how to solve quadratic functions, using quadratic formula

1 answer

To solve quadratic equations using the quadratic formula, you first need to know the standard form of a quadratic equation, which is:

\[ ax^2 + bx + c = 0 \]

Where:

  • \( a \), \( b \), and \( c \) are coefficients (with \( a \neq 0 \)),
  • \( x \) represents the variable.

The quadratic formula is:

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

Here’s a step-by-step guide on how to use the quadratic formula:

  1. Identify coefficients: Determine the values of \( a \), \( b \), and \( c \) from your quadratic equation.

  2. Calculate the discriminant: Compute the value of the discriminant, which is the expression under the square root in the formula: \[ D = b^2 - 4ac \]

  3. Determine the nature of the roots:

    • If \( D > 0 \): There are two distinct real roots.
    • If \( D = 0 \): There is exactly one real root (a repeated root).
    • If \( D < 0 \): There are no real roots (but two complex roots).
  4. Apply the quadratic formula:

    • Substitute \( a \), \( b \), and \( D \) into the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \]
  5. Calculate the roots:

    • If \( D \) is positive, calculate both roots: \[ x_1 = \frac{-b + \sqrt{D}}{2a} \] \[ x_2 = \frac{-b - \sqrt{D}}{2a} \]
    • If \( D = 0 \), calculate the single root: \[ x = \frac{-b}{2a} \]
    • If \( D < 0 \), express the roots in terms of imaginary numbers: \[ x_1 = \frac{-b + i\sqrt{-D}}{2a} \] \[ x_2 = \frac{-b - i\sqrt{-D}}{2a} \]

Example

Let's solve the quadratic equation:

\[ 2x^2 - 4x - 6 = 0 \]

  1. Identify coefficients: Here, \( a = 2 \), \( b = -4 \), and \( c = -6 \).

  2. Calculate the discriminant: \[ D = (-4)^2 - 4 \cdot 2 \cdot (-6) = 16 + 48 = 64 \]

  3. Determine the nature of the roots: Since \( D > 0 \), we will have two distinct real roots.

  4. Apply the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 2} = \frac{4 \pm 8}{4} \]

  5. Calculate the roots:

    • First root: \[ x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3 \]
    • Second root: \[ x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1 \]

So the solutions to the equation \( 2x^2 - 4x - 6 = 0 \) are \( x = 3 \) and \( x = -1 \).