Solving X + √(2x + 5) = 5: A Step-by-Step Guide

Hey everyone, let's dive into solving the equation x + √(2x + 5) = 5. This problem is a classic example of an equation involving a square root, and it requires a bit of algebra to isolate the variable, x. It's a fun challenge, and understanding how to solve it can really boost your math skills. So, grab your pencils, and let's get started. We'll break down the steps to find the solution set, which is essentially the set of all values of 'x' that make the equation true. Remember, when dealing with square roots, we need to be extra careful about extraneous solutions – solutions that appear to satisfy the equation but don't actually work when plugged back in. We will cover this aspect to ensure our solution set is accurate. This will be an exciting journey, and I’m sure you guys will find this easy. Let’s get started.

First things first, our main goal is to isolate the square root term. This will make it easier to get rid of the square root later on. To do this, we need to move the 'x' term to the other side of the equation. This is achieved by subtracting 'x' from both sides. We perform these kinds of operations to keep our equation balanced – whatever we do to one side, we must do to the other. This ensures that the equality remains valid throughout the solution process. So, after doing this, the equation transforms a bit. This step is about simplifying the equation to a form where we can deal with the square root more effectively. It’s like clearing the way before tackling the main problem. Keep in mind that these initial steps are crucial as they set the stage for the rest of the solution. Every mathematical operation must follow the rules of algebra.

After isolating the square root, you will be left with the equation in the form of √(2x + 5) = 5 - x. The next step involves getting rid of that pesky square root. We do this by squaring both sides of the equation. This is a fundamental technique in algebra. But hold on a sec, what does that really mean? Squaring means raising each side of the equation to the power of 2. So, when you square the left side, the square root vanishes, leaving you with just 2x + 5. When you square the right side (5 - x), you'll need to expand it using the distributive property (or FOIL method if you're familiar with that). This step is where things can get a little tricky, so pay close attention. Squaring both sides might introduce extraneous solutions. This is why we absolutely must check our final answers later. It’s like setting a trap; the equation might give us an answer that doesn't actually fit. That is why checking the answer is so important. Trust me, it’s a lifesaver. Keep the whole process in mind; it's going to be essential for us to solve more complicated equations.

Squaring Both Sides and Simplifying the Equation

Alright, now that we've isolated the square root term, let's square both sides of the equation √(2x + 5) = 5 - x. This is a crucial step to eliminate the square root and move towards solving for 'x'. As we discussed earlier, squaring both sides means raising each side of the equation to the power of 2. So, on the left side, (√(2x + 5))^2 simplifies to 2x + 5, as the square and square root cancel each other out. On the right side, we have (5 - x)^2. This needs to be expanded carefully. Using the distributive property, or the FOIL method, we get (5 - x)(5 - x) = 25 - 10x + x^2. This step is where many people can mess up, so make sure to double-check that you’ve expanded correctly. It's easy to make a small mistake here, which can mess up the rest of your solution.

Now, our equation looks like this: 2x + 5 = 25 - 10x + x^2. Notice that we now have a quadratic equation (an equation where the highest power of the variable is 2). This means we'll likely end up with two possible solutions for 'x'. To solve this, we need to rearrange the equation to set it equal to zero. This involves moving all terms to one side of the equation. Let’s do it step by step. First, subtract 2x from both sides, then subtract 5 from both sides. This will give us a new equation where the right side will equal zero. Be careful with those negative signs! After simplifying, the equation becomes x^2 - 12x + 20 = 0. This is the standard form of a quadratic equation, and it's now ready to be solved. We are almost there, guys, keep it up.

Solving quadratic equations can be done in a few different ways: factoring, completing the square, or using the quadratic formula. Let’s look at our options. Factoring is usually the easiest and fastest method, if the equation can be factored easily. Completing the square is another option, but it can be more complicated. The quadratic formula always works, but it can be a bit more time-consuming. In this case, our equation x^2 - 12x + 20 = 0 is indeed factorable. We need to find two numbers that multiply to 20 and add up to -12. These numbers are -2 and -10. So, we can factor the equation into (x - 2)(x - 10) = 0. This factored form tells us the potential solutions for 'x'. It's the moment of truth!

Finding Potential Solutions for x

Alright, we've successfully factored our quadratic equation into (x - 2)(x - 10) = 0. This form is fantastic because it allows us to easily find the potential solutions for 'x'. The Zero Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. Think of it like this: if you have two numbers multiplied together, and the result is zero, one (or both) of the numbers must be zero. Using this property, we can set each factor equal to zero and solve for x. First, let's set x - 2 = 0. Adding 2 to both sides gives us x = 2. This is our first potential solution. Now, let’s consider the other factor and set x - 10 = 0. Adding 10 to both sides gives us x = 10. This is our second potential solution. So, at this point, we have two possible answers: x = 2 and x = 10. But hold on, we're not done yet. Remember what we talked about earlier? Extraneous solutions? That’s where we're going to use this. We need to check these solutions in the original equation to make sure they actually work. It’s like double-checking your work to make sure you didn’t miss anything. This is a super important step, trust me.

Verifying the Solutions

We have two potential solutions: x = 2 and x = 10. Now comes the critical step: verifying these solutions to make sure they satisfy the original equation, x + √(2x + 5) = 5. As we've discussed, squaring both sides of an equation can sometimes introduce extraneous solutions. So, we've got to check our answers. Let’s begin by substituting x = 2 into the original equation. That means replacing every instance of 'x' with the number 2. We get 2 + √(22 + 5) = 5. Simplify the expression inside the square root: 22 + 5 = 4 + 5 = 9. So, the equation becomes 2 + √9 = 5. The square root of 9 is 3, so we have 2 + 3 = 5. And guess what? 2 + 3 does equal 5. So, x = 2 is a valid solution. Great job, guys!

Now, let's move on to the second potential solution, x = 10. We'll substitute x = 10 into the original equation: 10 + √(210 + 5) = 5. Simplify the expression inside the square root: 210 + 5 = 20 + 5 = 25. The equation becomes 10 + √25 = 5. The square root of 25 is 5, so we get 10 + 5 = 5. Uh oh… 10 + 5 equals 15, not 5. This means that x = 10 does not satisfy the original equation. It's an extraneous solution! So, we discard x = 10 as a valid solution. This is exactly why it’s so important to check our answers.

Determining the Solution Set

After all that hard work, we’re now ready to define our solution set. We’ve found that x = 2 is a valid solution, while x = 10 is an extraneous solution and, therefore, incorrect. The solution set represents all the values of 'x' that satisfy the original equation. In our case, only one value of 'x' works. So, our solution set is simply {2}. This means that the only value of 'x' that makes the equation x + √(2x + 5) = 5 true is 2. Congratulations! You've successfully solved the equation. Wasn't that fun?

This whole process demonstrates the importance of algebraic manipulations, especially when dealing with square roots. The careful squaring of both sides, the factoring of the quadratic equation, and the essential verification of the solutions—each step is crucial. This problem is a great example of how mathematical problem-solving requires us to be precise and always double-check our work. Extraneous solutions can be tricky, but by taking the time to verify our answers, we make sure our solution set is accurate. Now you can use this knowledge and approach to tackle other equations involving square roots with confidence. Keep practicing, and you'll become a master of these kinds of problems in no time. If you got any questions, don’t be afraid to ask, and keep up the great work! You guys rock!