Solve Quadratic Equations by Using the Quadratic Formula - Algebra-2

1. Fundamental Concepts

Definition: The quadratic formula is used to solve any quadratic equation of the form $$ax^2 + bx + c = 0$$ where $a \neq 0$ .
Formula: The solutions to the quadratic equation are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Discriminant: The discriminant, $\Delta = b^2 - 4ac$ , determines the nature of the roots:
- If $\Delta > 0$ , there are two distinct real roots.
- If $\Delta = 0$ , there is one real root (a repeated root).
- If $\Delta < 0$ , there are two complex conjugate roots.

2. Key Concepts

Standard Form: $$ax^2 + bx + c = 0$$

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

Application: Used in physics, engineering, and economics to model parabolic motion and other phenomena.

3. Examples

Example 1 (Basic)

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

Step-by-Step Solution:

Identify coefficients: $a = 1$ , $b = -5$ , $c = 6$
Calculate the discriminant: $$\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1$$
Apply the quadratic formula: $$x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$$
Solve for $x$ : $$x = \frac{5 + 1}{2} = 3$$ and $$x = \frac{5 - 1}{2} = 2$$

Validation: Substitute $x = 3$ and $x = 2$ into the original equation:

- For $x = 3$ : $3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$

- For $x = 2$ : $2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$

Example 2 (Intermediate)

Problem: Solve the quadratic equation $$2x^2 + 3x - 2 = 0$$

Step-by-Step Solution:

Identify coefficients: $a = 2$ , $b = 3$ , $c = -2$
Calculate the discriminant: $$\Delta = 3^2 - 4(2)(-2) = 9 + 16 = 25$$
Apply the quadratic formula: $$x = \frac{-3 \pm \sqrt{25}}{2(2)} = \frac{-3 \pm 5}{4}$$
Solve for $x$ : $$x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$$ and $$x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$$

Validation: Substitute $x = \frac{1}{2}$ and $x = -2$ into the original equation:

- For $x = \frac{1}{2}$ : $2\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) - 2 = 2\left(\frac{1}{4}\right) + \frac{3}{2} - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0$

- For $x = -2$ : $2(-2)^2 + 3(-2) - 2 = 2(4) - 6 - 2 = 8 - 6 - 2 = 0$

4. Problem-Solving Techniques

Identify Coefficients: Clearly identify $a$ , $b$ , and $c$ from the standard form of the quadratic equation.
Calculate the Discriminant: Determine the nature of the roots by calculating $\Delta = b^2 - 4ac$ .
Apply the Quadratic Formula: Use the quadratic formula to find the roots, ensuring to handle the $\pm$ sign correctly.
Check Solutions: Always substitute the solutions back into the original equation to verify their correctness.