Solving Quadratic Equations: A Step-by-Step Guide with Examples
Solving quadratic equations is a fundamental skill in algebra. Whether you prefer a straightforward approach or enjoy the challenge of finding alternative methods, there are multiple ways to tackle these problems. In this article, we will explore how to solve the quadratic equation (0.25x^2 - 1.45x 1 0) using both the quadratic formula and alternative methods. We will also discuss the underlying principles of each method and when you might choose to use one over the other.
Introduction to Quadratic Equations
A quadratic equation is an equation that can be written in the standard form (ax^2 bx c 0), where (a), (b), and (c) are real numbers, and (a eq 0). The solutions to these equations can be found using various methods, but the most common and direct approach is the quadratic formula. However, it is also possible to solve them by factoring, completing the square, or through substitution techniques. Letrsquo;s solve the given equation using both the quadratic formula and an alternative method.
Solving the Quadratic Equation Using the Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations and is given by:
$$x frac{-b pm sqrt{b^2 - 4ac}}{2a}$$In the equation (0.25x^2 - 1.45x 1 0), the coefficients are:
$$a 0.25, quad b -1.45, quad c 1$$Step 1: Calculate the Discriminant
The discriminant (D) is given by:
$$D b^2 - 4ac$$Substituting the values of (a), (b), and (c):
$$D (-1.45)^2 - 4 cdot 0.25 cdot 1 2.1025 - 1 1.1025$$Step 2: Apply the Quadratic Formula
Substitute (a), (b), and (D) into the quadratic formula:
$$x frac{-(-1.45) pm sqrt{1.1025}}{2 cdot 0.25}$$Calculate (sqrt{1.1025} 1.05). Now substitute this value back into the formula:
$$x frac{1.45 pm 1.05}{0.5}$$Now calculate the two possible values for (x):
First Solution:
$$x_1 frac{1.45 1.05}{0.5} frac{2.5}{0.5} 5$$Second Solution:
$$x_2 frac{1.45 - 1.05}{0.5} frac{0.4}{0.5} 0.8$$Final Solutions: The exact solutions to the equation (0.25x^2 - 1.45x 1 0) are (x_1 5) and (x_2 0.8).
Solving the Quadratic Equation Using Alternative Methods
There are several alternative methods to solve quadratic equations. In this section, we will explore two such methods: factoring and completing the square.
Factoring Method
Letrsquo;s start by factoring the quadratic equation (0.25x^2 - 1.45x 1 0). To make the calculations simpler, we can multiply the entire equation by 100:
$$25x^2 - 145x 100 0$$Next, we divide the equation by 5:
$$5x^2 - 29x 20 0$$Now, we factor the quadratic expression:
$$5x^2 - 25x - 4x 20 0$$We can factor by grouping:
$$5x(x - 5) - 4(x - 5) 0$$Factor out the common term (x - 5):
$$ (5x - 4)(x - 5) 0$$Solving for (x), we get:
$$x - 5 0 quad text{or} quad 5x - 4 0$$Therefore, the solutions are:
$$x 5 quad text{or} quad x frac{4}{5}$$Completing the Square Method
To solve the equation (5x^2 - 29x 20 0) using the completing the square method, we follow these steps:
Step 1: Isolate the quadratic and linear terms
$$x^2 - frac{29}{5}x 4 0$$Step 2: Move the constant term to the other side of the equation
$$x^2 - frac{29}{5}x -4$$Step 3: Add and subtract (left(frac{b}{2a}right)^2) to complete the square
$$x^2 - frac{29}{5}x left(frac{29}{10}right)^2 -4 left(frac{29}{10}right)^2$$Step 4: Simplify and write the equation in the form ((x - h)^2 k)
$$x^2 - frac{29}{5}x frac{841}{100} frac{489}{100}$$This can be rewritten as:
$$left(x - frac{29}{10}right)^2 frac{489}{100}$$Solving for (x), we take the square root of both sides:
$$x - frac{29}{10} pm sqrt{frac{489}{100}}$$Calculating the square root, we get:
$$x - frac{29}{10} pm frac{sqrt{489}}{10}$$Solving for (x), we find:
$$x frac{29 pm sqrt{489}}{10}$$The solutions are:
$$x 5 quad text{or} quad x frac{29 - sqrt{489}}{10} approx 0.8$$Therefore, the solutions are (x_1 5) and (x_2 0.8).
Conclusion
Solving quadratic equations can be approached in various ways, including the quadratic formula, factoring, and completing the square. Each method has its unique advantages and can be chosen based on the specific equation and personal preference. Understanding the underlying principles and practicing different techniques will enhance your problem-solving skills in algebra.