Solving The Equation Sqrt(x+1) = X-5 A Step-by-Step Guide

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Hey guys! Today, we're diving into a fun little algebraic adventure where we'll tackle the equation ${\sqrt{x+1} = x-5}$. This equation involves a square root, so we'll need to be a bit careful about how we approach it. Let's break it down step by step to make sure we not only find the solution but also understand why each step is necessary. Get ready to sharpen those pencils (or keyboards!) and let's get started!

Understanding the Challenge

When you first look at the equation ${\sqrt{x+1} = x-5}$, the square root might seem a bit intimidating. But don't worry, it's totally manageable! The key here is to remember that we need to isolate the variable x. However, the square root is kind of acting like a cage around the x on the left side. Our main goal is to free x from this cage. To do that, we're going to use the inverse operation of the square root, which is squaring. But before we jump into squaring both sides, it's super important to understand why we do what we do. Squaring both sides is a powerful tool, but it can sometimes introduce solutions that aren't actually solutions to the original equation. These sneaky little guys are called extraneous solutions, and we'll talk more about them later.

Think of it like this: imagine you're trying to find the secret entrance to a hidden room. The square root is like a complicated lock, and squaring is like using the right key to unlock it. But sometimes, the key might also unlock a door to a completely different room that you didn't intend to enter. That's what extraneous solutions are – they're like entering the wrong room. So, we need to be extra careful to check our answers at the end to make sure we're in the right room (i.e., that our solutions actually work in the original equation).

Another thing to consider is the domain of the square root function. Remember, the square root of a negative number isn't a real number. So, the expression inside the square root, which is ${x+1}$, must be greater than or equal to zero. This gives us a little side condition: ${x+1 \geq 0}$, which means ${x \geq -1}$. This is like setting a boundary for where we can look for solutions. We can't go wandering off into negative territory less than -1, because that would lead us into imaginary number land, which isn't what we want for this problem.

So, to recap, before we even start solving, we know two crucial things: we need to square both sides to get rid of the square root, and we need to remember to check for extraneous solutions. Plus, we know that x has to be greater than or equal to -1. With these things in mind, we're ready to start cracking this equation!

Step-by-Step Solution

Okay, let's get down to business and solve this equation step-by-step. Remember our equation: ${\sqrt{x+1} = x-5}$. The first thing we want to do, as we discussed, is to get rid of that pesky square root. And how do we do that? By squaring both sides! This is a fundamental algebraic technique – whatever you do to one side of the equation, you must do to the other to keep the equation balanced. It's like a seesaw; if you add weight to one side, you need to add the same weight to the other side to keep it level.

So, let's square both sides:

${(\sqrt{x+1})^2 = (x-5)^2}$

On the left side, the square root and the square cancel each other out. This is the magic of inverse operations! They undo each other, leaving us with just the expression inside the square root, which is ${x+1}$. On the right side, we have ${(x-5)^2}$. This means ${(x-5)}$ multiplied by itself. We need to expand this carefully. Remember, ${(a-b)^2}$ is not equal to ${a^2 - b^2}$. Instead, we need to use the formula ${(a-b)^2 = a^2 - 2ab + b^2}$. This is a common mistake, so let's make sure we get it right.

Expanding ${(x-5)^2}$, we get:

${(x-5)(x-5) = x^2 - 5x - 5x + 25 = x^2 - 10x + 25}$

So, our equation now looks like this:

${x+1 = x^2 - 10x + 25}$

Awesome! We've gotten rid of the square root, and we're left with a quadratic equation. Quadratic equations are equations of the form ${ax^2 + bx + c = 0}$, where a, b, and c are constants. To solve a quadratic equation, we usually want to set it equal to zero and then either factor it, use the quadratic formula, or complete the square. In this case, let's set our equation equal to zero.

To do that, we'll subtract x and 1 from both sides:

${0 = x^2 - 10x + 25 - x - 1}$

Simplifying, we get:

${0 = x^2 - 11x + 24}$

Now we have a quadratic equation in standard form. The next step is to try to factor it. Factoring is like finding the two numbers that multiply to give us the constant term (24 in this case) and add up to give us the coefficient of the x term (-11 in this case). Can you think of two numbers that do the trick?

The numbers -3 and -8 work perfectly! Because (-3) * (-8) = 24 and (-3) + (-8) = -11. So, we can factor the quadratic equation as follows:

${0 = (x - 3)(x - 8)}$

Now we have the equation factored, which is fantastic. The zero product property states that if the product of two factors is zero, then at least one of the factors must be zero. This is a super useful property for solving equations like this. It means that either ${x-3 = 0}$ or ${x-8 = 0}$.

Solving these two simple equations, we get:

${x = 3 \text{ or } x = 8}$

So, we have two potential solutions: x = 3 and x = 8. But remember our earlier warning about extraneous solutions? We're not out of the woods yet! We need to check if these solutions actually work in the original equation.

Checking for Extraneous Solutions

Alright, we've arrived at a crucial step in solving our equation: checking for extraneous solutions. Remember, extraneous solutions are those sneaky values that pop up during the solving process (in our case, squaring both sides) but don't actually satisfy the original equation. They're like imposters pretending to be solutions, and it's our job to unmask them!

We found two potential solutions: x = 3 and x = 8. To check them, we need to plug each one back into the original equation, which is ${\sqrt{x+1} = x-5}$, and see if the equation holds true. It's like testing a key in a lock – it either fits and opens the door (the solution works), or it doesn't (the solution is extraneous).

Let's start with x = 3. Plugging this into the equation, we get:

${\sqrt{3+1} = 3-5}$

Simplifying the left side:

${\sqrt{4} = 2}$

Simplifying the right side:

${3-5 = -2}$

So, we have:

${2 = -2}$

This is definitely not true! 2 does not equal -2. This means that x = 3 is an extraneous solution. It's an imposter! It doesn't belong in our solution set.

Now, let's check x = 8. Plugging this into the original equation, we get:

${\sqrt{8+1} = 8-5}$

Simplifying the left side:

${\sqrt{9} = 3}$

Simplifying the right side:

${8-5 = 3}$

So, we have:

${3 = 3}$

This is true! 3 does indeed equal 3. This means that x = 8 is a legitimate solution. It's the real deal!

So, after checking our potential solutions, we've found that x = 3 is an extraneous solution, and x = 8 is the only valid solution. This highlights the importance of always checking your solutions when you're dealing with equations that involve square roots (or other operations that can introduce extraneous solutions). It's like double-checking your work on a test – it can save you from making a mistake!

Final Answer

We've reached the end of our algebraic journey, guys! We started with the equation ${\sqrt{x+1} = x-5}$, and after a series of steps – squaring both sides, solving the resulting quadratic equation, and most importantly, checking for extraneous solutions – we've arrived at our final answer.

The only solution to the equation ${\sqrt{x+1} = x-5}$ is:

${x = 8}$

That's it! We've successfully navigated the square root, conquered the quadratic equation, and unmasked the extraneous solution. Give yourselves a pat on the back! This problem illustrates a really important concept in algebra: the need to be careful when performing operations that can introduce extraneous solutions. Squaring both sides of an equation is a powerful technique, but it comes with the responsibility of checking your answers.

Remember, mathematics is not just about finding the right answer; it's about understanding the process and the why behind each step. By understanding why we square both sides, why we need to check for extraneous solutions, and how to solve quadratic equations, you've gained valuable tools that you can use to tackle many other algebraic problems. So, keep practicing, keep exploring, and keep that mathematical curiosity alive!