Learn on PengiSaxon Algebra 2Chapter 6: Lessons 51-60, Investigation 6

Investigation 6: Deriving the Quadratic Formula

In this Grade 10 Saxon Algebra 2 investigation, students derive the quadratic formula by using algebra tiles to model completing the square on the general form ax² + bx + c = 0. Students explore perfect square trinomials, discover the relationship between coefficients b and c, and work through a step-by-step algebraic process — dividing by a, subtracting the constant, completing the square, and applying the Square Root Property — to arrive at the standard quadratic formula. The lesson builds conceptual understanding of why the formula works, not just how to apply it.

Section 1

📘 Deriving the Quadratic Formula

New Concept

The most common way of solving a quadratic equation is to use the quadratic formula

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

What’s next

Next, you’ll use the technique of completing the square to derive this powerful formula step-by-step, revealing how it was constructed.

Section 2

quadratic formula

Property

To find the roots of any quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0, you can use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

To solve 2x2+6x3=02x^2 + 6x - 3 = 0, identify a=2,b=6,c=3a=2, b=6, c=-3.
x=(6)±(6)24(2)(3)2(2)=6±36+244x = \frac{-(6) \pm \sqrt{(6)^2 - 4(2)(-3)}}{2(2)} = \frac{-6 \pm \sqrt{36 + 24}}{4}.
x=6±604=6±2154=3±152x = \frac{-6 \pm \sqrt{60}}{4} = \frac{-6 \pm 2\sqrt{15}}{4} = \frac{-3 \pm \sqrt{15}}{2}.

Think of the quadratic formula as the ultimate cheat code for solving these types of equations. Instead of wrestling with factoring or completing the square every single time, you can just plug in your aa, bb, and cc values. It’s a powerful shortcut that works for any quadratic equation you encounter.

Section 3

What is a Perfect Square

Property

A perfect square trinomial comes from squaring a binomial, like (x+n)2=x2+2nx+n2(x + n)^2 = x^2 + 2nx + n^2. The key relationship is that the constant term, cc, is the square of half the coefficient of the xx term, bb. In other words, the magic formula is c=(b2)2c = (\frac{b}{2})^2.

In x2+8x+16x^2 + 8x + 16, half of b=8b=8 is 44, and 42=164^2=16. This is a perfect square: (x+4)2(x+4)^2.
In x2+10x+25x^2 + 10x + 25, half of b=10b=10 is 55, and 52=255^2=25. This is a perfect square: (x+5)2(x+5)^2.
In x212x+36x^2 - 12x + 36, half of b=12b=-12 is 6-6, and (6)2=36(-6)^2=36. This is a perfect square: (x6)2(x-6)^2.

Imagine building a literal square with algebra tiles. A perfect square trinomial is an equation that forms a perfect, gap-free square. This special relationship between the bb and cc terms is the secret to making this happen, and it's the foundation for the 'completing the square' method you'll use later on.

Section 4

Completing the Square

Property

If you have an expression like x2+bxx^2 + bx, you can turn it into a perfect square by adding a specific value. This value is always (b2)2(\frac{b}{2})^2. To keep the equation balanced, you must add this same amount to both sides of the equals sign. This creates a solvable perfect square trinomial.

To solve x2+6x+1=0x^2 + 6x + 1 = 0, first move the constant: x2+6x=1x^2 + 6x = -1.
Add (62)2=9(\frac{6}{2})^2 = 9 to both sides: x2+6x+9=1+9x^2 + 6x + 9 = -1 + 9.
Factor the left side into a perfect square and solve: (x+3)2=8(x+3)^2 = 8.

What happens when an equation isn't a perfect square? You make it one! 'Completing the square' is a clever trick where you find the one missing number needed to create a perfect square on one side. It’s like finding the last puzzle piece that makes the whole picture snap into place.

Section 5

Deriving the Quadratic Formula

Property

The quadratic formula is derived by applying the 'completing the square' method to the general quadratic equation, ax2+bx+c=0ax^2 + bx + c = 0. By systematically isolating the variable xx through algebraic steps, the general solution is revealed. This process proves that the formula works for all quadratic equations.

Start with ax2+bx+c=0ax^2 + bx + c = 0 and divide by aa: x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0.
Move the constant term: x2+bax=cax^2 + \frac{b}{a}x = -\frac{c}{a}.
Add (b2a)2(\frac{b}{2a})^2 to both sides, which simplifies to (x+b2a)2=b24ac4a2(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}, leading to the formula.

The quadratic formula isn't just some random magic spell; it's pure mathematical logic. By taking the generic form of a quadratic equation and using the 'completing the square' technique on it, you can see exactly how the formula is built, step-by-step. It’s like seeing how a master key is designed to fit any lock.

Book overview

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Chapter 6: Lessons 51-60, Investigation 6

  1. Lesson 1

    Lesson 51: Using Synthetic Division

  2. Lesson 2

    Lesson 52: Using Two Special Right Triangles

  3. Lesson 3

    Lesson 53: Performing Compositions of Functions

  4. Lesson 4

    Lesson 54: Using Linear Programming

  5. Lesson 5

    Lesson 55: Finding Probability

  6. Lesson 6

    Lesson 56: Finding Angles of Rotation

  7. Lesson 7

    Lesson 57: Finding Exponential Growth and Decay

  8. Lesson 8

    Lesson 58: Completing the Square (Exploration: Modeling Completing the Square)

  9. Lesson 9

    Lesson 59: Using Fractional Exponents

  10. Lesson 10

    Lesson 60: Distinguishing Between Mutually Exclusive and Independent Events

  11. Lesson 11Current

    Investigation 6: Deriving the Quadratic Formula

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Deriving the Quadratic Formula

New Concept

The most common way of solving a quadratic equation is to use the quadratic formula

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

What’s next

Next, you’ll use the technique of completing the square to derive this powerful formula step-by-step, revealing how it was constructed.

