Solving (v-5)^2=6 A Step-by-Step Guide

This article delves into the step-by-step solution of the equation $(v-5)^2 = 6$, where $v$ represents a real number. Our primary goal is to find the value(s) of $v$ that satisfy this equation, simplifying the answer as much as possible. If multiple solutions exist, they will be presented, separated by commas, ensuring clarity and precision in our final result. The problem falls under the category of mathematics, specifically algebra, which involves manipulating equations to isolate the variable. This article will guide you through the process of solving such equations, providing a comprehensive understanding of the techniques involved. Understanding how to solve equations like this is fundamental in algebra and serves as a building block for more complex mathematical concepts. Whether you are a student learning algebra for the first time or someone looking to refresh your skills, this guide offers a clear and detailed explanation. By breaking down the problem into manageable steps, we aim to make the solution accessible to everyone, regardless of their mathematical background. The importance of this skill extends beyond academic settings, as algebraic problem-solving is crucial in various real-world applications, including engineering, finance, and computer science. Therefore, mastering the techniques discussed in this article can significantly enhance your analytical and problem-solving capabilities. Let’s embark on this mathematical journey together, unraveling the mysteries of this equation and discovering the values of $v$ that hold true.

Understanding the Equation $(v-5)^2 = 6$

The given equation is $(v-5)^2 = 6$. To effectively solve this equation, it is crucial to understand its structure. At its core, this is a quadratic equation, though it's presented in a slightly disguised form. The presence of the squared term, $(v-5)^2$, immediately signals its quadratic nature. Quadratic equations are equations of the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ is the variable. While our equation isn't in this standard form, we can transform it into one if necessary. However, there are more direct methods to tackle this specific equation, which we will explore shortly. The equation essentially states that the square of the quantity $(v-5)$ is equal to 6. This means that the quantity $(v-5)$ can be either the positive square root of 6 or the negative square root of 6. This is a critical concept in solving equations involving squares. When we take the square root of both sides of an equation, we must consider both the positive and negative roots, as both will satisfy the original equation. This understanding forms the foundation of our solution strategy. We will leverage the properties of square roots and algebraic manipulation to isolate the variable $v$ and determine its possible values. This equation is a prime example of how algebraic techniques can be applied to uncover the solutions hidden within mathematical expressions. By grasping the underlying principles and employing the appropriate methods, we can confidently navigate the world of quadratic equations and beyond.

Solving for $v$

To solve the equation $(v-5)^2 = 6$, we need to isolate $v$. The first step involves taking the square root of both sides of the equation. Remember, when taking the square root, we must consider both the positive and negative roots. This gives us: $\sqrt{(v-5)^2} = \pm\sqrt{6}$. This simplifies to $v-5 = \pm\sqrt{6}$. Now, we have two separate equations to consider: $v-5 = \sqrt{6}$ and $v-5 = -\sqrt{6}$. To solve for $v$ in each case, we simply add 5 to both sides of each equation. For the first equation, $v-5 = \sqrt{6}$, adding 5 to both sides gives us $v = 5 + \sqrt{6}$. For the second equation, $v-5 = -\sqrt{6}$, adding 5 to both sides gives us $v = 5 - \sqrt{6}$. Therefore, the two solutions for $v$ are $5 + \sqrt{6}$ and $5 - \sqrt{6}$. These are the exact solutions, expressed in simplest radical form. We can approximate these values using a calculator, but for the purpose of this problem, the exact solutions are preferred. The solutions represent the values of $v$ that, when substituted back into the original equation, will make the equation true. This process of solving for $v$ demonstrates the power of algebraic manipulation and the importance of considering all possible roots when dealing with square roots. By carefully applying these techniques, we can confidently solve a wide range of quadratic equations.

Simplifying the Answer

Having found the solutions $v = 5 + \sqrt{6}$ and $v = 5 - \sqrt{6}$, the next step is to simplify the answer as much as possible. In this case, the solutions are already in their simplest form. The square root of 6, $\sqrt{6}$, cannot be simplified further because 6 has no perfect square factors other than 1. Therefore, the expressions $5 + \sqrt{6}$ and $5 - \sqrt{6}$ are the simplest radical forms of the solutions. These solutions are irrational numbers, meaning they cannot be expressed as a simple fraction. They are exact solutions, representing the precise values of $v$ that satisfy the original equation. While we could use a calculator to approximate these values as decimals, it's generally preferred to leave the answers in their exact form unless a decimal approximation is specifically requested. Leaving the answers in simplest radical form maintains the accuracy and precision of the solutions. It also demonstrates a strong understanding of radical expressions and their properties. In many mathematical contexts, exact solutions are more valuable than approximations, especially when dealing with further calculations or theoretical analysis. Therefore, we can confidently state that the solutions $v = 5 + \sqrt{6}$ and $v = 5 - \sqrt{6}$ are fully simplified and represent the final answer to the problem. This process highlights the importance of not just finding the solutions, but also expressing them in the most concise and accurate form possible.

Presenting the Solutions

Now that we have found and simplified the solutions for $v$, which are $5 + \sqrt{6}$ and $5 - \sqrt{6}$, the final step is to present them in the requested format. The problem statement asks us to separate multiple solutions with commas. Therefore, the final answer should be written as: $v = 5 + \sqrt{6}, 5 - \sqrt{6}$. This format clearly and concisely presents both solutions, fulfilling the requirements of the problem. It's essential to pay close attention to the instructions in mathematical problems, as the way the answer is presented can be just as important as the answer itself. In this case, the comma serves as a clear separator, indicating that we have two distinct solutions for $v$. This method of presenting multiple solutions is common in mathematics and helps to avoid ambiguity. By adhering to the specified format, we ensure that our answer is easily understood and correctly interpreted. The solutions $5 + \sqrt{6}$ and $5 - \sqrt{6}$ represent the values of $v$ that make the original equation $(v-5)^2 = 6$ true. Substituting either of these values back into the equation will confirm that they indeed satisfy the equality. This final presentation of the solutions marks the successful completion of the problem. We have not only found the solutions but also expressed them in the required format, demonstrating a thorough understanding of the problem and its solution process. This meticulous approach is crucial in mathematics and ensures that our work is both accurate and clear.

Final Answer

$v = 5 + \sqrt{6}, 5 - \sqrt{6}$