Solving The Equation Sqrt(4y^2 - 5y + 28) = 2y + 1 A Step-by-Step Guide

Hey guys! Let's dive into a cool mathematical problem today. We're going to tackle the equation $\sqrt{4y^2 - 5y + 28} = 2y + 1$. This kind of problem is super common in algebra and it’s a great way to sharpen your equation-solving skills. So, grab your pencils, and let's get started!

Breaking Down the Problem: Our Initial Steps

When we look at an equation like $\sqrt{4y^2 - 5y + 28} = 2y + 1$, the first thing that probably jumps out at you is the square root. Square roots can be a little tricky to deal with directly, so our main goal is to get rid of it. How do we do that? Simple! We square both sides of the equation. Squaring both sides is a fundamental technique in algebra. It helps us eliminate radicals and simplify the equation into something more manageable. But remember, guys, whenever we square both sides, we have to be extra careful about potential extraneous solutions. These are solutions that pop up during the solving process but don’t actually work in the original equation. So, we’ll definitely need to check our answers later.

So, let's square both sides of our equation. On the left side, squaring $\sqrt{4y^2 - 5y + 28}$ just cancels out the square root, leaving us with $4y^2 - 5y + 28$. On the right side, we have $(2y + 1)^2$. Remember, guys, to expand this correctly, we need to use the formula $(a + b)^2 = a^2 + 2ab + b^2$. So, $(2y + 1)^2$ becomes $(2y)^2 + 2(2y)(1) + 1^2$, which simplifies to $4y^2 + 4y + 1$. Now our equation looks like this: $4y^2 - 5y + 28 = 4y^2 + 4y + 1$. See? Already, it looks a lot cleaner and easier to handle!

Simplifying the Equation: A Little Algebraic Magic

Alright, now that we've squared both sides, we’ve got a polynomial equation. The next step is to simplify it. The key here is to move all the terms to one side, setting the equation equal to zero. This is a classic move in algebra because it allows us to combine like terms and often leads to a simpler form that we can solve more easily. In our case, we’ve got $4y^2$ on both sides, which is super convenient because they'll cancel each other out when we subtract $4y^2$ from both sides. This significantly simplifies our equation.

So, let’s subtract $4y^2$ from both sides of $4y^2 - 5y + 28 = 4y^2 + 4y + 1$. This gives us $-5y + 28 = 4y + 1$. Cool, right? Now, we need to get all the ‘y’ terms on one side and the constants on the other. Let's add $5y$ to both sides. This will eliminate the ‘y’ term on the left side and give us $28 = 9y + 1$. Next, we want to isolate the term with ‘y’, so we’ll subtract 1 from both sides. This gives us $27 = 9y$. We're almost there, guys!

Now, to finally solve for ‘y’, we simply divide both sides by 9. This isolates ‘y’ and gives us our potential solution: $y = 3$. But remember what we talked about earlier? We need to check this solution in the original equation to make sure it’s not an extraneous solution. This is a crucial step, so don't skip it!

Verifying the Solution: The All-Important Check

We’ve arrived at a potential solution: $y = 3$. But before we celebrate, we absolutely have to check if this solution actually works in our original equation, $\sqrt{4y^2 - 5y + 28} = 2y + 1$. Plugging in our value is super important because, as we discussed earlier, squaring both sides of an equation can sometimes introduce solutions that aren't valid.

So, let's substitute $y = 3$ into our original equation. On the left side, we have $\sqrt{4(3)^2 - 5(3) + 28}$. Let's break this down step by step. First, $(3)^2$ is 9, so we have $\sqrt{4(9) - 5(3) + 28}$. Next, $4(9)$ is 36 and $5(3)$ is 15, giving us $\sqrt{36 - 15 + 28}$. Now, $36 - 15$ is 21, so we have $\sqrt{21 + 28}$, which simplifies to $\sqrt{49}$. And the square root of 49 is, of course, 7. So, the left side of our equation evaluates to 7.

Now, let’s check the right side of the equation. We have $2y + 1$, and substituting $y = 3$ gives us $2(3) + 1$. This is $6 + 1$, which equals 7. Fantastic! Both sides of the equation are equal when $y = 3$. This means our solution is valid, and we don't have any extraneous solutions to worry about. So, we can confidently say that $y = 3$ is the correct solution to our original equation.

Final Answer: Wrapping Things Up

Okay, guys, we've successfully navigated this problem from start to finish! We started with the equation $\sqrt{4y^2 - 5y + 28} = 2y + 1$, and through careful algebraic manipulation and a crucial verification step, we found that the solution is $y = 3$. This whole process highlights some really important techniques in algebra, such as squaring both sides to eliminate square roots, simplifying equations by combining like terms, and the absolute necessity of checking for extraneous solutions.

Remember, guys, the key to mastering these kinds of problems is practice. The more you work through equations like this, the more comfortable and confident you’ll become. So, keep practicing, and don't be afraid to tackle challenging problems. You've got this! And that’s a wrap on this mathematical adventure. Keep your eyes peeled for more exciting problems to solve together!