Session 02-03 - Quadratic & Biquadratic Equations

Section 02: Equations & Problem-Solving Strategies

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review of Essential Skills

10 minutes - individual work, then peer review

  1. Factor completely: \(x^2 - 7x + 12\)

  2. Factor by grouping: \(2x^3 - 6x^2 + x - 3\)

  3. Solve the system: \(\begin{cases} 2x + y = 10 \\ x - y = 2 \end{cases}\)

  4. Complete the square: \(x^2 + 6x + ?\)

  5. Identify \(a\), \(b\), \(c\) in: \(3x^2 - 2x + 5 = 0\)

These skills are essential for today’s methods!

Homework Presentations

Solutions from Tasks 02-02

30 minutes - presentation and discussion

  • Present your most challenging problem
  • Share alternative solution methods
  • Discuss any conceptual difficulties
  • Ask questions about problems you struggled with

Key Concepts

Equation Types Overview

Today’s new topics:

  • Linear: \(ax + b = 0\) → One solution
  • Quadratic: \(ax^2 + bx + c = 0\) → Up to two solutions
  • Biquadratic: \(ax^4 + bx^2 + c = 0\) → Up to four solutions

Why equal to zero?

Good question! Either we want to determine the intersection of the graph and the x-axis (hence y=0) or we try to make an equation equal to zero to determine the value of x easily.

Solving Equations

Zero Form & Linear Equations

The Zero Product Property

If \(A \cdot B = 0\), then \(A = 0\) or \(B = 0\)

Example: Solve \(3x - 6 = 0\)

  • Factor: \(3(x - 2) = 0\)
  • Apply property: \(x - 2 = 0\)
  • Solution: \(x = 2\)

This principle extends to all equation types!

Three Methods for Quadratics

Let’s solve the same equation three ways: \(x^2 - 5x + 6 = 0\)

When to use: Integer coefficients, factorable, fastest

  • Factor: \((x - 2)(x - 3) = 0\)
  • Apply Zero Product Property
  • Solutions: \(x = 2\) or \(x = 3\)

When to use: Always works, but is slower

  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  • \(a = 1\), \(b = -5\), \(c = 6\)
  • \(x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}\)
  • Solutions: \(x = 3\) or \(x = 2\)

When to use: Only in special cases (my recomendation)

  • \(x^2 - 5x = -6\)
  • \(x^2 - 5x + \frac{25}{4} = -6 + \frac{25}{4} = \frac{1}{4}\)
  • \((x - \frac{5}{2})^2 = \frac{1}{4}\)
  • \(x - \frac{5}{2} = \pm\frac{1}{2}\)
  • Solutions: \(x = 3\) or \(x = 2\)

The Discriminant

For \(ax^2 + bx + c = 0\), the discriminant \(\Delta = b^2 - 4ac\) tells us:

\(\Delta\) Value Solution Type Graph Behavior Factorability
\(\Delta > 0\) and perfect square Two rational solutions Crosses x-axis twice Easily factorable
\(\Delta > 0\) but not perfect square Two real (irrational) solutions Crosses x-axis twice Not factorable over integers
\(\Delta = 0\) One repeated real solution Touches x-axis once Perfect square factorization
\(\Delta < 0\) No real solutions Doesn’t touch x-axis Not factorable over reals

Method Selection Guide

Which method should you use?

Quadratic Equation: \(ax^2 + bx + c = 0\)

Calculate Δ = b² - 4ac
│
├─ Δ < 0 → No real solutions
│
├─ Δ = 0 → One solution: x = -b/(2a) (Perfect square trinomial)
│
└─ Δ > 0 → Two real solutions
           │
           └─ Is Δ a perfect square?
              │
              ├─ YES → Try factoring first
              │
              └─ NO → Use quadratic formula

Interested in more details and the origin of the quadratic formula? Head over here

Biquadratic Equations

Extending to fourth-degree

Form: \(ax^4 + bx^2 + c = 0\)

Strategy: Substitution!

