Solving For Y: A Step-by-Step Guide To Y = √(2y + 15)

Hey math enthusiasts! Ever stumbled upon an equation that looks a bit intimidating but is actually super fun to solve? Today, we're diving into one such equation: y = √(2y + 15). Don't worry, we'll break it down step-by-step so you can confidently tackle similar problems in the future. So, let's get started and find out the real value of y!

Understanding the Equation and Initial Steps

Before we jump into the nitty-gritty, let's quickly understand what we're dealing with. We have an equation where y is equal to the square root of a binomial expression involving y itself. This means we need to isolate y while keeping in mind the properties of square roots and real numbers. This involves algebraic manipulation to eliminate the square root and arrive at a solvable equation, typically a quadratic equation. Remember folks, when dealing with square roots, we need to be mindful of potential extraneous solutions, which are solutions that arise from the solving process but don't actually satisfy the original equation. This usually happens because squaring both sides of an equation can introduce solutions that were not there initially. It's crucial to check our solutions by plugging them back into the original equation to ensure they are valid. So, let's dive in and get this equation solved!

The first key step in solving for y in the equation y = √(2y + 15) is to eliminate the square root. How do we do that? By squaring both sides of the equation! Squaring both sides gets rid of the square root on the right side, making the equation much easier to work with. When we square the left side (y), we get y². When we square the right side (√(2y + 15)), the square root disappears, leaving us with just 2y + 15. So, after this step, our equation transforms from a square root equation into a quadratic equation, which we know how to handle. Remember, squaring both sides is a fundamental technique in algebra for dealing with square roots, but it's essential to remember that this can sometimes introduce extraneous solutions, which we'll need to check later. Now that we've eliminated the square root, we can proceed with rearranging the equation into a standard quadratic form. This sets us up nicely for using methods like factoring or the quadratic formula to find the possible values of y. So, let’s move on to the next step and see how we can transform our equation into a familiar format.

Transforming into a Quadratic Equation

Now that we've squared both sides, we've got ourselves a quadratic equation brewing. To solve it effectively, we need to get it into the standard quadratic form, which is ax² + bx + c = 0. This form is super helpful because it allows us to easily identify the coefficients a, b, and c, which we'll need for methods like factoring or the quadratic formula. To get our equation into this form, we'll need to rearrange the terms. This typically involves moving all terms to one side of the equation, leaving zero on the other side. In our case, we need to move the terms 2y and 15 from the right side to the left side. This means subtracting 2y and 15 from both sides of the equation. Once we do this, we'll have a quadratic equation in the standard form, ready to be solved. Remember, the order of terms is important in the standard form, with the x² term first, followed by the x term, and then the constant term. Keeping the equation in this order helps prevent mistakes when applying solution methods. So, let's rearrange the terms and see what our quadratic equation looks like in its standard form!

After squaring both sides of the original equation y = √(2y + 15) and rearranging the terms, we arrive at the quadratic equation: y² - 2y - 15 = 0. See how we've moved everything to one side, setting the equation equal to zero? This is the standard form, and it's our gateway to solving for y. Now, we have a couple of options for tackling this quadratic: we can try factoring, or we can use the quadratic formula. Factoring is like a puzzle – we try to find two binomials that multiply together to give us our quadratic expression. If we can find those binomials, we can easily find the roots (or solutions) of the equation. The quadratic formula, on the other hand, is a reliable workhorse that always gives us the solutions, even when factoring is tricky. It might look a bit intimidating at first glance, but it's a powerful tool to have in your mathematical toolkit. Before we decide which method to use, let's take a look at our equation and see if factoring seems like a viable option. Sometimes, with a bit of practice, you can spot the factors quickly, saving you some time and effort. So, let's explore both factoring and the quadratic formula to see which one works best for us in this case.

Solving the Quadratic Equation by Factoring

Okay, let's see if we can crack this quadratic equation by factoring. Factoring is a neat trick when it works, and it can often be quicker than using the quadratic formula. The idea behind factoring is to rewrite the quadratic expression as a product of two binomials. For our equation, y² - 2y - 15 = 0, we need to find two numbers that multiply to -15 (the constant term) and add up to -2 (the coefficient of the y term). This might sound like a bit of a puzzle, but with a little practice, you'll get the hang of it! Think about the factors of -15: we have pairs like 1 and -15, -1 and 15, 3 and -5, and -3 and 5. Which of these pairs adds up to -2? Aha! It's 3 and -5. So, those are the numbers we'll use to factor our quadratic. Once we find these numbers, we can rewrite the quadratic equation in its factored form. This form gives us a direct path to finding the solutions for y. Remember, the factored form is just another way of expressing the same quadratic equation, but it's in a format that makes the solutions much clearer. So, let's put these numbers into action and see how our quadratic equation looks when it's factored!

