Introduction to the Quadratic Formula

In mathematics, the quadratic formula is used to solve quadratic equations. A quadratic equation has the form:

$$ ax^2 + bx + c = 0 $$

where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. The quadratic formula allows us to find the values of \( x \) that satisfy this equation:

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

How to Use the Quadratic Formula

To use the formula, identify the values of \( a \), \( b \), and \( c \) from the equation. Then substitute them into the formula. Don't forget to consider both the plus and minus signs (\( \pm \)), which give the two possible solutions for \( x \).

Example

Let's solve the equation:

$$ 2x^2 - 4x - 6 = 0 $$

Step 1: Identify \( a \), \( b \), and \( c \):

  • \( a = 2 \)
  • \( b = -4 \)
  • \( c = -6 \)

Step 2: Substitute into the quadratic formula:

$$ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2} $$

Step 3: Simplify:

$$ x = \frac{4 \pm \sqrt{16 + 48}}{4} $$

$$ x = \frac{4 \pm \sqrt{64}}{4} $$

$$ x = \frac{4 \pm 8}{4} $$

This gives us two solutions:

  • \( x = \frac{4 + 8}{4} = 3 \)
  • \( x = \frac{4 - 8}{4} = -1 \)

Therefore, the solutions are \( x = 3 \) and \( x = -1 \).

Practice Questions

  1. Solve the equation \( x^2 + 6x + 5 = 0 \).
    Solution

    Identify \( a = 1 \), \( b = 6 \), and \( c = 5 \).

    Using the quadratic formula:

    $$ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} $$

    $$ x = \frac{-6 \pm \sqrt{36 - 20}}{2} $$

    $$ x = \frac{-6 \pm \sqrt{16}}{2} $$

    $$ x = \frac{-6 \pm 4}{2} $$

    • \( x = \frac{-6 + 4}{2} = -1 \)
    • \( x = \frac{-6 - 4}{2} = -5 \)

    The solutions are \( x = -1 \) and \( x = -5 \).

  2. Solve the equation \( 3x^2 - 12x + 9 = 0 \).
    Solution

    Identify \( a = 3 \), \( b = -12 \), and \( c = 9 \).

    Using the quadratic formula:

    $$ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 3 \cdot 9}}{2 \cdot 3} $$

    $$ x = \frac{12 \pm \sqrt{144 - 108}}{6} $$

    $$ x = \frac{12 \pm \sqrt{36}}{6} $$

    $$ x = \frac{12 \pm 6}{6} $$

    • \( x = \frac{12 + 6}{6} = 3 \)
    • \( x = \frac{12 - 6}{6} = 1 \)

    The solutions are \( x = 3 \) and \( x = 1 \).

→Completing the Square