Quadratic Formula Derivation — Step-by-Step CAS Verification

Introduction

This tutorial demonstrates the mode=solve feature by deriving the quadratic formula with step-by-step CAS verification. You'll learn why mode=solve is essential for equation-solving derivations and how it detects lost or extraneous solutions.

Why mode=solve?

The quadratic formula derivation involves operations that transform solution sets rather than just expressions. Dividing both sides by a is NOT an algebraic equivalence (the expressions differ by factor a), but the equations have the same solutions when a ≠ 0.

Understanding the Difference

Let's understand why mode=chain fails for equation-solving derivations:

Mode	What It Checks	Use For
chain	Algebraic equivalence: `simplify(expr1 - expr2) == 0`	Factoring, expanding, simplifying
solve	Solution-set equivalence: `solveset(eq1) == solveset(eq2)`	Equation solving, isolating variables

Consider the step "divide both sides by a":

ax² + bx and x² + (b/a)x are NOT algebraically equivalent
But the equations ax² + bx = -c and x² + (b/a)x = -c/a have the SAME solution set

The Derivation

Step 1: Start with the General Quadratic Equation

We begin with the standard form of a quadratic equation where a ≠ 0:

General Quadratic Equation

The starting point of our derivation

ax^2 + bx + c = 0

where $a \neq 0$ .

Step 2: Subtract c from Both Sides

Subtract c

Moving the constant term to the right side

Not Verified

Assumptions: a != 0

1

ax^2 + bx + c = 0

2

ax^2 + bx = -c

Expected Result: ✅ EQUIV (same solution set)

Step 3: Divide Both Sides by a

This is where mode=solve shines. This step would FAIL with mode=chain because the expressions are NOT algebraically equivalent.

Divide by a

This step requires mode=solve because expressions differ by factor a

Not Verified

Assumptions: a != 0

1

ax^2 + bx = -c

2

x^2 + \frac{b}{a}x = -\frac{c}{a}

Expected Result: ✅ EQUIV (same solution set when a ≠ 0)

Why mode=chain Would Fail Here

mode=chain checks if ax² + bx equals x² + (b/a)x. These are NOT equal (they differ by factor a). But mode=solve checks if the equations have the same solutions, which they do.

Step 4: Complete the Square

Add (b/2a)² to both sides:

Complete the Square

Adding (b/2a)² to both sides

Not Verified

Assumptions: a != 0

1

x^2 + \frac{b}{a}x = -\frac{c}{a}

2

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}

Expected Result: ✅ EQUIV

Step 5a: Simplify the Right Side

Combine the fractions on the right side:

Simplify Right Side

Combining fractions

Not Verified

Assumptions: a != 0

1

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}

2

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}

Expected Result: ✅ EQUIV

Step 5b: Factor the Left Side as a Perfect Square

Factor as Perfect Square

The left side is a perfect square trinomial

Not Verified

Assumptions: a != 0

1

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}

2

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Expected Result: ✅ EQUIV

Step 6: Take the Square Root of Both Sides

This is a critical step. We must use ± to capture both solutions:

Take Square Root

Using ± to preserve both solutions

Not Verified

Assumptions: a != 0

1

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

2

x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}

Expected Result: ✅ EQUIV (the ± captures both solutions)

What Happens Without ±

If we wrote x + b/(2a) = sqrt(b² - 4ac)/(2a) (without ±), the CAS would return NARROWS because we'd lose one solution.

Step 7: Solve for x

Solve for x

The final step - isolating x

Not Verified

Assumptions: a != 0

1

x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}

2

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Expected Result: ✅ EQUIV — We've derived the quadratic formula!

Full Derivation Chain

Here's the complete derivation verified in one block:

Complete Quadratic Formula Derivation

All steps verified together

Not Verified

Assumptions: a != 0

1

ax^2 + bx + c = 0

2

ax^2 + bx = -c

3

x^2 + \frac{b}{a}x = -\frac{c}{a}

4

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}

5

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

6

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Expected Result: ✅ EQUIV for all steps

Demonstrating NARROWS and WIDENS

The mode=solve feature can detect when derivation steps lose or introduce solutions.

Example: NARROWS (Lost Solutions)

When we take a square root without ±, we lose solutions:

Lost Solutions Example

Taking square root without ± loses the negative solution

Not Verified

1

x^2 = 4

2

x = 2

Expected Result: ⚠️ NARROWS — The first equation has solutions {-2, 2}, but the second only has {2}. We lost the x = -2 solution.

