Solving Quadratic Equations: Step-by-Step Guide

by Alex Johnson 48 views

Welcome! Let's dive into solving the quadratic equation: (y+5)^2 = 2y^2 + 17y + 31. This guide will walk you through the process step-by-step, ensuring you understand each element and arrive at the correct solution(s). We'll break down the equation, simplify it, and use the quadratic formula to find the values of y. Whether you're a student brushing up on algebra or just curious, this guide aims to make solving quadratic equations clear and straightforward.

Step 1: Expand and Simplify the Equation

Our first step is to expand the equation. We'll begin by expanding (y+5)^2. This means multiplying (y+5) by itself: (y+5)(y+5). When we perform this multiplication, we get y^2 + 5y + 5y + 25, which simplifies to y^2 + 10y + 25. Now, our equation looks like this: y^2 + 10y + 25 = 2y^2 + 17y + 31. The key here is to be meticulous with the expansion, ensuring that each term is correctly multiplied and that no terms are missed.

Next, we need to bring all the terms to one side of the equation to set it equal to zero. To do this, we'll subtract y^2, 10y, and 25 from both sides. This gives us: 0 = 2y^2 - y^2 + 17y - 10y + 31 - 25. Combining like terms simplifies this to: 0 = y^2 + 7y + 6. At this stage, it's extremely important to double-check the calculations. A small error in the expansion or simplification can lead to incorrect solutions. A common mistake is to forget to distribute the negative sign when moving terms across the equals sign, so always be mindful of this! The objective here is to transform the equation into a standard quadratic form: ax^2 + bx + c = 0. Always ensure that the equation is in this form before proceeding.

Once we have our quadratic equation set to zero, we are ready to move on to the next step, where we can choose to solve it by either factoring or applying the quadratic formula, and in this case, we will be applying the factoring method. Before proceeding to the next step, take some time to review each step and ensure your understanding is firm. You can also utilize online calculators to verify if your expansion and simplification are correct. Doing so helps build confidence, and confidence is vital when tackling more complex mathematical problems. This detailed approach enables us to tackle such equations efficiently, making sure we haven’t overlooked any step. It also promotes a solid foundation for future mathematical endeavors. Remember, practice is key, and the more you practice these steps, the more comfortable and confident you'll become in solving similar equations.

Step 2: Factor the Quadratic Equation

Now, let's factor the quadratic equation we derived: y^2 + 7y + 6 = 0. Factoring involves finding two numbers that multiply to give us the constant term (6) and add up to give us the coefficient of the linear term (7). In this case, the numbers are 6 and 1, since 6 * 1 = 6 and 6 + 1 = 7. Therefore, we can factor the quadratic equation into (y + 6)(y + 1) = 0. This factored form is essential because it allows us to find the roots (or solutions) of the equation easily. The roots are the values of y that make the equation true.

To find the roots, we set each factor equal to zero and solve for y. First, we have y + 6 = 0. Subtracting 6 from both sides gives us y = -6. Second, we have y + 1 = 0. Subtracting 1 from both sides gives us y = -1. Therefore, the solutions to the equation are y = -6 and y = -1. It’s crucial to understand the logic behind factoring. The principle is that if the product of two factors is zero, then at least one of the factors must be zero. This principle is fundamental in solving factored quadratic equations.

Always double-check your factoring to avoid common errors. Sometimes, a quadratic equation may not be easily factorable; in those cases, the quadratic formula would be a more suitable method. Practicing factoring different types of quadratic equations will enhance your ability to recognize patterns and choose the most efficient solution method. Also, remember to write the solution set clearly, especially if the problem asks for multiple solutions. Properly formatted answers and well-structured work increase clarity and aid in understanding. Always confirm your answers by substituting the calculated y-values back into the original equation to verify that they satisfy the equality. This is a good practice to avoid errors and increase your confidence. Once you have mastered this step, you will be well-equipped to tackle more complex quadratic equations.

Step 3: Finding the Solutions

After factoring, we have the expression (y + 6)(y + 1) = 0. This step is about identifying the actual solutions for y. As previously mentioned, we set each factor equal to zero: y + 6 = 0 and y + 1 = 0. Now, let's solve each of these simple linear equations. For y + 6 = 0, to isolate y, we subtract 6 from both sides. This gives us y = -6. For y + 1 = 0, we subtract 1 from both sides, which gives us y = -1. Thus, we have two solutions for y: -6 and -1. The beauty of this method lies in its simplicity. By breaking down the quadratic equation into its factors, we transform a complex problem into two straightforward linear equations, which are simple to solve.

The solutions represent the points where the quadratic equation intersects the x-axis when graphed. In other words, they are the roots of the equation. Understanding the graphical interpretation can deepen your comprehension of quadratic equations and their solutions. Furthermore, once you find the solutions, it is always a good practice to plug these values back into the original equation to verify that the equation holds true. This is an essential step to ensure the accuracy of the solutions and to minimize the risk of making arithmetic errors along the way. Doing this builds trust in your results and helps reinforce the concepts of quadratic equations. Take a moment to think about the different solution methods and when they are best used. Practice with different types of quadratic equations to become versatile in applying factoring and using the quadratic formula. With consistency and practice, these problem-solving skills will become second nature.

Step 4: Verification of the Solutions

To ensure the solutions are correct, we'll substitute each value of y back into the original equation: (y + 5)^2 = 2y^2 + 17y + 31. First, let's substitute y = -6:

  • (-6 + 5)^2 = 2(-6)^2 + 17(-6) + 31
  • (-1)^2 = 2(36) - 102 + 31
  • 1 = 72 - 102 + 31
  • 1 = 1

The equation holds true, so y = -6 is a correct solution. Next, let's substitute y = -1:

  • (-1 + 5)^2 = 2(-1)^2 + 17(-1) + 31
  • (4)^2 = 2(1) - 17 + 31
  • 16 = 2 - 17 + 31
  • 16 = 16

The equation also holds true, so y = -1 is a correct solution. These verification steps are crucial; they confirm that our calculated solutions are accurate and satisfy the original equation. It is also an excellent practice to ensure that no calculation errors were made during the process. By confirming both solutions, we can be confident in the accuracy of our answer. Verification not only validates the solutions but also reinforces the understanding of how quadratic equations function. The process of substituting and confirming solutions also aids in improving your problem-solving skills.

By following this method, you can effectively solve various quadratic equations. The important thing is to take your time, understand each step, and verify your answers. Through this methodical approach, you'll not only solve the equation but also strengthen your understanding of quadratic equations.

Conclusion

We successfully solved the quadratic equation (y+5)^2 = 2y^2 + 17y + 31 and found that y = -6 and y = -1. This was achieved by expanding the equation, simplifying it, factoring it, and verifying the solutions. The process reinforces the importance of meticulous calculation and methodical problem-solving in mathematics. The quadratic formula, although not used in this specific example, is an alternative method and a fundamental tool in solving quadratic equations. Mastering both the factoring method and the use of the quadratic formula will enable you to solve a wide range of quadratic equations. Keep practicing, and you'll find yourself becoming increasingly proficient in algebra.

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