  • Let \(u = x^2\)
  • Solve \(au^2 + bu + c = 0\)
  • Back-substitute to find \(x\)

Solving Biquadratic Equations

Let’s Look at an Example

Example: \(x^4 - 5x^2 + 4 = 0\)

  • Let \(u = x^2\): \(u^2 - 5u + 4 = 0\)
  • Factor: \((u - 1)(u - 4) = 0\)
  • So \(u = 1\) or \(u = 4\)
  • If \(x^2 = 1\): \(x = \pm 1\)
  • If \(x^2 = 4\): \(x = \pm 2\)
  • Four solutions: \(x = -2, -1, 1, 2\)

Guided Practice

Individual Exercises

Work independently, then we’ll discuss

  1. Solve: \((2x - 6)(x + 4) = 0\)

  2. Solve: \(5x - 15 = 0\)

  3. Solve by factoring: \(x^2 + 7x + 10 = 0\)

  4. Use quadratic formula: \(2x^2 - 3x - 2 = 0\)

  5. Complete the square: \(x^2 - 4x - 5 = 0\)

  6. Solve: \(x^4 - 13x^2 + 36 = 0\)

Break - 10 Minutes

Practice Session

Practice Set A

10 minutes - Fundamentals

  1. Solve for \(x\):
  1. \(4x - 12 = 0\)
  2. \(-3x + 15 = 0\)
  3. \(\frac{2x - 8}{4} = 3\)
  1. Without solving, determine the number of real solutions:
  1. \(x^2 + 4x + 4 = 0\)
  2. \(x^2 - 3x + 5 = 0\)
  3. \(3x^2 - 12x + 9 = 0\)

Practice Set B: Core Skills

5 minutes - Individually

Solve each using the most efficient method and justify your choice:

  1. \(x^2 - 11x + 30 = 0\)
  2. \(2x^2 + 5x - 3 = 0\)
  3. \(x^2 - 6x + 9 = 0\)
  4. \(3x^2 - 7x + 1 = 0\)

Practice Set C: Challenge Problems

  1. Solve these equations:
  1. \((x^2 - 4)(x^2 - 9) = 0\)
  2. \(x^3 - 4x^2 - 5x + 20 = 0\)
  3. \((x - 1)^2(x + 3) = 0\)
  1. A factory produces custom widgets. The cost function is \(C = x^2 - 40x + 500\) and the revenue function is \(R = 20x\), where \(x\) is the number of units.
  1. Find the profit function \(P = R - C\)
  2. At what production levels does the factory break even?

Practice Set D: Patterns

5 minutes - Individually

Solve these related equations and find the pattern:

  1. \(x^2 - 5x + 6 = 0\)
  2. \(x^2 - 5x + 4 = 0\)
  3. \(x^2 - 5x + 0 = 0\)
  4. \(x^2 - 5x - 6 = 0\)

What do you notice about the solutions as the constant term changes?

Application & Extension

Break Even

Real-world quadratic application

A company’s profit function1 \(P = -2x^2 + 120x - 1600\)

Find break-even points.

  • Break-even: Solve \(-2x^2 + 120x - 1600 = 0\)
  • Divide by -2: \(x^2 - 60x + 800 = 0\)
  • Using formula: \(x = \frac{60 \pm \sqrt{3600 - 3200}}{2} = \frac{60 \pm 20}{2}\)
  • Break-even at \(x = 20\) or \(x = 40\) (2,000 or 4,000 units)

Collaborative Problem-Solving

Market Analysis Challenge

Work in groups

A new product’s market share \(M\) after \(t\) months follows: \[M = -2t^2 + 12t\]

  1. Find when market share is zero (factor completely)
  2. Graph the market lifecycle

We’ll explore finding the maximum profit point when we study quadratic functions in Section 03.

Coffee Break - 15 Minutes

Final Assessment

5 minutes - individual work

Solve using the most efficient method:

  1. \(x^2 - 8x + 15 = 0\)

  2. \(3x^2 + 2x - 1 = 0\)

  3. \(x^4 - 10x^2 + 9 = 0\)

Wrap-up & Synthesis

Key Takeaways

Essential skills mastered today

  • Zero Product Property is fundamental to all equation solving
  • Three methods for quadratics - each has its place
  • Discriminant predicts solution behavior
  • Biquadratic equations use substitution strategy
  • Business applications often involve quadratic models

Next Session Preview

Session 02-04: Fractional, Radical & Cubic Equations

  • Solving rational equations
  • Domain restrictions
  • Asymptotic behavior
  • Business applications with rates