Solving The Equation Sqrt(-2x-5)-4=x A Step-by-Step Guide

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Hey guys! Let's dive into solving this equation together. It looks a bit tricky with the square root, but we'll break it down step by step to make it super clear. Our mission is to find the value(s) of x that make the equation $\sqrt{-2x-5}-4=x$ true. We'll explore how to isolate the square root, deal with potential extraneous solutions, and ultimately nail the correct answer.

Understanding the Equation

At the heart of our equation, the square root $\sqrt{-2x-5}$ is the key element we need to tackle first. Remember, the expression inside the square root, which is "-2x-5", has to be greater than or equal to zero because we can't take the square root of a negative number and get a real result. This gives us a crucial constraint that will help us later when we check our solutions. Also, we have the term "-4" subtracting from the square root and "x" on the other side of the equation, setting up an intriguing algebraic puzzle.

Isolating the square root is the crucial first step. Think of it as peeling away the layers to get to the core. To do this, we need to get the square root term all by itself on one side of the equation. We'll add 4 to both sides, which will effectively cancel out the "-4" on the left and move it to the right. This gives us a new, simplified equation that's much easier to work with. By isolating the square root, we set the stage for squaring both sides, which is the next big move in our solution strategy.

Now, after isolating, we have the equation $\sqrt{-2x-5} = x + 4$. Squaring both sides is where the magic happens! This eliminates the square root, but it's super important to remember that this step can sometimes introduce solutions that don't actually work in the original equation. These are called extraneous solutions, and we'll need to watch out for them. When we square both sides, we're essentially undoing the square root, but we also need to be careful about the algebra on the right side. Squaring "x + 4" means multiplying it by itself, which gives us a quadratic expression. This sets us up to solve a quadratic equation, which we can do by getting everything on one side and either factoring or using the quadratic formula.

Solving the Equation

Let's dive into squaring both sides. As we mentioned, squaring $\sqrt{-2x-5}$ simply gives us "-2x-5". But squaring "x + 4" requires a little more attention. We need to remember that (x + 4)² is the same as (x + 4)(x + 4). Using the FOIL method (First, Outer, Inner, Last) or the distributive property, we get x² + 8x + 16. So, our equation now transforms into -2x - 5 = x² + 8x + 16. See how squaring both sides gets rid of the square root but also introduces a quadratic term? That's a classic move in these types of problems.

Now, we need to rearrange the equation into the standard quadratic form, which is ax² + bx + c = 0. This makes it much easier to solve. To do this, we'll move all the terms from the left side to the right side. We'll add 2x to both sides and add 5 to both sides. This way, we get a zero on the left and a quadratic expression on the right. This step is all about getting our equation into a recognizable format that we know how to deal with. Once we have it in standard form, we can think about factoring, using the quadratic formula, or even completing the square if we're feeling ambitious!

After rearranging, our equation looks like this: 0 = x² + 10x + 21. This is a quadratic equation in standard form, and now we're ready to factor the quadratic. Factoring is like reverse-multiplying. We're looking for two numbers that multiply to 21 and add up to 10. If you think about the factors of 21, you'll quickly realize that 3 and 7 fit the bill perfectly! So, we can factor the quadratic expression as (x + 3)(x + 7). This gives us a super useful form of the equation because now we can use the zero-product property.

The zero-product property is a fundamental concept when solving equations by factoring. It basically says that if the product of two factors is zero, then at least one of the factors must be zero. In our case, we have (x + 3)(x + 7) = 0. This means that either x + 3 = 0 or x + 7 = 0 (or both!). To find the possible values of x, we simply set each factor equal to zero and solve. This gives us two potential solutions for x, which we'll need to check later to make sure they actually work in the original equation. Solving each of these simple equations gives us our candidate solutions for x.

Setting each factor to zero, we get x + 3 = 0, which gives us x = -3, and x + 7 = 0, which gives us x = -7. So, we have two potential solutions: x = -3 and x = -7. But hold on! We're not done yet. We need to remember the crucial step of checking for extraneous solutions. Just because we found these values algebraically doesn't mean they actually satisfy the original equation with the square root. This is where the careful work we did at the beginning, understanding the constraints of the square root, really pays off. We need to plug each of these values back into the original equation and see if they make it true. If they don't, then they're extraneous and we need to discard them.

Checking for Extraneous Solutions

This is a critical step! We have two potential solutions, x = -3 and x = -7, but we need to plug them back into the original equation $\sqrt{-2x-5}-4=x$ to see if they actually work. This is like the final exam for our solutions – they need to pass the test to be considered valid. Remember, squaring both sides can sometimes introduce solutions that don't fit the original equation, so this check is not optional.

Let's start with checking x = -3. Plugging it into the equation, we get $\sqrt{-2(-3)-5}-4=-3$. Simplifying inside the square root, we have $\sqrt{6-5}-4=-3$, which becomes $\sqrt{1}-4=-3$. Since the square root of 1 is 1, we have 1 - 4 = -3, which simplifies to -3 = -3. This is a true statement! So, x = -3 is a valid solution. It passes the test and gets to stay in our solution set.

Now, let's check x = -7. Plugging it into the original equation, we get $\sqrt{-2(-7)-5}-4=-7$. Simplifying inside the square root, we have $\sqrt{14-5}-4=-7$, which becomes $\sqrt{9}-4=-7$. The square root of 9 is 3, so we have 3 - 4 = -7, which simplifies to -1 = -7. This is a false statement! So, x = -7 is an extraneous solution. It doesn't work in the original equation, and we have to discard it. This highlights why checking for extraneous solutions is so important.

The Solution

After carefully solving the equation and checking for extraneous solutions, we've arrived at our final answer. We found two potential solutions, x = -3 and x = -7, but only one of them, x = -3, actually works in the original equation. The other solution, x = -7, turned out to be extraneous and had to be discarded. This process demonstrates the importance of not only solving the equation algebraically but also verifying our solutions to ensure they are valid.

Therefore, the correct answer is C. -3. Great job, everyone! We tackled a tricky equation with a square root, handled potential extraneous solutions, and nailed the final answer. Remember, math is all about breaking down problems into manageable steps and carefully checking our work along the way. You guys got this!