Solving $(v-8)^2-20=0$: A Step-by-Step Guide

Let's dive into solving the equation $(v-8)^2 - 20 = 0$, where $v$ represents a real number. This equation might look a bit daunting at first, but don't worry! We'll break it down step-by-step, making it super easy to understand. Our main goal here is to isolate $v$ and find its possible values. We'll be using some basic algebraic principles, which are like the secret ingredients to solving mathematical puzzles. So, grab your thinking cap, and let’s get started on this mathematical adventure together! Remember, math can be fun, especially when you tackle problems one step at a time. By the end of this guide, you'll not only know how to solve this specific equation but also gain confidence in handling similar problems in the future. Let's turn this equation from a challenge into a triumph!

1. Understanding the Equation and Initial Steps

The given equation is $(v-8)^2 - 20 = 0$. The first key observation is the squared term, $(v-8)^2$. This suggests that we might be dealing with a quadratic equation in disguise. To solve for $v$, we want to isolate the squared term first. This is like peeling an onion – we need to remove the outer layers to get to the core. So, our initial step involves adding 20 to both sides of the equation. This maintains the balance of the equation and moves us closer to isolating the term with $v$. Think of it as a balancing scale; whatever you do to one side, you must do to the other to keep it level. This simple addition is a crucial move in simplifying the equation and paving the way for the next steps. By understanding this initial maneuver, we set ourselves up for a smoother solution process. Let's perform this step and see what our equation looks like now.

2. Isolating the Squared Term

Adding 20 to both sides of the equation $(v-8)^2 - 20 = 0$ gives us $(v-8)^2 = 20$. Now, the squared term is nicely isolated on one side. This is a significant milestone because it allows us to address the square. The next natural step is to get rid of the square by taking the square root of both sides. But, a word of caution! When we take the square root, we need to consider both the positive and negative roots. Why? Because both a positive and a negative number, when squared, yield a positive result. For example, both $5^2$ and $(-5)^2$ equal 25. This is a critical concept in algebra, and understanding it ensures we don't miss any possible solutions. By acknowledging both positive and negative roots, we ensure that we capture all potential values for $v$ that satisfy the original equation. Let's proceed with taking the square root and see what possibilities unfold.

3. Taking the Square Root and Considering Both Solutions

Taking the square root of both sides of $(v-8)^2 = 20$, we get $v-8 = \pm\sqrt{20}$. It’s crucial to remember the $\pm$ symbol, which indicates both positive and negative square roots. This is where many people can make a mistake, so let’s emphasize it: always consider both possibilities! Now we have two potential paths to follow, one for the positive root and one for the negative root. Before we split these paths, we can simplify $\sqrt{20}$. Notice that 20 can be factored into $4 \times 5$, and since 4 is a perfect square, we can simplify $\sqrt{20}$ to $2\sqrt{5}$. This simplification makes the numbers easier to work with and leads to a cleaner final answer. Simplifying radicals is a useful skill in algebra, and it often makes further calculations more manageable. So, let's rewrite our equation with the simplified radical and then separate the two solution paths.

4. Simplifying the Radical and Separating Solution Paths

Substituting the simplified radical, we have $v-8 = \pm 2\sqrt{5}$. Now we have two separate equations to solve:

1. $v - 8 = 2\sqrt{5}$
2. $v - 8 = -2\sqrt{5}$

This separation is a pivotal moment because it clarifies the two distinct possibilities for $v$. Each of these equations represents a different solution branch, and we need to solve each one independently. To isolate $v$ in each case, we simply need to add 8 to both sides of the equation. This is a straightforward algebraic step, but it's essential to perform it accurately to arrive at the correct solutions. Think of these two paths as forks in a road; each leads to a different destination, and we need to explore both to get the complete picture. By carefully handling each equation, we ensure that we find all possible values of $v$ that satisfy our original equation. Let's proceed with isolating $v$ in each equation and revealing our solutions.

5. Isolating 'v' and Finding the Solutions

For the first equation, $v - 8 = 2\sqrt{5}$, adding 8 to both sides gives us $v = 8 + 2\sqrt{5}$. This is one solution for $v$. For the second equation, $v - 8 = -2\sqrt{5}$, adding 8 to both sides gives us $v = 8 - 2\sqrt{5}$. This is our second solution for $v$. So, we have two distinct real number solutions for $v$. These solutions involve the irrational number $\sqrt{5}$, which means they cannot be expressed as simple fractions. This is perfectly normal and highlights the richness of the real number system. These solutions are exact values, and while we could approximate them using a calculator, it's often preferable to leave them in this form for precision. We've successfully navigated the equation and found the values of $v$ that make it true. Let's summarize our findings and celebrate our mathematical victory!

6. Summarizing the Solutions

The solutions to the equation $(v-8)^2 - 20 = 0$ are:

• $v = 8 + 2\sqrt{5}$
• $v = 8 - 2\sqrt{5}$

These are the two real numbers that, when substituted for $v$ in the original equation, will make the equation true. We arrived at these solutions by systematically isolating the squared term, taking the square root (remembering both positive and negative roots), simplifying the radical, and finally, isolating $v$. This step-by-step approach is a powerful strategy for solving algebraic equations, and it's a technique you can apply to many similar problems. We started with a seemingly complex equation and, through careful manipulation, arrived at clear and concise solutions. This is the essence of mathematical problem-solving: breaking down challenges into manageable steps and applying logical principles. Congratulations on making it to the end! You've successfully solved this equation. For further exploration of quadratic equations and their solutions, you can visit Khan Academy's Quadratic Equations Section.