Section 2

quadratic formula

Property

To find the roots of any quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0, you can use the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

To solve 2x2+6x3=02x^2 + 6x - 3 = 0, identify a=2,b=6,c=3a=2, b=6, c=-3.
x=(6)±(6)24(2)(3)2(2)=6±36+244x = \frac{-(6) \pm \sqrt{(6)^2 - 4(2)(-3)}}{2(2)} = \frac{-6 \pm \sqrt{36 + 24}}{4}.
x=6±604=6±2154=3±152x = \frac{-6 \pm \sqrt{60}}{4} = \frac{-6 \pm 2\sqrt{15}}{4} = \frac{-3 \pm \sqrt{15}}{2}.

Think of the quadratic formula as the ultimate cheat code for solving these types of equations. Instead of wrestling with factoring or completing the square every single time, you can just plug in your aa, bb, and cc values. It’s a powerful shortcut that works for any quadratic equation you encounter.

Section 3

What is a Perfect Square

Property

A perfect square trinomial comes from squaring a binomial, like (x+n)2=x2+2nx+n2(x + n)^2 = x^2 + 2nx + n^2. The key relationship is that the constant term, cc, is the square of half the coefficient of the xx term, bb. In other words, the magic formula is c=(b2)2c = (\frac{b}{2})^2.

In x2+8x+16x^2 + 8x + 16, half of b=8b=8 is 44, and 42=164^2=16. This is a perfect square: (x+4)2(x+4)^2.
In x2+10x+25x^2 + 10x + 25, half of b=10b=10 is 55, and 52=255^2=25. This is a perfect square: (x+5)2(x+5)^2.
In x212x+36x^2 - 12x + 36, half of b=12b=-12 is 6-6, and (6)2=36(-6)^2=36. This is a perfect square: (x6)2(x-6)^2.

Imagine building a literal square with algebra tiles. A perfect square trinomial is an equation that forms a perfect, gap-free square. This special relationship between the bb and cc terms is the secret to making this happen, and it's the foundation for the 'completing the square' method you'll use later on.

Section 4

Completing the Square

Property

If you have an expression like x2+bxx^2 + bx, you can turn it into a perfect square by adding a specific value. This value is always (b2)2(\frac{b}{2})^2. To keep the equation balanced, you must add this same amount to both sides of the equals sign. This creates a solvable perfect square trinomial.

To solve x2+6x+1=0x^2 + 6x + 1 = 0, first move the constant: x2+6x=1x^2 + 6x = -1.
Add (62)2=9(\frac{6}{2})^2 = 9 to both sides: x2+6x+9=1+9x^2 + 6x + 9 = -1 + 9.
Factor the left side into a perfect square and solve: (x+3)2=8(x+3)^2 = 8.

What happens when an equation isn't a perfect square? You make it one! 'Completing the square' is a clever trick where you find the one missing number needed to create a perfect square on one side. It’s like finding the last puzzle piece that makes the whole picture snap into place.

Section 5

Deriving the Quadratic Formula

Property

The quadratic formula is derived by applying the 'completing the square' method to the general quadratic equation, ax2+bx+c=0ax^2 + bx + c = 0. By systematically isolating the variable xx through algebraic steps, the general solution is revealed. This process proves that the formula works for all quadratic equations.

Start with ax2+bx+c=0ax^2 + bx + c = 0 and divide by aa: x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0.
Move the constant term: x2+bax=cax^2 + \frac{b}{a}x = -\frac{c}{a}.
Add (b2a)2(\frac{b}{2a})^2 to both sides, which simplifies to (x+b2a)2=b24ac4a2(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}, leading to the formula.

The quadratic formula isn't just some random magic spell; it's pure mathematical logic. By taking the generic form of a quadratic equation and using the 'completing the square' technique on it, you can see exactly how the formula is built, step-by-step. It’s like seeing how a master key is designed to fit any lock.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 6: Lessons 51-60, Investigation 6

  1. Lesson 1

    Lesson 51: Using Synthetic Division

  2. Lesson 2

    Lesson 52: Using Two Special Right Triangles

  3. Lesson 3

    Lesson 53: Performing Compositions of Functions

  4. Lesson 4

    Lesson 54: Using Linear Programming

  5. Lesson 5

    Lesson 55: Finding Probability

  6. Lesson 6

    Lesson 56: Finding Angles of Rotation

  7. Lesson 7

    Lesson 57: Finding Exponential Growth and Decay

  8. Lesson 8

    Lesson 58: Completing the Square (Exploration: Modeling Completing the Square)

  9. Lesson 9

    Lesson 59: Using Fractional Exponents

  10. Lesson 10

    Lesson 60: Distinguishing Between Mutually Exclusive and Independent Events

  11. Lesson 11Current

    Investigation 6: Deriving the Quadratic Formula