x - 4)$

• Method 2: The Algebraic Fraction Method:

  • Set up a fraction with $a$ in the denominator and $ax$ in both binomials:         $\frac{(ax + m)(ax + n)}{a}$
  • Insert the factors $m$ and $n$ (where $m \times n = ac$ and $m + n = b$).
  • Factorise out the common factor from the numerator to cancel the denominator $a$.
  • Example for $5x^2 + 13x - 6$ ($ac = -30$, factors $15$ and $-2$):         $\frac{(5x - 2)(5x + 15)}{5} = \frac{(5x - 2) \cdot 5(x + 3)}{5} = (5x - 2)(x + 3)$

• Method 3: The Cross Method (Enrichment):

  • Decompose $ax^2$ into factors in the first vertical column and $c$ into factors in the second vertical column.
  • Cross-multiply diagonally and sum the products to match $bx$.

• Example for $6x^2 - x - 15$:     Vertical factors of $6x^2$: $2x$ and $3x$     Vertical factors of $-15$: $3$ and $-5$     Cross-product check: $(2x)(-5) + (3x)(3) = -10x + 9x = -1x$     Binomial factors: $(2x + 3)(3x - 5)$

Simplifying Algebraic Fractions: Multiplication and Division

• Core Rule of Simplification:

  • Cancel only common factors, never individual terms.
  • Incorrect Cancelling: $\frac{2x + 4}{2} \neq 2x + 2$
  • Correct Cancelling: $\frac{2x + 4}{2} = \frac{2(x + 2)}{2} = x + 2$

• Protocol for Multiplication:

  1. Completely factorise all polynomials in both numerators and denominators.
  2. Cancel common factors between any numerator and denominator.
  3. Multiply the remaining numerators and denominators:         $\frac{A}{B} \times \frac{C}{D} = \frac{A \cdot C}{B \cdot D}$

• Protocol for Division:

  1. Invert the fraction following the division symbol to obtain its reciprocal.
  2. Convert the division operation into multiplication:         $\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C}$
  3. Factorise, cancel common factors, and multiply.

• Worked Examples (Examples 15 & 16):

  • Single fractions:
    • $\frac{3(x + 2)(x - 4)}{6(x - 4)} = \frac{x + 2}{2}$
    • $\frac{20 - 5x}{8 - 2x} = \frac{5(4 - x)}{2(4 - x)} = \frac{5}{2}$
    • $\frac{x^2 - 4}{x + 2} = \frac{(x + 2)(x - 2)}{x + 2} = x - 2$
  • Multiplication:
    • $\frac{3(x - 1)}{x + 2} \times \frac{4(x + 2)}{9(x - 1)(x - 7)} = \frac{4}{3(x - 7)}$
  • Division:
    • $\frac{(x - 3)(x + 4)}{x(x + 7)} \div \frac{3(x + 4)}{x + 7} = \frac{(x - 3)(x + 4)}{x(x + 7)} \times \frac{x + 7}{3(x + 4)} = \frac{x - 3}{3x}$
  • Multiplication with Quadratic Factoring:
    • $\frac{x^2 - 4}{25} \times \frac{5x + 5}{x^2 - x - 2} = \frac{(x - 2)(x + 2)}{25} \times \frac{5(x + 1)}{(x - 2)(x + 1)} = \frac{x + 2}{5}$

Simplifying Algebraic Fractions: Addition and Subtraction

• Procedure:

  1. Find the Lowest Common Denominator (LCD).
  2. Convert each fraction into an equivalent fraction with the LCD.
  3. Combine numerators over the common denominator.
  4. Expand and collect like terms in the numerator.
  5. Factorise the resulting numerator to check for cancellation.

