Solve $\sqrt{x+2}-15=-3$: A Step-by-Step Guide

Hey guys! Ever stared at a math problem and felt that little spark of curiosity, wondering "what's the solution to $\sqrt{x+2}-15=-3$?" Well, you've come to the right place! Today, we're diving deep into this radical equation, breaking it down piece by piece, and uncovering the mystery behind the value of 'x'. This isn't just about finding an answer; it's about understanding the process, the logic that leads us to the correct solution. We'll go through each step with clarity, ensuring that by the end of this article, you'll not only know the answer but also feel confident tackling similar problems on your own. Whether you're a math whiz looking for a quick refresher or a student grappling with algebraic concepts, this guide is designed for you. So, grab your favorite beverage, settle in, and let's get this math party started! We'll explore the common pitfalls to avoid and the techniques that make solving equations like this a breeze. Remember, the journey of a thousand miles begins with a single step, and in math, that first step is often understanding the problem itself. Let's demystify this equation and find that elusive value of 'x' together.

Isolating the Radical: The First Crucial Step

The first major hurdle in solving an equation like $\sqrt{x+2}-15=-3$ is to get the square root part of the equation all by itself on one side. Think of it like unwrapping a present – you need to get to the main gift before you can really figure out what it is. Right now, our radical, $\sqrt{x+2}$, is being 'attacked' by that '-15'. Our mission is to eliminate that '-15' from the left side of the equation. How do we do that? With the magic of inverse operations! The opposite of subtracting 15 is adding 15. So, we're going to add 15 to both sides of the equation. It's super important to remember that whatever you do to one side of an equation, you must do to the other side to keep it balanced. If you don't, you're essentially throwing off the whole equilibrium, and your solution will be wrong. So, let's add 15 to both sides:

$\sqrt{x+2}-15 + 15 = -3 + 15$

On the left side, the '-15' and '+15' cancel each other out, leaving us with just our beloved radical: $\sqrt{x+2}$. On the right side, we perform the addition: $-3 + 15 = 12$. So, our equation now looks much simpler and more manageable:

$\sqrt{x+2} = 12$

See? We've successfully isolated the square root. This is a huge win! This step is fundamental because it sets us up for the next move, which is to get rid of the square root entirely. Without isolating it first, we wouldn't be able to proceed effectively. It’s all about a systematic approach, building upon each correct step. This initial isolation is where many students gain confidence, realizing that the problem isn't as complex as it initially appeared. We've taken a potentially intimidating equation and transformed it into something much more approachable, paving the way for the final solution.

Eliminating the Square Root: Squaring Both Sides

Now that we have our radical isolated, $\sqrt{x+2} = 12$, the next logical step is to completely remove the square root symbol. The inverse operation of taking a square root is squaring a number. So, to eliminate the square root, we need to square both sides of the equation. Again, the golden rule applies: whatever you do to one side, you do to the other. Squaring the left side will cancel out the square root, and squaring the right side will give us a numerical value. Let's do it:

$(\sqrt{x+2})^2 = (12)^2$

On the left side, $(\sqrt{x+2})^2$ simplifies to just $x+2$, because squaring a square root perfectly undoes the operation. On the right side, $12^2$ means 12 multiplied by 12, which equals 144. So, our equation transforms into:

$x+2 = 144$

This is fantastic! We've now gotten rid of the radical, and we're left with a simple linear equation. This is the goal when dealing with radical equations – to simplify them step-by-step until you reach a form that's easy to solve. This squaring step is critical and is where many potential errors can occur if not applied correctly to both sides. Sometimes, students might forget to square the entire number on the other side, or they might square incorrectly. But by carefully applying the operation to both sides, we ensure accuracy. We're so close to finding the value of 'x' now. This simplification makes the final step a piece of cake.

Solving for 'x': The Final Calculation

We've reached the home stretch, guys! Our equation has been simplified beautifully to $x+2 = 144$. Now, all that stands between us and the solution is that '+2' on the left side. To isolate 'x' completely, we need to perform the inverse operation of adding 2, which is subtracting 2. And, you guessed it, we do this to both sides of the equation to maintain balance.

$x+2 - 2 = 144 - 2$

On the left side, the '+2' and '-2' cancel each other out, leaving us with just 'x'. On the right side, we perform the subtraction: $144 - 2 = 142$. And there it is!

