10 Example Of Quadratic Equation With Solution

10 Examples of Quadratic Equations with Solutions: A Comprehensive Guide

Quadratic equations, those elegant algebraic expressions of the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0, are fundamental to many areas of mathematics and its applications. Understanding how to solve them is crucial for anyone studying algebra and beyond. This comprehensive guide will explore ten diverse examples of quadratic equations, demonstrating various solution methods and highlighting key concepts. We'll delve into factoring, the quadratic formula, and completing the square, providing a robust understanding of this essential mathematical concept.

Understanding Quadratic Equations

Before we dive into specific examples, let's solidify our understanding of the basic components. A quadratic equation always contains a squared term (x²), a linear term (bx), and a constant term (c). The coefficient 'a' determines the parabola's orientation (opens upwards if a > 0, downwards if a < 0), while 'b' influences the parabola's vertex position and 'c' represents the y-intercept.

Solving Quadratic Equations: The Methods

Several methods exist to solve quadratic equations. The most common are:

• Factoring: This method involves expressing the quadratic equation as a product of two linear factors. Setting each factor to zero allows us to find the roots (solutions).
• Quadratic Formula: A universally applicable formula, it provides the solutions directly, regardless of whether the equation is factorable. The formula is: x = [-b ± √(b² - 4ac)] / 2a
• Completing the Square: This method involves manipulating the equation to create a perfect square trinomial, making it easily solvable by taking the square root.

10 Example Quadratic Equations and Their Solutions

Let's tackle ten diverse examples, showcasing different techniques and equation characteristics:

Example 1: Simple Factoring

Equation: x² - 5x + 6 = 0

Solution: This equation factors easily: (x - 2)(x - 3) = 0. Therefore, the solutions are x = 2 and x = 3.

Example 2: Factoring with a Leading Coefficient

Equation: 2x² + 7x + 3 = 0

Solution: This requires a bit more factoring skill: (2x + 1)(x + 3) = 0. This gives us solutions x = -1/2 and x = -3.

Example 3: Using the Quadratic Formula (Easy Case)

Equation: x² + 4x + 1 = 0

Solution: This equation doesn't factor easily. Applying the quadratic formula (a=1, b=4, c=1):

x = [-4 ± √(16 - 4)] / 2 = [-4 ± √12] / 2 = -2 ± √3

The solutions are approximately x ≈ -0.268 and x ≈ -3.732

Example 4: Using the Quadratic Formula (Negative Discriminant)

Equation: x² + 2x + 5 = 0

Solution: Applying the quadratic formula (a=1, b=2, c=5):

x = [-2 ± √(4 - 20)] / 2 = [-2 ± √(-16)] / 2 = -1 ± 2i

Here, we encounter complex roots due to the negative discriminant (b² - 4ac < 0). The solutions are x = -1 + 2i and x = -1 - 2i.

Example 5: Completing the Square

Equation: x² + 6x + 5 = 0

Solution: Completing the square:

x² + 6x + 9 = 4 (Add and subtract (b/2)² = (6/2)² = 9)

(x + 3)² = 4

x + 3 = ±2

x = -3 ± 2

Therefore, x = -1 and x = -5.

Example 6: Equation with a Fractional Coefficient

Equation: (1/2)x² - 3x + 4 = 0

Solution: Multiply by 2 to eliminate the fraction: x² - 6x + 8 = 0. This factors to (x - 2)(x - 4) = 0, giving solutions x = 2 and x = 4.

Example 7: Equation with a Negative Leading Coefficient

Equation: -x² + 4x - 3 = 0

Solution: Multiply by -1 to make the leading coefficient positive: x² - 4x + 3 = 0. This factors to (x - 1)(x - 3) = 0, yielding solutions x = 1 and x = 3.

Example 8: Word Problem Application

Problem: A rectangular garden has a length 3 feet more than its width. If the area is 70 square feet, find the dimensions.

Solution: Let width = x. Then length = x + 3. Area = x(x + 3) = 70. This gives the quadratic equation x² + 3x - 70 = 0. Factoring gives (x - 7)(x + 10) = 0. Since width cannot be negative, the width is 7 feet and the length is 10 feet.

Example 9: Equation with Repeated Roots

Equation: x² - 6x + 9 = 0

Solution: This factors to (x - 3)(x - 3) = (x - 3)² = 0. This has a repeated root: x = 3.

Example 10: Equation with Irrational Coefficients

Equation: √2x² + 2x - √2 = 0

Solution: This equation can be solved using the quadratic formula, although the solutions will involve irrational numbers. Applying the quadratic formula (a = √2, b = 2, c = -√2):

x = [-2 ± √(4 - 4(√2)(-√2))] / (2√2) = [-2 ± √8] / (2√2) = [-2 ± 2√2] / (2√2) = (-1 ± √2) / √2

Simplifying, we get x = -1/√2 + 1 and x = -1/√2 -1.

Conclusion: Mastering Quadratic Equations

These ten examples demonstrate the versatility of quadratic equations and the various methods used to solve them. From simple factoring to the more complex quadratic formula and completing the square, understanding these techniques is key to mastering quadratic equations. Remember to always check your solutions by substituting them back into the original equation. The ability to solve quadratic equations is a cornerstone of algebra, opening doors to more advanced mathematical concepts and their diverse applications in science, engineering, and beyond. Consistent practice and understanding the underlying principles will solidify your understanding and make solving these equations second nature.