Example: WIDENS (Extraneous Solutions)

When we square both sides, we can introduce extraneous solutions:

Extraneous Solutions Example

Squaring both sides introduces an extra solution

Not Verified

1

x = 2

2

x^2 = 4

Expected Result: ⚠️ WIDENS — The first equation has solution {2}, but the second has {-2, 2}. We added the extraneous solution x = -2.

Example: Correct Use of ±

Using ± preserves the solution set:

Preserving Solutions with ±

Using ± keeps both solutions

Not Verified

1

x^2 = 4

2

x = \pm 2

Expected Result: ✅ EQUIV — Both equations have the same solution set {-2, 2}.

Domain Options

The domain attribute specifies where to look for solutions:

Domain	Description	Example
R (default)	Real numbers	x² = 4 → {-2, 2}
C	Complex numbers	x² = -1 → {i, -i}
Z	Integers	2x = 4 → {2}
N	Natural numbers	x = 2 → {2}

Example: Complex Domain

Complex Domain Example

Finding solutions in the complex numbers

Not Verified

1

x^2 = -1

2

x^2 + 1 = 0

Expected Result: ✅ EQUIV — Both have solutions {i, -i} in the complex domain.

When to Use Each Mode

Scenario	Mode	Reason
Verify identity: sin²(x) + cos²(x) = 1	equivalence	Expression identity
Factor: x² - 1 = (x-1)(x+1)	chain	Expression rewrite
Simplify: (x²-1)/(x-1) = x+1	chain	Algebraic simplification
Solve: ax² + bx + c = 0 → quadratic formula	solve	Equation derivation
Isolate variable: 2x + 3 = 7 → x = 2	solve	Solution-set transformation

Status Indicators for mode=solve

Status	Meaning	Example
✅ EQUIV	Same solution set (valid transformation)	2x = 4 → x = 2
⚠️ NARROWS	Lost solutions (e.g., forgot ± when taking square root)	x² = 4 → x = 2
⚠️ WIDENS	Extraneous solutions (e.g., squared both sides)	x = 2 → x² = 4
❓ UNKNOWN	Cannot determine relationship	Complex expressions
❌ ERROR	Parse or solve failed	Invalid LaTeX

Multiple Solutions Syntax

Use \pm or \text{or} to express multiple solutions:

Multiple Solutions Syntax

Two ways to express multiple solutions

Using ±: $x = \pm 2$

Using "or": $x = 2 \,\,\text{or}\,\, x = -2$

Both are equivalent to the solution set {-2, 2}.

Best Practices

Always Use mode=solve for Equation Derivations

When your derivation involves dividing both sides, taking roots, or any operation that might change the solution set, use mode=solve.

Include the a ≠ 0 Assumption

For the quadratic formula derivation, always include assumptions="a != 0" since we divide by a.

Use ± When Taking Square Roots

Always use \pm when taking square roots of both sides to preserve all solutions. The CAS will warn you with NARROWS if you forget.

Specify the Solve Variable

Use solve_var=x to explicitly tell the CAS which variable you're solving for. This helps when you have multiple variables like a, b, c, and x.

Next Steps

Now that you understand mode=solve, explore more CAS verification examples in the CAS Verification Demo, which covers mode=equivalence and mode=chain for expression verification.

Quadratic Formula Derivation — Step-by-Step CAS Verification

Introduction

Understanding the Difference

The Derivation

Step 1: Start with the General Quadratic Equation

General Quadratic Equation

Step 2: Subtract c from Both Sides

Subtract c

Step 3: Divide Both Sides by a

Divide by a

Step 4: Complete the Square

Complete the Square

Step 5a: Simplify the Right Side

Simplify Right Side

Step 5b: Factor the Left Side as a Perfect Square

Factor as Perfect Square

Step 6: Take the Square Root of Both Sides

Take Square Root

Step 7: Solve for x

Solve for x

Full Derivation Chain

Complete Quadratic Formula Derivation

Demonstrating NARROWS and WIDENS

Example: NARROWS (Lost Solutions)

Lost Solutions Example

Example: WIDENS (Extraneous Solutions)

Extraneous Solutions Example

Example: Correct Use of ±

Preserving Solutions with ±

Domain Options

Example: Complex Domain

Complex Domain Example

When to Use Each Mode

Status Indicators for mode=solve

Multiple Solutions Syntax

Multiple Solutions Syntax

Best Practices

Next Steps

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