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Maths I — Week 2: Polynomials, Quadratics, and Coordinate Geometry

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Roots, discriminants, Vieta's formulas, and coordinate geometry including section formula, distance, slopes, and perpendicular bisectors. # Week 2: Polynomials and Quadratics ## 1.

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Week 2: Polynomials and Quadratics

1. Anatomy of a Polynomial

A polynomial is an expression of the form anxn+an−1xn−1+⋯+a1x+a0a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 where all exponents are non-negative integers.
  • Degree: The highest power. The degree controls the maximum number of roots.
  • Leading Coefficient: ana_n (the coefficient of the highest power).

2. The Quadratic Equation

Standard form: ax2+bx+c=0ax^2 + bx + c = 0, where a≠0a \neq 0.
The roots are given by the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The Discriminant Δ=b2−4ac\Delta = b^2 - 4ac

The discriminant completely determines the nature of the roots before solving:
Δ\DeltaNature of RootsGraph Behaviour
Δ>0\Delta > 02 distinct real rootsParabola crosses x-axis twice
Δ=0\Delta = 01 repeated real rootParabola touches x-axis (tangent)
Δ<0\Delta < 02 complex conjugate rootsParabola never crosses x-axis
Trap: If Δ<0\Delta < 0 and someone asks "find the roots using the formula", the answer will involve negative\sqrt{\text{negative}} — a complex (imaginary) number, not a real root. Always check Δ\Delta first.

3. Vieta's Formulas (Root Relationships without Finding Roots)

For a quadratic ax2+bx+c=0ax^2 + bx + c = 0 with roots α\alpha and β\beta: α+β=−ba\alpha + \beta = -\frac{b}{a} α⋅β=ca\alpha \cdot \beta = \frac{c}{a}
This lets you reconstruct the equation from the roots or find properties of roots without solving the full quadratic.

4. Factorization Strategies

  1. Factor out GCD: 6x2+9x=3x(2x+3)6x^2 + 9x = 3x(2x + 3).
  2. Product-Sum Method: Find two numbers that multiply to acac and add to bb.
  3. Difference of Squares: a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b).
  4. Perfect Square Trinomial: a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2.

5. Flashcards

<Flashcard front="What does a negative discriminant indicate?" back="No real roots. The parabola does not cross the x-axis. Roots are complex conjugates." /> <Flashcard front="State Vieta's formulas for a quadratic ax² + bx + c = 0." back="Sum of roots = -b/a. Product of roots = c/a." /> <Flashcard front="What is the vertex form of a quadratic?" back="f(x) = a(x - h)² + k, where (h, k) is the vertex." />

6. Problem Solving Combinations & Solutions

Type 1: Find equation given roots

Q: Find a quadratic with roots 33 and −2-2. Solution: Using Vieta's — Sum =1= 1, Product =−6= -6. Equation: x2−x−6=0x^2 - x - 6 = 0.

Type 2: Find properties from equation without solving

Q: For 2x2−5x+3=02x^2 - 5x + 3 = 0, find α2+β2\alpha^2 + \beta^2. Solution: α+β=5/2\alpha + \beta = 5/2, αβ=3/2\alpha\beta = 3/2. α2+β2=(α+β)2−2αβ=25/4−3=13/4\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 25/4 - 3 = 13/4.

Type 3: Nature of roots analysis

Q: For what values of kk does kx2+4x+1=0kx^2 + 4x + 1 = 0 have equal roots? Solution: Equal roots   ⟹  Δ=0  ⟹  16−4k=0  ⟹  k=4\implies \Delta = 0 \implies 16 - 4k = 0 \implies k = 4.

Common Traps

  • a=0a = 0 makes it linear, not quadratic.
  • Repeated roots (Δ=0\Delta=0) means only ONE distinct root, but algebraically it counts as two.
  • Complex roots always come in conjugate pairs for real-coefficient polynomials.

7. Coordinate Geometry Toolkit (Week 2 Graded Focus)

Week 2's graded assignment heavily tests coordinate geometry alongside polynomials. Master these tools.

7.1 Key Formulas

FormulaExpression
Distance between (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2)(x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
Midpoint(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)
Section Formula (Internal)(mx2+nx1m+n,my2+ny1m+n)\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right) for ratio m:nm:n
Slope of a linem=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}
Slope of perpendicularm⊥=−1mm_\perp = -\frac{1}{m}
Point-slope formy−y1=m(x−x1)y - y_1 = m(x - x_1)
Tip
Perpendicular Slope Rule: If the slope of a line is 33, the perpendicular slope is −1/3-1/3. If slope is −1/2-1/2, perpendicular is 22. Always negate and flip.

7.2 Pattern — Perpendicular Bisector

What to recognize: Given two points AA and BB, find the perpendicular bisector.

Abstract Solution (Strategy)

  1. [Find midpoint]: The perpendicular bisector passes through the midpoint of ABAB.
  2. [Find slope of AB]: mABm_{AB}.
  3. [Perpendicular slope]: m⊥=−1/mABm_\perp = -1/m_{AB}.
  4. [Point-slope form]: Write the line through the midpoint with slope m⊥m_\perp.
Worked Example:
Perpendicular bisector of segment joining (1,0)(1, 0) and (5,4)(5, 4).
  • Midpoint: (3,2)(3, 2).
  • Slope of segment: 4−05−1=1\frac{4-0}{5-1} = 1.
  • Perpendicular slope: −1-1.
  • Line: y−2=−1(x−3)  ⟹  y=−x+5y - 2 = -1(x - 3) \implies y = -x + 5.

8. Additional Flashcards

<Flashcard front="State the Internal Section Formula." back="Point dividing AB in ratio m:n is ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))." /> <Flashcard front="If two lines are perpendicular, how are their slopes related?" back="Their slopes are negative reciprocals: m₁ × m₂ = -1." /> <Flashcard front="How do you find α² + β² from Vieta's?" back="(α+β)² - 2αβ. Use sum and product from Vieta's without solving for the roots." />

9. Practice Targets

  • Solve Graded Assignment 2 (Coordinate Geometry + Polynomials).
  • Find α3+β3\alpha^3 + \beta^3 using Vieta's (hint: factor the sum of cubes identity).
  • Determine nature of roots for 5 random quadratics using discriminant only.
  • Prove that the diagonals of a square are perpendicular bisectors using the section and slope formulas.

10. Connections

Connects toHow
Week 1 — FunctionsQuadratics are the first non-linear functions we analyze.
Week 3 — LimitsLimit of a quadratic at a removable discontinuity is resolved by factoring (the same quadratic methods).
Statistics IThe normal distribution curve is bell-shaped — a Gaussian, which is related to e−x2e^{-x^2}.
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