• Denominators with Numerals:

  • Example: $\frac{x}{4} - \frac{2x}{5} = \frac{5x}{20} - \frac{8x}{20} = -\frac{3x}{20}$
  • Example: $\frac{7x}{3} + \frac{x}{6} = \frac{14x}{6} + \frac{x}{6} = \frac{15x}{6} = \frac{5x}{2}$

• Denominators with Pronumerals:

  • Distinct variable factors ($\text{LCD of } x \text{ and } 2x \text{ is } 2x$):         $\frac{2}{x} - \frac{5}{2x} = \frac{4}{2x} - \frac{5}{2x} = -\frac{1}{2x}$
  • Powers of variables ($\text{LCD of } x \text{ and } x^2 \text{ is } x^2$):         $\frac{2}{x} + \frac{3}{x^2} = \frac{2x}{x^2} + \frac{3}{x^2} = \frac{2x + 3}{x^2}$

• Compound Numerators and the Subtraction Precaution:

  • Warning: When subtracting a fraction with multiple terms in the numerator, enclose the entire numerator in brackets to distribute the negative sign properly:         $\frac{x - 1}{3} - \frac{x + 4}{5} = \frac{5(x - 1) - 3(x + 4)}{15} = \frac{5x - 5 - 3x - 12}{15} = \frac{2x - 17}{15}$Common Error: Writing $-3(x + 4) = -3x + 12$ rather than $-3x - 12$.         $\frac{2x - 3}{6} - \frac{3 - x}{5} = \frac{5(2x - 3) - 6(3 - x)}{30} = \frac{10x - 15 - 18 + 6x}{30} = \frac{16x - 33}{30}$

• Binomial Denominators:

  • Distinct linear factors ($\text{LCD } = (x + 1)(x - 2)$):         $\frac{4}{x + 1} + \frac{3}{x - 2} = \frac{4(x - 2) + 3(x + 1)}{(x + 1)(x - 2)} = \frac{4x - 8 + 3x + 3}{(x + 1)(x - 2)} = \frac{7x - 5}{(x + 1)(x - 2)}$
  • Repeated squared binomial factors ($\text{LCD } = (x - 1)^2$):         $\frac{3}{(x - 1)^2} - \frac{2}{x - 1} = \frac{3 - 2(x - 1)}{(x - 1)^2} = \frac{3 - 2x + 2}{(x - 1)^2} = \frac{5 - 2x}{(x - 1)^2}$

Equations Involving Algebraic Fractions

• Solution Strategies:

  • Strategy 1 (Multiply by the LCD): Determine the LCD across all fractions on both sides of the equation. Multiply every single term on both sides by the LCD to eliminate all denominators.
  • Strategy 2 (Cross-Multiplication): If an equation consists of a single rational fraction on each side ($\frac{A}{B} = \frac{C}{D}$), cross-multiply: $A \cdot D = B \cdot C$.

• Worked Examples (Examples 21 & 22):

  • Linear numerators with constant denominators:         $\frac{2x}{3} + \frac{x}{2} = 7$         Multiply every term by $\text{LCD} = 6$:         $2(2x) + 3(x) = 7(6) \implies 4x + 3x = 42 \implies 7x = 42 \implies x = 6$
  • Binomial numerators with subtraction:         $\frac{x - 2}{5} - \frac{x - 1}{3} = 1$         Multiply every term by $\text{LCD} = 15$:         $3(x - 2) - 5(x - 1) = 15 \implies 3x - 6 - 5x + 5 = 15 \implies -2x - 1 = 15 \implies -2x = 16 \implies x = -8$
  • Algebraic terms in denominators:         $\frac{5}{2x} - \frac{4}{3x} = 2$         Multiply every term by $\text{LCD} = 6x$:         $3(5) - 2(4) = 2(6x) \implies 15 - 8 = 12x \implies 7 = 12x \implies x = \frac{7}{12}$
  • Binomial expressions in denominators:         $\frac{3}{x + 1} = \frac{2}{x + 4}$         Cross-multiply:         $3(x + 4) = 2(x + 1) \implies 3x + 12 = 2x + 2 \implies x = -10$

• Literal Equations (Solving for $x$ in terms of constants):

  • $\frac{x}{a} - \frac{x}{2a} = b \implies \frac{2x - x}{2a} = b \implies \frac{x}{2a} = b \implies x = 2ab$
  • $\frac{ax}{b} - \frac{cx}{2} = d \implies x\left(\frac{2a - bc}{2b}\right) = d \implies x = \frac{2bd}{2a - bc}$
  • $\frac{x - a}{b} = \frac{x}{c} \implies c(x - a) = bx \implies cx - ac = bx \implies x(c - b) = ac \implies x = \frac{ac}{c - b}$

Applied Problems, Mathematical Investigations, and Challenges

• Ice-Cream Vendor Profit Optimization:

