Learn on PengienVision, Algebra 1Chapter 9: Solving Quadratic Equations

Lesson 6: The Quadratic Formula and the Discriminant

In this Grade 11 enVision Algebra 1 lesson from Chapter 9, students learn to apply the quadratic formula to solve any quadratic equation in standard form, including deriving the formula by completing the square. The lesson also introduces the discriminant as a tool for determining the number of real solutions a quadratic equation has. Students practice solving real-world problems, such as modeling projectile motion, by substituting values of a, b, and c into the quadratic formula.

Section 1

Quadratic Formula

Property

The solutions to a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, where a0a \neq 0 are given by the formula:

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

To solve a quadratic equation using the Quadratic Formula:
Step 1. Write the quadratic equation in standard form, ax2+bx+c=0ax^2 + bx + c = 0. Identify the values of aa, bb, and cc.
Step 2. Write the Quadratic Formula. Then substitute in the values of aa, bb, and cc.
Step 3. Simplify.
Step 4. Check the solutions.

Examples

  • To solve 2x2+5x3=02x^2 + 5x - 3 = 0, we identify a=2,b=5,c=3a=2, b=5, c=-3. Substituting into the formula gives x=5±524(2)(3)2(2)=5±25+244=5±494=5±74x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}. The solutions are x=12x = \frac{1}{2} and x=3x = -3.
  • To solve 3x2+10x+5=03x^2 + 10x + 5 = 0, we have a=3,b=10,c=5a=3, b=10, c=5. The formula gives x=10±1024(3)(5)2(3)=10±100606=10±406=10±2106=5±103x = \frac{-10 \pm \sqrt{10^2 - 4(3)(5)}}{2(3)} = \frac{-10 \pm \sqrt{100 - 60}}{6} = \frac{-10 \pm \sqrt{40}}{6} = \frac{-10 \pm 2\sqrt{10}}{6} = \frac{-5 \pm \sqrt{10}}{3}.
  • To solve x2+2x+10=0x^2 + 2x + 10 = 0, we have a=1,b=2,c=10a=1, b=2, c=10. The formula gives x=2±224(1)(10)2(1)=2±4402=2±362=2±6i2=1±3ix = \frac{-2 \pm \sqrt{2^2 - 4(1)(10)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 40}}{2} = \frac{-2 \pm \sqrt{-36}}{2} = \frac{-2 \pm 6i}{2} = -1 \pm 3i.

Explanation

The Quadratic Formula is a powerful tool derived from completing the square on the general quadratic equation. It provides a direct solution for any quadratic equation, saving you from repeating the steps of completing the square every time.

Section 2

The quadratic formula

Property

The solutions of the equation

ax2+bx+c=0,a0ax^2 + bx + c = 0, \quad a \neq 0

are given by the formula

Section 3

The Discriminant

Property

In the Quadratic Formula, x=b±b24ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, the quantity b24acb^2 - 4ac is called the discriminant.
For a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, a0a \neq 0:

  • If b24ac>0b^2 - 4ac > 0, the equation has 2 real solutions.
  • If b24ac=0b^2 - 4ac = 0, the equation has 1 real solution.
  • If b24ac<0b^2 - 4ac < 0, the equation has no real solutions.

Examples

Book overview

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Chapter 9: Solving Quadratic Equations

  1. Lesson 1

    Lesson 1: Solving Quadratic Equations Using Graphs and Tables

  2. Lesson 2

    Lesson 2: Solving Quadratic Equations by Factoring

  3. Lesson 3

    Lesson 3: Rewriting Radical Expressions

  4. Lesson 4

    Lesson 4: Solving Quadratic Equations Using Square Roots

  5. Lesson 5

    Lesson 5: Completing the Square

  6. Lesson 6Current

    Lesson 6: The Quadratic Formula and the Discriminant

  7. Lesson 7

    Lesson 7: Solving Nonlinear Systems of Equations

Lesson overview

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Section 1

Quadratic Formula

Property

The solutions to a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, where a0a \neq 0 are given by the formula:

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

To solve a quadratic equation using the Quadratic Formula:
Step 1. Write the quadratic equation in standard form, ax2+bx+c=0ax^2 + bx + c = 0. Identify the values of aa, bb, and cc.
Step 2. Write the Quadratic Formula. Then substitute in the values of aa, bb, and cc.
Step 3. Simplify.
Step 4. Check the solutions.

Examples

  • To solve 2x2+5x3=02x^2 + 5x - 3 = 0, we identify a=2,b=5,c=3a=2, b=5, c=-3. Substituting into the formula gives x=5±524(2)(3)2(2)=5±25+244=5±494=5±74x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}. The solutions are x=12x = \frac{1}{2} and x=3x = -3.
  • To solve 3x2+10x+5=03x^2 + 10x + 5 = 0, we have a=3,b=10,c=5a=3, b=10, c=5. The formula gives x=10±1024(3)(5)2(3)=10±100606=10±406=10±2106=5±103x = \frac{-10 \pm \sqrt{10^2 - 4(3)(5)}}{2(3)} = \frac{-10 \pm \sqrt{100 - 60}}{6} = \frac{-10 \pm \sqrt{40}}{6} = \frac{-10 \pm 2\sqrt{10}}{6} = \frac{-5 \pm \sqrt{10}}{3}.
  • To solve x2+2x+10=0x^2 + 2x + 10 = 0, we have a=1,b=2,c=10a=1, b=2, c=10. The formula gives x=2±224(1)(10)2(1)=2±4402=2±362=2±6i2=1±3ix = \frac{-2 \pm \sqrt{2^2 - 4(1)(10)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 40}}{2} = \frac{-2 \pm \sqrt{-36}}{2} = \frac{-2 \pm 6i}{2} = -1 \pm 3i.

Explanation

The Quadratic Formula is a powerful tool derived from completing the square on the general quadratic equation. It provides a direct solution for any quadratic equation, saving you from repeating the steps of completing the square every time.

Section 2

The quadratic formula

Property

The solutions of the equation

ax2+bx+c=0,a0ax^2 + bx + c = 0, \quad a \neq 0

are given by the formula

Section 3

The Discriminant

Property

In the Quadratic Formula, x=b±b24ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}, the quantity b24acb^2 - 4ac is called the discriminant.
For a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, a0a \neq 0:

  • If b24ac>0b^2 - 4ac > 0, the equation has 2 real solutions.
  • If b24ac=0b^2 - 4ac = 0, the equation has 1 real solution.
  • If b24ac<0b^2 - 4ac < 0, the equation has no real solutions.

Examples

Book overview

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

Continue this chapter

Chapter 9: Solving Quadratic Equations

  1. Lesson 1

    Lesson 1: Solving Quadratic Equations Using Graphs and Tables

  2. Lesson 2

    Lesson 2: Solving Quadratic Equations by Factoring

  3. Lesson 3

    Lesson 3: Rewriting Radical Expressions

  4. Lesson 4

    Lesson 4: Solving Quadratic Equations Using Square Roots

  5. Lesson 5

    Lesson 5: Completing the Square

  6. Lesson 6Current

    Lesson 6: The Quadratic Formula and the Discriminant

  7. Lesson 7

    Lesson 7: Solving Nonlinear Systems of Equations