$x = 142$

We have found our solution! So, the answer to "what is the solution of $\sqrt{x+2}-15=-3$?" is $x = 142$. It's always a good practice, especially in exams or when you want to be absolutely sure, to check your answer. Let's plug $x=142$ back into the original equation to see if it holds true:

$\sqrt{142+2} - 15 = -3$ $\sqrt{144} - 15 = -3$ $12 - 15 = -3$ $-3 = -3$

It works! The left side perfectly equals the right side, confirming that $x=142$ is indeed the correct solution. This checking step is invaluable. It confirms your work and builds confidence. It's like double-checking your work before submitting an important assignment. We've successfully navigated the process of solving a radical equation, from isolating the radical to squaring both sides and finally solving for 'x'. This systematic approach ensures accuracy and understanding. So next time you see a similar problem, you’ll know exactly what to do!

Understanding Potential Solutions and Extraneous Roots

When we work with equations involving square roots, there's a sneaky little concept we need to be aware of: extraneous roots. Don't let the fancy name scare you, guys. It's actually pretty straightforward. An extraneous root is a solution that we find through our algebraic steps, but when we plug it back into the original equation, it doesn't work. It's like finding a key that fits the lock but doesn't actually open the door – it looks like it should work, but it's not the real solution.

Why do these extraneous roots pop up? They often appear because of the squaring step we performed. When you square a number, you lose the sign information. For example, both $5^2 = 25$ and $(-5)^2 = 25$. So, when we square both sides of an equation, we might inadvertently introduce solutions that satisfy the squared equation but not the original one where the sign mattered. In our problem, $\sqrt{x+2} = 12$, we squared both sides. If, hypothetically, our isolated radical had ended up equaling a negative number (which is impossible for a principal square root), squaring it would have made it positive, potentially leading to a valid-looking solution that was actually extraneous.

For instance, imagine we had an equation that simplified to $\sqrt{x+2} = -12$. If we then squared both sides, we'd get $x+2 = 144$, leading to $x = 142$. However, if we plug $x=142$ back into $\sqrt{x+2} = -12$, we get $\sqrt{142+2} = \sqrt{144} = 12$. And $12$ is definitely not equal to $-12$. So, in that hypothetical case, $x=142$ would be an extraneous root.

This is precisely why the checking step is not just recommended, but absolutely essential when solving radical equations. It's your safety net. By substituting our calculated value of 'x' back into the original equation, we confirm that it satisfies all conditions, including the implicit ones that the square root must result in a non-negative value (the principal root). In our specific problem, $\sqrt{x+2}-15=-3$, we found $x=142$. When we checked it, we got $12 - 15 = -3$, which is $-3 = -3$. This solution is valid because it works in the original equation. If we had ended up with a solution that didn't satisfy the original equation, we would have concluded that there is no solution for that particular problem, even though our algebraic steps seemed to lead somewhere. Understanding extraneous roots helps us to be more critical and thorough mathematicians, ensuring our answers are not just algebraically derived but also logically sound within the constraints of the original problem.

Final Answer and Options Review

After meticulously working through the problem $\sqrt{x+2}-15=-3$, we've arrived at a definitive solution: $x = 142$. We've followed a clear, step-by-step process: first isolating the radical by adding 15 to both sides, transforming the equation into $\sqrt{x+2} = 12$. Then, we eliminated the square root by squaring both sides, resulting in $x+2 = 144$. Finally, we solved for 'x' by subtracting 2 from both sides, yielding $x = 142$. Crucially, we also performed a check by substituting $x=142$ back into the original equation: $\sqrt{142+2}-15 = \sqrt{144}-15 = 12-15 = -3$. Since $-3 = -3$, our solution is confirmed as correct.

Now, let's look at the options provided:

A. x=142 B. x=232 C. x=322 D. no solution

Our calculated and verified solution, $x = 142$, directly matches option A. The other options, $x=232$ and $x=322$, would not satisfy the original equation. For instance, if we were to check $x=232$, we'd get $\sqrt{232+2}-15 = \sqrt{234}-15$, which is approximately $15.3 - 15 = 0.3$, clearly not equal to -3. Similarly, $x=322$ would also lead to an incorrect result. Since we found a valid solution, option D, "no solution," is also incorrect for this particular problem.

Therefore, the correct answer to the question "What is the solution of $\sqrt{x+2}-15=-3$?" is A. x=142. It’s always satisfying when the answer neatly aligns with one of the given choices, isn't it? This confirms our methodical approach was spot on. Keep practicing these types of problems, guys, and you'll become masters of radical equations in no time!