Session 02-02 - Quadratic, Biquadratic & Cubic Equations

Section 02: Equations & Problem-Solving Strategies

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz - 10 Minutes

Quick Review of Essential Skills

10 minutes - individual work, then peer review

Factor completely: \(x^2 - 7x + 12\)
Factor by grouping: \(2x^3 - 6x^2 + x - 3\)
Solve the system: \(\begin{cases} 2x + y = 10 \\ x - y = 2 \end{cases}\)
Complete the square: \(x^2 + 6x + ?\)
Identify \(a\), \(b\), \(c\) in: \(3x^2 - 2x + 5 = 0\)

. . .

These skills are essential for today’s methods!

Homework Discussion - 15 Minutes

Solution from Tasks 02-01

15 minutes - presentation and discussion

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

Learning Objectives

By the end of this session, you can:

Solve quadratic equations by factoring, formula, and completing the square
Use the discriminant to predict the number and type of solutions
Reduce biquadratic equations with the substitution \(u = x^2\)
Factor and solve cubic equations with integer roots

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
Cubic: \(ax^3 + bx^2 + cx + d = 0\) → Up to three 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
\(\Delta > 0\), perfect square	Two rational	Crosses twice	Easily factorable
\(\Delta > 0\), not a square	Two real (irrational)	Crosses twice	Not over integers
\(\Delta = 0\)	One repeated	Touches once	Perfect square
\(\Delta < 0\)	No real solutions	Misses the x-axis	Not over reals

Seeing the Discriminant

. . .

The parabola’s position relative to the x-axis is the discriminant story: crossing twice, touching once, or missing it entirely.

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: \(x = -2, -1, 1, 2\)

Calculator: Accessing the Equation Solver

From MENU:

Press MENU → A (Gleichung/Funkt)
Choose your equation type:
- 1: Gleichungssyst. = Systems of linear equations
- 2: Polynom-Gleich. = Polynomial equations

Calculator: Solving Quadratic Equations

Solve \(x^2 + 2x - 2 = 0\)

Press MENU → A → 2 (Polynom-Gleich.)
Select Grad: 2
Enter coefficients: a = 1, b = 2, c = -2
Press = to see solutions

Calculator: Interpreting Polynomial Solutions

The calculator shows:

\(x_1 = -1 + \sqrt{3}\)
\(x_2 = -1 - \sqrt{3}\)

Press = repeatedly to cycle through all solutions.

. . .

After the two x-values, it also shows you the vertex!

Cubic Equations

From Degree Two to Degree Three

One step up: cubic equations

Form: \(ax^3 + bx^2 + cx + d = 0\)

For simple cubics, try small integer values:

Test \(x = 0, \pm 1, \pm 2, \pm 3\)…
If \(x = a\) works, then \((x - a)\) is a factor
Use long division to find the other factor

For \(x^3 + px^2 + qx + r\):

Look for common factors first
Try grouping in pairs
Example: \(x^3 + 2x^2 - x - 2 = x^2(x + 2) - 1(x + 2) = (x + 2)(x^2 - 1)\)

Recap: Sum & Difference of Cubes

You already know these from Session 01-04

Sum of cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
Difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

Example: Solve \(x^3 - 27 = 0\)

Factor: \((x - 3)(x^2 + 3x + 9) = 0\)
Second factor: \(\Delta = 9 - 36 < 0\) → no real solutions
Only real solution: \(x = 3\)

. . .

The full derivation and practice for these formulas is in Session 01-04 - revisit it there if this recap felt fast.

Solving Cubic Equations

Step-by-step approach

Solve: \(x^3 - 6x^2 + 11x - 6 = 0\)

Try simple values: Test \(x = 1\): \(1 - 6 + 11 - 6 = 0\) ✓
So \((x - 1)\) is a factor! Now we need to find what it multiplies with
Divide: We know \(x^3 - 6x^2 + 11x - 6 = (x - 1) \times ?\)
By inspection or long division: \((x - 1)(x^2 - 5x + 6) = 0\)
Factor the quadratic: \((x - 1)(x - 2)(x - 3) = 0\)
Solutions: \(x = 1, 2, 3\)

Guided Practice

Individual Exercises

Work independently, then we’ll discuss

Solve: \((2x - 6)(x + 4) = 0\)
Solve: \(5x - 15 = 0\)
Solve by factoring: \(x^2 + 7x + 10 = 0\)
Use quadratic formula: \(2x^2 - 3x - 2 = 0\)
Complete the square: \(x^2 - 4x - 5 = 0\)
Solve: \(x^4 - 13x^2 + 36 = 0\)

Spot the Error

Two student solutions - both wrong!

Solution 1: \(\quad x^2 = 5x \;\Rightarrow\; x = 5\)

Solution 2: \(\quad 3x^2 - 12 = 0 \;\Rightarrow\; x^2 - 12 = 0 \;\Rightarrow\; x = \pm\sqrt{12}\)

Find and fix both errors!

Break - 10 Minutes

Practice Session

Practice Set A

10 minutes - Fundamentals

Solve for \(x\):

\(4x - 12 = 0\)
\(-3x + 15 = 0\)
\(\frac{2x - 8}{4} = 3\)

Without solving, determine the number of real solutions:

\(x^2 + 4x + 4 = 0\)
\(x^2 - 3x + 5 = 0\)
\(3x^2 - 12x + 9 = 0\)

Practice Set B: Core Skills

5 minutes - Individually

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

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

Practice Set C: Challenge Problems

Solve these equations:

\((x^2 - 4)(x^2 - 9) = 0\)
\(x^3 - 4x^2 - 5x + 20 = 0\)
\((x - 1)^2(x + 3) = 0\)

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.

Find the profit function \(P = R - C\)
At what production levels does the factory break even?

Practice Set D: Patterns

5 minutes - Individually

Solve these related equations and find the pattern:

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

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

Practice Set E: FSP-Style

10 minutes - individually, full written solutions

Solve completely: \(x^4 - 29x^2 + 100 = 0\)
Determine \(c\) so that \(x^2 - 10x + c = 0\) has exactly one solution, and state it.

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: \[M = -2t^2 + 12t\]

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

. . .

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

Coffee Break - 15 Minutes

Wrap-Up & Key Takeaways

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
Cubic equations: find one root, then factor down to a quadratic
Business applications often involve quadratic models

Final Assessment

5 minutes - individual work

Solve using the most efficient method:

\(x^2 - 8x + 15 = 0\)
\(3x^2 + 2x - 1 = 0\)
\(x^4 - 10x^2 + 9 = 0\)
\(x^3 - 4x = 0\)

Next Session Preview

Session 02-03: Fractional, Radical, Exponential & Logarithmic Equations

Rational equations and domain restrictions
Radical equations and extraneous solutions
Exponential and logarithmic equations
Business applications with rates and growth

. . .

Homework Assignment

Complete Tasks 02-02!

Footnotes

Where \(x\) = units sold (in hundreds).↩︎