Solving $x^2-36=5x$: Finding The Positive Solution

Hey guys, let's dive into a cool math problem today: finding the positive solution to the equation $x^2 - 36 = 5x$. This might look a bit intimidating at first glance, but trust me, it's totally manageable once we break it down. We're not just going to crunch numbers; we're going to understand why we do each step, making sure you guys can tackle similar problems with confidence. So, grab your favorite thinking cap, and let's get this algebraic adventure started!

Understanding Quadratic Equations

Before we jump into solving, it's super important to get a handle on what we're dealing with here. The equation $x^2 - 36 = 5x$ is what we call a quadratic equation. What makes it quadratic? That $x^2$ term, my friends! It means the highest power of our variable, $x$, is 2. Quadratic equations generally have the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are numbers, and $a$ isn't zero (otherwise, it wouldn't be quadratic anymore!). Our goal today is to rearrange $x^2 - 36 = 5x$ into this standard form. Why do we do this? Because this standard form unlocks some powerful methods for solving these types of equations, like factoring or using the quadratic formula. It's like getting the secret decoder ring for quadratic puzzles. So, the first mission, should you choose to accept it, is to move all the terms to one side of the equals sign, setting the whole thing equal to zero. This is a fundamental step that prepares our equation for whatever solving strategy we decide to use. We'll take the $5x$ term from the right side and subtract it from both sides of the equation. This keeps the equation balanced, which is key in algebra. So, $x^2 - 36 - 5x = 5x - 5x$, which simplifies to $x^2 - 5x - 36 = 0$. Boom! We've successfully transformed our original equation into the standard quadratic form. Now that it's in $ax^2 + bx + c = 0$ format, where $a=1$, $b=-5$, and $c=-36$, we're ready for the next phase: finding those elusive solutions.

Factoring the Quadratic Equation

Alright, so we've got our equation in the sweet spot: $x^2 - 5x - 36 = 0$. Now, one of the most common and, dare I say, elegant ways to solve quadratic equations is through factoring. This method relies on finding two binomials that, when multiplied together, give us our original quadratic. Think of it like a reverse FOIL (First, Outer, Inner, Last). We're looking for two numbers that multiply to give us our constant term ($c$, which is -36 in our case) and add up to give us our middle coefficient ($b$, which is -5). This is where the detective work really kicks in, guys! We need to list out pairs of factors for -36. Let's brainstorm: (1, -36), (-1, 36), (2, -18), (-2, 18), (3, -12), (-3, 12), (4, -9), (-4, 9), (6, -6). Now, for each of these pairs, we need to check their sum. Which pair adds up to -5? Let's see... 1 + (-36) = -35, -1 + 36 = 35, 2 + (-18) = -16, -2 + 18 = 16, 3 + (-12) = -9, -3 + 12 = 9, 4 + (-9) = -5! Bingo! We found our magic pair: 4 and -9. These are the numbers that will help us factor our quadratic. So, we can rewrite $x^2 - 5x - 36 = 0$ as $(x + 4)(x - 9) = 0$. This is super cool because if the product of two things is zero, it means at least one of those things must be zero. This is known as the Zero Product Property. Therefore, either $(x + 4) = 0$ or $(x - 9) = 0$. To find the solutions, we just solve these two simple linear equations. If $x + 4 = 0$, then subtracting 4 from both sides gives us $x = -4$. If $x - 9 = 0$, then adding 9 to both sides gives us $x = 9$. So, the solutions to our quadratic equation are $x = -4$ and $x = 9$. Factoring really is a neat trick when it works out nicely!

