🎯 Polynomial Roots & Vieta

Master polynomial root problems using Vieta’s formulas and symmetric sum techniques.

🎯 Recognition Cues

Look for these patterns:

  • Root relationships: “sum of roots”, “product of roots”
  • Symmetric expressions: $r_1^2 + r_2^2$, $r_1^3 + r_2^3$
  • Parameter problems: “find $k$ such that…”
  • Polynomial construction: “find polynomial with roots…”
  • Coefficient relationships: “if $ax^2 + bx + c = 0$ has roots…”

📋 Template Solution (4 Steps)

  1. Identify the polynomial and its roots
  2. Apply Vieta’s formulas to find basic relationships
  3. Use identities to find desired expressions
  4. Check your answer makes sense

🔍 Common Patterns

Pattern 1: Basic Vieta’s Application

Template: For $ax^2 + bx + c = 0$ with roots $r_1, r_2$:

  • Sum: $r_1 + r_2 = -\frac{b}{a}$
  • Product: $r_1 r_2 = \frac{c}{a}$

Example: If $x^2 - 5x + 6 = 0$ has roots $r$ and $s$, find $r^2 + s^2$

Solution:

  1. Vieta’s: $r + s = 5$, $rs = 6$
  2. Identity: $r^2 + s^2 = (r+s)^2 - 2rs$
  3. Substitute: $r^2 + s^2 = 5^2 - 2(6) = 25 - 12 = 13$

Pattern 2: Higher Degree Polynomials

Template: For $ax^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$:

  • Sum: $r_1 + r_2 + r_3 = -\frac{b}{a}$
  • Sum of products: $r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{c}{a}$
  • Product: $r_1 r_2 r_3 = -\frac{d}{a}$

Example: If $x^3 - 6x^2 + 11x - 6 = 0$ has roots $a, b, c$, find $a^2 + b^2 + c^2$

Solution:

  1. Vieta’s: $a + b + c = 6$, $ab + ac + bc = 11$, $abc = 6$
  2. Identity: $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$
  3. Substitute: $a^2 + b^2 + c^2 = 6^2 - 2(11) = 36 - 22 = 14$

Pattern 3: Parameter Problems

Template: Use Vieta’s to find parameter values

Example: Find $k$ such that $x^2 + kx + 12 = 0$ has roots whose sum is 7

Solution:

  1. Vieta’s: Sum of roots = $-k = 7$
  2. Solve: $k = -7$
  3. Check: Product = $12$ ✓

Pattern 4: Symmetric Sums

Template: Use identities to express higher powers in terms of basic sums

Example: If $r$ and $s$ are roots of $x^2 - 3x + 1 = 0$, find $r^3 + s^3$

Solution:

  1. Vieta’s: $r + s = 3$, $rs = 1$
  2. Identity: $r^3 + s^3 = (r+s)^3 - 3rs(r+s)$
  3. Substitute: $r^3 + s^3 = 3^3 - 3(1)(3) = 27 - 9 = 18$

🎯 AMC-Style Worked Example

Problem: If $x^3 - 4x^2 + 5x - 2 = 0$ has roots $a, b, c$, find the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

Solution:

  1. Vieta’s formulas:

    • $a + b + c = 4$
    • $ab + ac + bc = 5$
    • $abc = 2$

  2. Find common denominator:

    • $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc}$

  3. Substitute known values:

    • $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{5}{2}$

Answer: $\frac{5}{2}$

📊 Common Identities

Quadratic Identities

  • $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2$
  • $r_1^3 + r_2^3 = (r_1 + r_2)^3 - 3r_1 r_2(r_1 + r_2)$
  • $r_1^4 + r_2^4 = (r_1^2 + r_2^2)^2 - 2(r_1 r_2)^2$

Cubic Identities

  • $r_1^2 + r_2^2 + r_3^2 = (r_1 + r_2 + r_3)^2 - 2(r_1 r_2 + r_1 r_3 + r_2 r_3)$
  • $r_1^3 + r_2^3 + r_3^3 = (r_1 + r_2 + r_3)^3 - 3(r_1 + r_2 + r_3)(r_1 r_2 + r_1 r_3 + r_2 r_3) + 3r_1 r_2 r_3$

Reciprocal Sums

  • $\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}$
  • $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{r_1 r_2 + r_1 r_3 + r_2 r_3}{r_1 r_2 r_3}$

⚠️ Common Pitfalls

Pitfall: Wrong signs in Vieta’s formulas

Fix: Remember the negative sign: sum of roots = $-\frac{b}{a}$.

Pitfall: Forgetting to check if roots exist

Fix: Ensure discriminant $\geq 0$ for real roots.

Pitfall: Wrong identity application

Fix: Double-check your algebraic manipulations.

Pitfall: Confusing sum and product

Fix: Sum = $-\frac{b}{a}$, Product = $\frac{c}{a}$ for $ax^2 + bx + c = 0$.

📋 Quick Reference

Vieta’s Formulas (Quadratic)

For $ax^2 + bx + c = 0$ with roots $r_1, r_2$:

  • Sum: $r_1 + r_2 = -\frac{b}{a}$
  • Product: $r_1 r_2 = \frac{c}{a}$

Vieta’s Formulas (Cubic)

For $ax^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$:

  • Sum: $r_1 + r_2 + r_3 = -\frac{b}{a}$
  • Sum of products: $r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{c}{a}$
  • Product: $r_1 r_2 r_3 = -\frac{d}{a}$

Essential Identities

  • $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2$
  • $r_1^3 + r_2^3 = (r_1 + r_2)^3 - 3r_1 r_2(r_1 + r_2)$
  • $\frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}$

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