Solving Sqrt(x) + 2 = Sqrt(x + 11) A Step-by-Step Guide

Introduction

Hey guys! Let's dive into a fun mathematical challenge today. We're going to tackle the equation √x + 2 = √(x + 11). This type of equation, involving square roots, might seem a bit daunting at first, but don't worry! We'll break it down step by step, making sure everyone understands the process. Our goal is not just to find the solution, but also to understand the underlying principles that make it work. So, grab your thinking caps, and let’s get started! This equation falls under the category of radical equations, which are equations that contain variables within radical expressions (like square roots, cube roots, etc.). Solving these requires a bit of algebraic manipulation, and careful attention to detail to avoid common pitfalls. One of the key things to remember when dealing with square roots is that the expression inside the square root (the radicand) must be non-negative, since the square root of a negative number is not a real number. This means we need to keep in mind that both x and (x + 11) must be greater than or equal to zero. Also, when we square both sides of the equation, we need to be cautious about introducing extraneous solutions – values that satisfy the transformed equation but not the original one. We'll check for these at the end to make sure our answer is correct. So, with these considerations in mind, let's jump into the solution process!

Step-by-Step Solution

1. Isolating the Square Root (Initial Setup)

Our initial equation is √x + 2 = √(x + 11). Before we can start getting rid of those square roots, we need to think about the best way to approach this. Notice that we have square roots on both sides of the equation. A good strategy here is to try and isolate one of the square root terms. In this case, it looks like the equation is already set up nicely for us, with the terms spread out. We have √x on the left side, a constant 2 also on the left, and √(x + 11) on the right. There's no immediate need to rearrange anything. This is a crucial first step because it sets us up for the next operation: squaring both sides. If we didn't have the square roots somewhat isolated, squaring could lead to a more complex expression due to cross-terms. By having the equation in this form, we can proceed more cleanly and efficiently. So, remember, when you encounter a radical equation, take a moment to assess the layout and see if any isolation steps are needed before you jump into squaring. This can save you from unnecessary complications down the road. Now that we've got our equation in a good starting position, let's move on to the next step, which involves eliminating those pesky square roots!

2. Squaring Both Sides (Eliminating Square Roots)

Now comes the fun part: getting rid of those square roots! We do this by squaring both sides of the equation. Remember, whatever you do to one side of an equation, you have to do to the other to keep things balanced. So, we're taking (√x + 2) and squaring it, and we're also taking √(x + 11) and squaring it. Let’s break this down. On the left side, we have (√x + 2)². This is like multiplying (√x + 2) by itself. If we use the FOIL method (First, Outer, Inner, Last), or the formula for squaring a binomial (a + b)² = a² + 2ab + b², we get:

• (√x)² + 2(√x)(2) + 2² = x + 4√x + 4

On the right side, we have (√(x + 11))². Squaring a square root is pretty straightforward – they essentially cancel each other out, leaving us with just the expression inside the square root, which is (x + 11). So, our equation now looks like this:

• x + 4√x + 4 = x + 11

We've successfully eliminated one square root, but notice we still have a square root term (4√x) on the left side. Don't worry, we're not done yet! We've made progress, and we're on the right track. The next step will involve isolating this remaining square root term so we can square again and finally get rid of all the radicals. This process might seem a bit repetitive, but it's a standard technique for solving radical equations. So, take a deep breath, and let’s move on to the next step!

3. Simplifying and Isolating the Remaining Square Root

Okay, so after squaring both sides, we landed at the equation x + 4√x + 4 = x + 11. Now, we need to isolate that remaining square root term, which is 4√x. To do this, we'll perform some algebraic acrobatics – specifically, we'll subtract terms from both sides of the equation to get the 4√x term by itself. First, notice that we have an x term on both sides of the equation. We can subtract x from both sides, and they'll cancel out:

• x + 4√x + 4 - x = x + 11 - x
• 4√x + 4 = 11

Now, we want to get rid of that +4 on the left side. So, we'll subtract 4 from both sides:

• 4√x + 4 - 4 = 11 - 4
• 4√x = 7

Look at that! We've successfully isolated the square root term. We now have 4√x = 7. This is great progress! We're one step closer to getting rid of the square root entirely. Before we square both sides again, it might be helpful to divide both sides by 4 to simplify the equation a bit further. This will make the numbers smaller and easier to work with in the next step. So, let's divide both sides by 4:

• (4√x) / 4 = 7 / 4
• √x = 7/4

Now we have a much simpler equation with the square root isolated. This makes the next step, squaring both sides again, a lot cleaner and less prone to errors. So, we've simplified and isolated – fantastic work! Let's move on to the next step and finally eliminate that last square root!

4. Squaring Again and Solving for x

Alright, we've reached a point where our equation looks much cleaner: √x = 7/4. We've successfully isolated the square root, so now it's time to square both sides again to eliminate it completely. Remember, squaring a square root just cancels it out, leaving us with the expression inside. So, when we square √x, we simply get x.

On the right side, we have (7/4)². This means we need to square both the numerator (7) and the denominator (4). So,

• 7² = 49
• 4² = 16

Therefore, (7/4)² = 49/16. Now our equation looks like this:

• x = 49/16

Wow! It seems like we've found a solution! We've isolated x, and we have a value for it. But before we get too excited and declare victory, there's a crucial step we need to take: checking for extraneous solutions. Remember, whenever we square both sides of an equation, there's a possibility that we might introduce solutions that don't actually work in the original equation. These are called extraneous solutions, and we need to make sure our answer isn't one of them. So, we'll take this value of x, 49/16, and plug it back into the original equation to see if it holds true. This is a critical step in solving radical equations, so let's do it carefully!

5. Checking for Extraneous Solutions

Okay, we've arrived at a potential solution: x = 49/16. But as we discussed, we need to be absolutely sure this solution works in the original equation. This is where the check for extraneous solutions comes in. It's like our final exam, making sure we've done everything correctly.

Our original equation was √x + 2 = √(x + 11). Let's substitute x = 49/16 into this equation and see what happens.

Left side:

• √(49/16) + 2

We know that √(49/16) is the same as √49 / √16, which is 7/4. So, the left side becomes:

• 7/4 + 2

To add these, we need a common denominator. We can rewrite 2 as 8/4, so we have:

• 7/4 + 8/4 = 15/4

So, the left side simplifies to 15/4.

Right side:

• √(49/16 + 11)

First, we need to add 49/16 and 11. We can rewrite 11 as a fraction with a denominator of 16: 11 = 176/16. So, we have:

• √(49/16 + 176/16) = √(225/16)

Now, we take the square root of both the numerator and the denominator:

• √225 / √16 = 15/4

So, the right side also simplifies to 15/4.

Comparing both sides:

• Left side: 15/4
• Right side: 15/4

They are equal! This means that x = 49/16 does satisfy the original equation. It's not an extraneous solution. We've successfully checked our work, and we can confidently say that we've found the solution.

Conclusion

Woohoo! We did it! We successfully solved the equation √x + 2 = √(x + 11). The solution is x = 49/16. But more importantly, we've walked through the entire process step by step, understanding why each step is necessary. We talked about isolating square roots, squaring both sides, simplifying equations, and, crucially, checking for extraneous solutions. This last step is so important because it ensures that our answer is valid and not just a byproduct of our algebraic manipulations. Solving radical equations like this one can seem tricky at first, but with practice and a systematic approach, you can conquer them! Remember, the key is to break down the problem into smaller, manageable steps, and to always double-check your work. Keep practicing, and you'll become a radical equation-solving pro in no time! If you guys have any questions or want to try another example, just let me know. Happy solving!