Solving Quadratic Equations: A Step-by-Step Guide

Hey guys! Let's dive into the world of quadratic equations and figure out how to find their solution sets. Don't worry, it's not as scary as it sounds. We'll break down the process step by step, making it super easy to understand. In this article, we'll focus on two specific examples, providing you with a solid understanding of the methods involved. Get ready to sharpen your math skills and become a quadratic equation solver extraordinaire! We will explore two examples of quadratic equations and the strategies to determine their solution sets. The goal is to provide a clear and comprehensive guide, breaking down each step so you can confidently solve similar problems on your own. By the end of this tutorial, you'll have a strong grasp of how to manipulate equations, factor expressions, and arrive at the correct solutions. This knowledge will not only help you with your current math assignments but also build a solid foundation for more advanced mathematical concepts. Let's get started and unlock the secrets of quadratic equations!

Example 1: Solving $6x^2 + 27x + 21 = 0$

Alright, let's tackle our first equation: $6x^2 + 27x + 21 = 0$. The primary goal here is to determine the values of 'x' that satisfy this equation. There are several ways to solve this, but we'll start with factoring, as it's often the most straightforward method. First, we need to check if there's a common factor among all the terms. In this case, we can see that 3 divides evenly into 6, 27, and 21. So, let's factor out a 3 from the equation. This simplifies our equation to: $3(2x^2 + 9x + 7) = 0$. Next, we'll focus on the quadratic expression inside the parentheses: $2x^2 + 9x + 7$. Now, we need to factor this quadratic expression. We're looking for two binomials that, when multiplied together, give us $2x^2 + 9x + 7$. This step often involves a bit of trial and error, but we can find the correct factorization. After some thinking, we realize that $(2x + 7)(x + 1)$ is the factored form of $2x^2 + 9x + 7$. Thus, the equation becomes: $3(2x + 7)(x + 1) = 0$. To find the solutions, we set each factor that contains 'x' equal to zero and solve for 'x'. Remember that the constant factor 3 doesn't affect the solutions. So, we'll set $2x + 7 = 0$ and $x + 1 = 0$. Solving these two simple linear equations gives us our solutions. From $2x + 7 = 0$, we subtract 7 from both sides, getting $2x = -7$, and then divide by 2, which results in $x = -7/2$ or -3.5. From $x + 1 = 0$, we subtract 1 from both sides, and we get $x = -1$. Therefore, the solution set for the equation $6x^2 + 27x + 21 = 0$ is $x = -3.5$ and $x = -1$. These are the values of 'x' that make the original equation true. Remember that these steps are very important to master. Practice this often to get a better grasp.

This is an important step in understanding how to determine the solution set of the quadratic equation. Mastering the factoring method allows you to simplify the equation into a form that is easy to solve. Practice this process with different quadratic equations to enhance your skills and get a better grasp of the methods. The ability to factor effectively is a cornerstone of algebra, enabling you to solve a wide variety of equations and simplify complex expressions. With consistent practice, you'll become adept at recognizing patterns and quickly determining the correct factors. Don't hesitate to revisit this example and try it again on your own to solidify your understanding.

Factoring and Solution Derivation

To summarize, the key steps in solving this quadratic equation using factoring are:

• Identify the equation: $6x^2 + 27x + 21 = 0$.
• Factor out the common factor: $3(2x^2 + 9x + 7) = 0$.
• Factor the quadratic expression: $3(2x + 7)(x + 1) = 0$.
• Set each factor containing 'x' to zero and solve for 'x': $2x + 7 = 0$ and $x + 1 = 0$.
• Find the solutions: $x = -7/2$ and $x = -1$.

These solutions represent the x-values where the original quadratic equation equals zero. Graphically, these are the x-intercepts of the parabola represented by the equation.

Example 2: Solving $8x^2 + 32x + 28 = 0$

Now, let's move on to the second example: $8x^2 + 32x + 28 = 0$. Just like before, we'll start by looking for a common factor among all the terms. In this case, the greatest common factor is 4. Factoring out a 4 simplifies our equation to: $4(2x^2 + 8x + 7) = 0$. Next, we need to factor the quadratic expression inside the parentheses: $2x^2 + 8x + 7$. This one is a bit trickier. Unlike our first example, this quadratic expression doesn't factor neatly into two binomials with integer coefficients. When faced with such a situation, we have a few options. One is to use the quadratic formula, which is a reliable method for solving any quadratic equation. Another approach is to complete the square. In this example, we will use the quadratic formula. The quadratic formula is x = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a}. For the quadratic equation $2x^2 + 8x + 7 = 0$, we have $a = 2$, $b = 8$, and $c = 7$. Plugging these values into the formula, we get: x = rac{-8 ext{±} ext{√}(8^2 - 4 * 2 * 7)}{2 * 2}. Simplifying further, we have x = rac{-8 ext{±} ext{√}(64 - 56)}{4}, which simplifies to x = rac{-8 ext{±} ext{√}8}{4}. Since the square root of 8 can be simplified to $2 ext{√}2$, we can rewrite the solutions as x = rac{-8 ext{±} 2 ext{√}2}{4}. Finally, we can simplify this by dividing each term by 2. So we have x = -2 ext{±} rac{ ext{√}2}{2}. This means our two solutions are x = -2 + rac{ ext{√}2}{2} and x = -2 - rac{ ext{√}2}{2}. Thus, the solution set for the equation $8x^2 + 32x + 28 = 0$ consists of these two values, expressed in terms of a square root.

The quadratic formula is an extremely versatile tool for solving quadratic equations, especially when factoring becomes difficult or impossible. The formula provides a direct and systematic way to find the solutions, regardless of whether the roots are real, complex, or irrational. Understanding the quadratic formula is critical for success in algebra and for many other advanced math topics.

Quadratic Formula Application

In this case, the key steps were:

• Identify the equation: $8x^2 + 32x + 28 = 0$.
• Factor out the common factor: $4(2x^2 + 8x + 7) = 0$.
• Apply the quadratic formula: x = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a}.
• Substitute and simplify: x = rac{-8 ext{±} ext{√}(8^2 - 4 * 2 * 7)}{2 * 2}.
• Solve for x: x = -2 ext{±} rac{ ext{√}2}{2}.

These are the x-values where the original quadratic equation equals zero. This is one of the most important tools in algebra. Remember that a strong understanding of the quadratic formula will benefit you for more advanced mathematical concepts.

Conclusion

Great job, guys! We've covered two examples, each solved using a different approach. In the first example, we used factoring, which is often the simplest method when the quadratic expression can be easily factored. In the second example, we turned to the quadratic formula, which is a powerful tool that works for all quadratic equations, especially when factoring is not straightforward. Remember, practice is key. The more you practice solving quadratic equations using both factoring and the quadratic formula, the more comfortable and confident you'll become. Try working through more examples on your own. You can find plenty of practice problems online or in your textbook. Don't be afraid to make mistakes; they're a natural part of the learning process. If you get stuck, review the steps we've covered, or look up additional examples and explanations. With perseverance and practice, you'll master quadratic equations in no time! Keep up the excellent work, and keep exploring the fascinating world of mathematics! Now you can tackle a variety of problems.

By understanding these examples and the methods used, you have equipped yourself with the skills to solve similar problems with confidence. Remember to review and practice these steps until they become second nature. Keep up the excellent work, and keep exploring the fascinating world of mathematics!