The quadratic equation is a fundamental concept in algebra. It deals with equations of the form:

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

where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. The equation is "quadratic" because the highest power of \( x \) is 2 (also called "degree 2").

### 1. **The Standard Form of a Quadratic Equation**
   The general form \( ax^2 + bx + c = 0 \) represents a parabola on a graph. Each term has a specific role:
   - \( ax^2 \): The "quadratic" term that makes the equation nonlinear.
   - \( bx \): The "linear" term.
   - \( c \): The "constant" term.

### 2. **The Quadratic Formula**
   To solve for \( x \) in a quadratic equation, you can use the **quadratic formula**:

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

   This formula gives you the values of \( x \) that satisfy the equation, also known as the "roots" or "solutions" of the equation.

### 3. **Understanding the Discriminant**
   The discriminant is the part of the quadratic formula under the square root:

   \[
   \Delta = b^2 - 4ac
   \]

   The value of the discriminant \( \Delta \) determines the nature of the roots:
   - If \( \Delta > 0 \): The equation has **two distinct real roots**.
   - If \( \Delta = 0 \): The equation has **one real root** (also called a "double root").
   - If \( \Delta < 0 \): The equation has **two complex roots** (they are not real numbers).

### 4. **Factoring the Quadratic Equation**
   In cases where a quadratic equation can be factored, you can solve it by rewriting it as a product of two binomials. For example:

   \[
   x^2 - 5x + 6 = 0
   \]

   can be factored as:

   \[
   (x - 2)(x - 3) = 0
   \]

   Then, setting each factor equal to zero gives the solutions: \( x = 2 \) and \( x = 3 \).

### 5. **Completing the Square**
   Another method to solve quadratic equations is **completing the square**. This method involves rewriting the equation in a form that allows you to find the solution by taking the square root.

### Example Problem
Suppose we want to solve \( 2x^2 + 3x - 5 = 0 \).

1. Identify \( a = 2 \), \( b = 3 \), \( c = -5 \).
2. Plug these into the quadratic formula:

   \[
   x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2}
   \]

3. Simplify under the square root:

   \[
   x = \frac{-3 \pm \sqrt{9 + 40}}{4}
   \]

4. Simplify further:

   \[
   x = \frac{-3 \pm \sqrt{49}}{4} = \frac{-3 \pm 7}{4}
   \]

5. Calculate the two solutions:

   \[
   x = \frac{-3 + 7}{4} = 1 \quad \text{and} \quad x = \frac{-3 - 7}{4} = -2.5
   \]

So, the solutions are \( x = 1 \) and \( x = -2.5 \).

Would you like more detailed explanations or have any questions?

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### Practice Questions
1. Solve \( x^2 + 6x + 9 = 0 \) by factoring.
2. Find the roots of \( x^2 - 4x + 4 = 0 \) using the quadratic formula.
3. Determine the nature of the roots for \( x^2 + 2x + 5 = 0 \) using the discriminant.
4. Solve \( 3x^2 - 12x + 12 = 0 \) by completing the square.
5. Solve \( x^2 - 2x - 15 = 0 \) using any preferred method.

#### Quick Tip
Always start by checking if the quadratic equation can be factored easily before using the quadratic formula—it saves time on simple problems!

Explore the quadratic equation with detailed explanations and examples. Learn the standard form, the quadratic formula, discriminant, and methods like factoring and completing the square. Perfect for Grades 8-10 students seeking to understand key algebra concepts.

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