Using the Quadratic Formula

Now, what if factoring seems a bit tricky, or maybe the numbers don't seem to have nice whole number factors? No sweat, guys! That's where the quadratic formula swoops in like a superhero. This formula is a universal key that works for any quadratic equation in the standard form $ax^2 + bx + c = 0$. It's a lifesaver when factoring gets complicated. The formula is: x = rac{-b eq ext{sqrt}(b^2 - 4ac)}{2a}. Remember, $a$, $b$, and $c$ are the coefficients from our standard form equation. In our case, $x^2 - 5x - 36 = 0$, we have $a = 1$, $b = -5$, and $c = -36$. Let's plug these values into the formula and see what happens. First, let's handle the part under the square root, often called the discriminant: $b^2 - 4ac$. This is $(-5)^2 - 4(1)(-36)$. Squaring -5 gives us 25. Then, $-4 imes 1 imes -36$ becomes $+144$. So, the discriminant is $25 + 144 = 169$. Since 169 is a perfect square (it's $13^2$), this tells us that our quadratic equation can be factored, which we already confirmed. But let's keep going with the formula! Now we substitute this back into the main formula: x = rac{-(-5) eq ext{sqrt}(169)}{2(1)}. This simplifies to x = rac{5 eq 13}{2}. The '$eq$' sign here means we have two possible solutions. For the first solution, we use the plus sign: x_1 = rac{5 + 13}{2} = rac{18}{2} = 9. For the second solution, we use the minus sign: x_2 = rac{5 - 13}{2} = rac{-8}{2} = -4. And voilà! We get the exact same solutions we found through factoring: $x = 9$ and $x = -4$. The quadratic formula is incredibly powerful because it guarantees you'll find the solutions, whether they're nice integers, fractions, or even irrational numbers. It's a reliable tool in any mathematician's arsenal.

Identifying the Positive Solution

Okay, guys, we've done the heavy lifting! We've transformed the equation, factored it, and even used the quadratic formula. Both methods have given us two potential solutions: $x = 9$ and $x = -4$. Now, the question specifically asks for the positive solution. Looking at our two answers, which one is positive? That's right, it's $x = 9$. The number 9 is greater than zero, making it our positive solution. The other solution, $x = -4$, is negative, so we set it aside for this particular question. It's important to read the question carefully and make sure you're answering exactly what's being asked. Sometimes, problems might ask for the negative solution, the sum of the solutions, or perhaps the product. In this case, our positive solution is 9. Always double-check your work and ensure your final answer directly addresses the prompt. So, if you were asked for the positive solution of $x^2 - 36 = 5x$, you'd confidently say it's 9. This is the culmination of our algebraic journey today, proving that with a systematic approach and the right tools, even seemingly complex equations can be solved.

Verifying the Solution

Before we wrap this up, let's do something super important in math: verification. This means plugging our solution back into the original equation to make sure it actually works. It's like a final check to ensure we haven't made any silly mistakes along the way. Remember our original equation: $x^2 - 36 = 5x$. We found our positive solution to be $x = 9$. Let's substitute 9 for every $x$ in the equation and see if the left side equals the right side. So, on the left side, we have $(9)^2 - 36$. That's $81 - 36$, which equals $45$. Now, let's look at the right side: $5x$. Substituting 9 gives us $5 imes 9$, which also equals $45$. Since the left side (45) equals the right side (45), our solution $x = 9$ is indeed correct! How awesome is that? This verification step is invaluable. It builds confidence in your answers and helps catch errors. Imagine you're building something; you wouldn't just assume it's perfect, right? You'd check your measurements, tighten the bolts, and make sure everything is secure. Math is no different! By verifying our solution, we're ensuring the integrity of our work. Let's quickly check the other solution, $x = -4$, just for practice. Original equation: $x^2 - 36 = 5x$. Substitute $x = -4$: Left side: $(-4)^2 - 36 = 16 - 36 = -20$. Right side: $5 imes (-4) = -20$. Both sides are equal (-20), so $x = -4$ is also a valid solution to the equation. However, since the question specifically asked for the positive solution, we stick with $x = 9$. This process of verification is a powerful habit to cultivate in your mathematical journey, guys. It solidifies your understanding and ensures accuracy, making you a more skilled and confident problem-solver. Keep this practice in your toolkit!