Quadratic equations play a crucial role in algebra and are frequently encountered in diverse mathematical problems. A quadratic equation is a polynomial equation of the second degree that involves a single variable and is typically represented as:

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

where (a), (b), and (c) are constants, and (x) represents the variable. The coefficient (a) must not be zero.

The Standard Form of a Quadratic Equation

The general form of a quadratic equation is:

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

• (a): The coefficient of$ (x^2)$, known as the quadratic coefficient.
• (b): The coefficient of $(x)$, known as the linear coefficient.
• (c): The constant term.

Methods to Solve Quadratic Equations

Quadratic equations can be solved using different methods:

1. Factoring: Expressing the quadratic equation as a product of two binomials.
2. Completing the Square: Rewriting the equation in the form $((x – p)^2 = q)$.
3. Quadratic Formula: Applying the formula

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

4. Graphical Method: Plotting the quadratic equation and identifying where the curve intersects the x-axis.

Let’s dive into some detailed examples of solving quadratic equations.

Solved Examples

Example 1: Factoring

Problem: Solve $(x^2 – 5x + 6 = 0).$

Step 1: Factor the quadratic equation.

$[ x^2 – 5x + 6 = (x – 2)(x – 3) ]$

Step 2: Set each factor equal to zero.

$[ x – 2 = 0 \quad \text{or} \quad x – 3 = 0 ]$

Step 3: Solve for (x).

$[ x = 2 \quad \text{or} \quad x = 3 ]$

So, the solutions are $(x = 2) $and $(x = 3).$

Example 2: Using the Quadratic Formula

Problem: Solve $(2x^2 – 4x – 6 = 0).$

Step 1: Identify the coefficients (a), (b), and (c).

Here,$ (a = 2)$, $(b = -4)$, and $(c = -6).$

Step 2: Substitute the values into the quadratic formula.

$[ x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)} ]$

Step 3: Simplify the equation.

$[ x = \frac{4 \pm \sqrt{16 + 48}}{4} ]$

$[ x = \frac{4 \pm \sqrt{64}}{4} ]$

$[ x = \frac{4 \pm 8}{4} ]$

$So, (x = 3) or (x = -1).$

Practice Problems with Detailed Solutions

Example 3: Factoring

Problem: Solve$ (x^2 – x – 12 = 0).$

Step 1: Factor the quadratic equation.

$[ x^2 – x – 12 = (x – 4)(x + 3) ]$

Step 2: Set each factor equal to zero.

$[ x – 4 = 0 \quad \text{or} \quad x + 3 = 0 ]$

Step 3: Solve for (x).

$[ x = 4 \quad \text{or} \quad x = -3 ]$

So, the solutions are $(x = 4) and (x = -3).$

Example 4: Completing the Square

Problem: Solve $(x^2 + 6x + 5 = 0).$

Step 1: Move the constant term to the right side of the equation.

$[ x^2 + 6x = -5 ]$

Step 2: Add $((\frac{b}{2})^2)$ to both sides to complete the square.

$ x^2 + 6x + 9 = 4 $

(since $ \left(\frac{6}{2}\right)^2 = 9 $)

Step 3: Rewrite the left side as a perfect square and solve.

$[ (x + 3)^2 = 4 ]$

$[ x + 3 = \pm 2 ]$

$[ x = -3 \pm 2 ]$

$So, (x = -1) or (x = -5).$

Practice Questions

Try solving these quadratic equations on your own:

1. $(x^2 + 5x + 6 = 0)$
2. $(3x^2 – x – 2 = 0)$
3. $(x^2 – 4x – 5 = 0)$
4. $(2x^2 + 3x + 1 = 0)$
5. $(x^2 – 10x + 21 = 0)$

Additional Practice Problems

1. Problem: Solve $(x^2 – 7x + 10 = 0).$

• Solution: Factor as $((x – 5)(x – 2) = 0).$
• Answer: $(x = 5)$ or $(x = 2).$

1. Problem: Solve $(4x^2 – 12x + 9 = 0).$

• Solution: Factor as $((2x – 3)^2 = 0).$
• Answer: $(x = \frac{3}{2}).$

Quadratic equations are a cornerstone of algebra, and understanding how to solve them will build a strong foundation for more advanced mathematical concepts. Learn trigonometry-formulas