Using the numbers 3 and -5, we can factor the quadratic equation y² - 2y - 15 = 0 into (y + 3)(y - 5) = 0. See how the numbers 3 and -5 fit perfectly into the binomial factors? This is the beauty of factoring! Now that we have the equation in this form, we can use the zero-product property to find the solutions for y. The zero-product property is a fundamental concept in algebra, and it states that if the product of two factors is zero, then at least one of the factors must be zero. In our case, this means that either (y + 3) must be zero, or (y - 5) must be zero. This gives us two separate equations to solve, each of which is very simple. We just need to isolate y in each equation to find the possible solutions. So, we'll set each factor equal to zero and solve for y. This will give us two potential values for y, but remember, we still need to check these solutions in the original equation to make sure they're not extraneous. Factoring has made it relatively straightforward to find these potential solutions, and now we're just one step away from identifying the actual values of y that satisfy our equation. So, let's solve each of these mini-equations and see what we get!

Setting each factor to zero gives us two simple equations: y + 3 = 0 and y - 5 = 0. Solving y + 3 = 0 is straightforward – just subtract 3 from both sides, and we get y = -3. Similarly, solving y - 5 = 0 involves adding 5 to both sides, which gives us y = 5. So, we have two potential solutions for y: -3 and 5. But before we declare victory and circle these as our final answers, we need to do a crucial step: check for extraneous solutions. Remember how we talked about squaring both sides of the equation potentially introducing solutions that don't actually work in the original equation? This is where that comes into play. We need to plug each of these values back into the original equation y = √(2y + 15) to see if they hold true. If a value makes the equation true, it's a valid solution. If it doesn't, it's an extraneous solution and we need to discard it. This check is a critical part of solving equations involving square roots, so don't skip it! It's like the final quality control step in our problem-solving process. So, let's put these potential solutions to the test and see which ones make the cut.

Checking for Extraneous Solutions

Alright, we've got our potential solutions, y = -3 and y = 5, but we're not done yet! This is the crucial step where we check for extraneous solutions. We need to plug each value back into the original equation, y = √(2y + 15), and see if it holds true. Let's start with y = -3. Substituting this into the equation, we get -3 = √(2(-3) + 15). Simplifying the right side, we have -3 = √(-6 + 15), which simplifies further to -3 = √9. The square root of 9 is 3, so we end up with -3 = 3. This is definitely not true! So, y = -3 is an extraneous solution and we need to discard it. It doesn't satisfy the original equation, even though it popped out when we solved the quadratic. This is why checking for extraneous solutions is so important. Now, let's test our other potential solution, y = 5. Substituting this into the original equation, we get 5 = √(2(5) + 15). Simplifying the right side, we have 5 = √(10 + 15), which becomes 5 = √25. The square root of 25 is 5, so we have 5 = 5. This is a true statement! So, y = 5 is a valid solution. It satisfies the original equation, and we can confidently include it in our final answer. This process of checking solutions is like a detective ensuring that we only keep the true suspects and eliminate the imposters. So, with this check completed, we're finally ready to state our solution.

Stating the Solution

After carefully solving the equation y = √(2y + 15) and checking for extraneous solutions, we've arrived at our final answer. We found two potential solutions, y = -3 and y = 5, but after plugging them back into the original equation, we discovered that y = -3 was an extraneous solution. This means it doesn't actually satisfy the original equation, so we had to discard it. The other solution, y = 5, did check out perfectly. When we substituted y = 5 back into the original equation, both sides were equal, confirming that it is indeed a valid solution. So, we can confidently say that the solution to the equation y = √(2y + 15) is y = 5. This is the only real number that makes the equation true. We've gone through the entire process, from squaring both sides to eliminate the square root, to solving the resulting quadratic equation, to the crucial step of checking for extraneous solutions. Each step was important in ensuring that we arrived at the correct answer. So, if you encounter a similar equation in the future, remember these steps, and you'll be well-equipped to solve it. Great job, everyone! We've successfully tackled this math problem together. Now, go forth and conquer more mathematical challenges!