Question

how to solve quadratic functions, using quadratic formula

Answer 1

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 \( x = -1 \).