The quadratic equation is a polynomial equation of the form: $ax^2 + bx + c = 0$ where $a$, $b$, and $c$ are constants, with $a \neq 0$. The goal is to find the values of $x$ that satisfy the equation, called roots.

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ This formula provides the solutions (roots) for any quadratic equation.

The term $b^2 - 4ac$ is called the discriminant:

• If $b^2 - 4ac > 0$, there are two distinct real roots.
• If $b^2 - 4ac = 0$, there is one real root.
• If $b^2 - 4ac < 0$, there are two complex roots.

Example 1: Two Real Roots

Problem: Solve $2x^2 - 4x - 6 = 0$

1. Identify $a$, $b$, and $c$: Here, $a = 2$, $b = -4$, $c = -6$.
2. Calculate the Discriminant: $b^2 - 4ac = (-4)^2 - 4(2)(-6) = 16 + 48 = 64$.
3. Use the Quadratic Formula: $x = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 2} = \frac{4 \pm 8}{4}$
4. Solve for $x$:
   • $x = \frac{4 + 8}{4} = 3$
   • $x = \frac{4 - 8}{4} = -1$

Solution: $x = 3$ and $x = -1$

Example 2: One Real Root

Problem: Solve $x^2 - 6x + 9 = 0$

1. Identify $a$, $b$, and $c$: $a = 1$, $b = -6$, $c = 9$.
2. Calculate the Discriminant: $b^2 - 4ac = (-6)^2 - 4(1)(9) = 36 - 36 = 0$.
3. Use the Quadratic Formula: $x = \frac{-(-6) \pm \sqrt{0}}{2 \cdot 1} = \frac{6}{2} = 3$

Solution: $x = 3$ (one root, or a "double root")

Example 3: Two Complex Roots

Problem: Solve $x^2 + 4x + 8 = 0$

1. Identify $a$, $b$, and $c$: $a = 1$, $b = 4$, $c = 8$.
2. Calculate the Discriminant: $b^2 - 4ac = (4)^2 - 4(1)(8) = 16 - 32 = -16$.
3. Use the Quadratic Formula: $x = \frac{-4 \pm \sqrt{-16}}{2 \cdot 1} = \frac{-4 \pm 4i}{2}$
4. Simplify:
   • $x = \frac{-4 + 4i}{2} = -2 + 2i$
   • $x = \frac{-4 - 4i}{2} = -2 - 2i$

Solution: $x = -2 + 2i$ and $x = -2 - 2i$

Example 4: Standard Quadratic Equation

Problem: Solve $x^2 - 5x + 6 = 0$

1. Identify $a$, $b$, and $c$: $a = 1$, $b = -5$, $c = 6$.
2. Calculate the Discriminant: $b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1$.
3. Use the Quadratic Formula: $x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 1} = \frac{5 \pm 1}{2}$
4. Solve for $x$:
   • $x = \frac{5 + 1}{2} = 3$
   • $x = \frac{5 - 1}{2} = 2$

Solution: $x = 3$ and $x = 2$

Example 5: Solving by Factoring

Problem: Solve $x^2 - 7x + 12 = 0$

1. Identify factors of 12 that add up to -7: $-3$ and $-4$.
2. Rewrite as a factored expression: $(x - 3)(x - 4) = 0$
3. Solve for $x$ by setting each factor to zero:
   • $x - 3 = 0 \Rightarrow x = 3$
   • $x - 4 = 0 \Rightarrow x = 4$

Solution: $x = 3$ and $x = 4$

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Summary

Using these examples, we covered:

• Two real roots (Example 1 and 4)
• One real root (Example 2)
• Two complex roots (Example 3)
• Solving by factoring (Example 5)

Would you like more details on any of these examples, or have any questions?

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Follow-up Questions

1. How does the discriminant affect the types of roots in a quadratic equation?
2. Can all quadratic equations be solved by factoring?
3. What are some real-world applications of quadratic equations?
4. How can quadratic equations be visualized on a graph?
5. Why is the quadratic formula derived in this specific way?

Tip

To check your solutions, substitute the values of $x$ back into the original equation to see if they satisfy it!