  • Profit function: $P = -120x^2 + 840x - 1200$, where $x$ is the selling price in dollars (\).
  • If given away free ($x = 0$): $P = -\$1200$ (fixed overhead costs).
  • Specific price evaluations:
    • At $x = 3$: $P = -120(9) + 840(3) - 1200 = \$240$
    • At $x = 4.5$: $P = -120(20.25) + 840(4.5) - 1200 = \$150$
  • Maximum profit price: The axis of symmetry occurs at $x = 3.5$:         $P(3.5) = -120(3.5)^2 + 840(3.5) - 1200 = \$270$
  • Factorisation and break-even points:         $P = -120(x^2 - 7x + 10) = -120(x - 2)(x - 5)$         A profit ($P > 0$) requires $(x - 2)(x - 5) < 0$, meaning the vendor must set prices strictly between $\$2.00$ and $\$5.00$.

• Vegetable Patch Optimization with Fixed Fencing:

  • Perimeter constraint: $8\,\text{m}$ of wire netting borders three sides of a rectangular plot adjacent to an existing brick wall.
  • Dimensions: Breadth $x\,\text{m}$, Length $L = 8 - 2x\,\text{m}$.
  • Area function: $A = x(8 - 2x) = 8x - 2x^2\,\text{m}^2$.
  • Physical boundaries: $x > 0$ and $8 - 2x > 0 \implies 0 < x < 4$.
  • Maximum area: Symmetrical peak occurs at breadth $x = 2\,\text{m}$, producing length $L = 4\,\text{m}$ and maximum area $A = 8\,\text{m}^2$.

• Square Castle Moat Investigation:

*   A square castle measuring $40\,\text{m} \times 40\,\text{m}$ has area $1600\,\text{m}^2$.
*   Surrounded by a moat of uniform breadth $x\,\text{m}$, the total outer side length is $40 + 2x$.
*   Total occupied area: $(40 + 2x)^2 = 1600 + 160x + 4x^2\,\text{m}^2$.
*   Water surface area of the moat alone:

        $\text{Moat Area} = (1600 + 160x + 4x^2) - 1600 = 4x^2 + 160x = 4x(40 + x)$ * Threshold condition: Moat area must be at least $4$ times the castle area:         $4x(40 + x) \ge 4(1600) \implies 4x^2 + 160x \ge 6400 \implies x^2 + 40x - 1600 \ge 0$ * Exact boundary root:         $x = \frac{-40 + \sqrt{40^2 - 4(1)(-1600)}}{2} = -20 + 20\sqrt{5} \approx 24.72\,\text{m}$         The smallest integer width satisfying the defensive requirement is $x = 25\,\text{m}$.

• Number Theory and Proof Challenges:
  • Difference of Consecutive Squares:$(n + 1)^2 - n^2 = (n + 1 - n)(n + 1 + n) = 2n + 1 = 97 \implies 2n = 96 \implies n = 48$         The two numbers are $48$ and $49$.
  • Divisibility Proof for Prime $n > 3$:
    • Factorisation: $n^2 - 1 = (n - 1)(n + 1)$.
    • Because any prime $n > 3$ is odd, $(n - 1)$ and $(n + 1)$ are consecutive even integers. One must be a multiple of $2$ and the other a multiple of $4$, making their product divisible by $8$ (and therefore divisible by $4$).
    • In any sequence of three consecutive integers $(n - 1), n, (n + 1)$, one integer must be divisible by $3$. Because $n$ is prime and $n > 3$, $n$ cannot be divisible by $3$. Thus, either $(n - 1)$ or $(n + 1)$ must be divisible by $3$.
    • Because $(n^2 - 1)$ is divisible by both $4$ and $3$, and $\gcd(4, 3) = 1$, it is necessarily divisible by $12$ (and in fact by $24$).
  • Identity Challenge Proof:         Show that the product of the rational chain equals $1$:         $\frac{2x^2 - 8}{5x^2 - 5} \div \frac{x - 2}{5x - 5} \div \frac{2x^2 - 10x - 28}{x^2 - 6x - 7} = 1$Proof:$= \frac{2(x - 2)(x + 2)}{5(x - 1)(x + 1)} \times \frac{5(x - 1)}{x - 2} \times \frac{(x - 7)(x + 1)}{2(x - 7)(x + 2)} = 1$