Solving 2x^2 + 3x - 7 = X^2 + 5x + 39 A Step-by-Step Guide

Hey guys! Today, we're diving into a classic algebra problem: solving for x in a quadratic equation. Specifically, we're tackling the equation 2x² + 3x - 7 = x² + 5x + 39. Don't worry, it might look intimidating at first, but we'll break it down step by step. Think of it as a puzzle – we just need to rearrange the pieces until we find our solution. So, grab your pencils and let's get started!

Understanding Quadratic Equations

Before we jump into the solution, let's quickly recap what quadratic equations are all about. In essence, quadratic equations are polynomial equations of the second degree. This means the highest power of the variable (x in our case) is 2. The standard form of a quadratic equation is ax² + bx + c = 0, where a, b, and c are constants (numbers), and a is not equal to zero (otherwise, it wouldn't be a quadratic equation anymore!). Quadratic equations pop up everywhere in math and real-world applications, from physics problems involving projectile motion to engineering designs and even financial modeling. They're incredibly versatile, which is why mastering them is so crucial.

Now, you might be wondering, why the fuss about the 'squared' term? Well, that x² term introduces a curve into the equation's graph, resulting in a parabola. This parabolic shape is what gives quadratic equations their unique properties and makes them so useful for modeling curved paths and relationships. Solving a quadratic equation essentially means finding the x-values where the parabola intersects the x-axis (these points are also called roots or zeros of the equation). There are a few ways to crack these equations, and we'll primarily focus on one method today: rearranging the equation into standard form and then either factoring or using the quadratic formula. Factoring is a great technique when the equation can be neatly broken down into two binomial expressions, but the quadratic formula is our trusty fallback that works every time, no matter how messy the equation gets. So, with that little refresher out of the way, let's get back to our specific problem and see how we can solve for x.

Step 1: Rearrange the Equation

Our main goal here is to get the equation into that beautiful standard form we talked about earlier: ax² + bx + c = 0. This means we need to shuffle all the terms to one side, leaving zero on the other. Looking at our initial equation, 2x² + 3x - 7 = x² + 5x + 39, we can see terms on both sides. The first step is to eliminate the terms on the right side of the equation. To do this, we will subtract x² from both sides. This gives us 2x² - x² + 3x - 7 = x² - x² + 5x + 39, which simplifies to x² + 3x - 7 = 5x + 39. Remember, whatever you do to one side of the equation, you must do to the other to maintain the balance. Think of it like a seesaw – if you remove weight from one side, you need to remove the same weight from the other to keep it level.

Next, we'll subtract 5x from both sides to get the x terms together. Our equation now becomes x² + 3x - 5x - 7 = 5x - 5x + 39, which simplifies to x² - 2x - 7 = 39. We're getting closer! Finally, to get that '0' on the right side, we subtract 39 from both sides. This leaves us with x² - 2x - 7 - 39 = 39 - 39, which beautifully simplifies to our standard quadratic form: x² - 2x - 46 = 0. So, that's it! We've successfully rearranged our original equation into the familiar ax² + bx + c = 0 format. Now we know that a = 1, b = -2, and c = -46. This is a crucial step because it sets us up perfectly for the next stage, which involves choosing the right method to actually solve for x. We could try factoring, but with that -46 term, it might be a bit tricky. So, let's keep our options open and see what the quadratic formula has in store for us.

Step 2: Applying the Quadratic Formula

Alright, now that we have our equation in standard form (x² - 2x - 46 = 0), it's time to unleash the mighty quadratic formula! This formula is our trusty sidekick when factoring proves difficult or impossible. The quadratic formula states that for an equation in the form ax² + bx + c = 0, the solutions for x are given by: x = (-b ± √(b² - 4ac)) / 2a. See? It might look a bit intimidating at first glance, but it's actually quite straightforward once you get the hang of plugging in the values.

Remember those values we identified earlier? a = 1, b = -2, and c = -46. Now, all we need to do is carefully substitute these values into the formula. Let's break it down. First, we have -b, which becomes -(-2), which is simply 2. Then, we have the plus-or-minus symbol (±), which means we'll actually have two solutions – one where we add and one where we subtract. Under the square root, we have b², which is (-2)² = 4. Next, we have -4ac, which is -4 * 1 * -46 = 184. So, inside the square root, we have 4 + 184 = 188. Finally, in the denominator, we have 2a, which is 2 * 1 = 2. Putting it all together, our solutions for x are: x = (2 ± √188) / 2.

We're not quite done yet! We can simplify that square root a bit. Notice that 188 has a factor of 4 (188 = 4 * 47), and the square root of 4 is 2. So, we can rewrite √188 as √(4 * 47) = 2√47. Now our solutions look like this: x = (2 ± 2√47) / 2. We can further simplify by dividing both terms in the numerator by 2, giving us our final solutions: x = 1 ± √47. So, there you have it! We've successfully used the quadratic formula to find the two solutions for x in our equation. One solution is x = 1 + √47, and the other is x = 1 - √47. These are the x-values where the parabola represented by our equation intersects the x-axis. Pretty cool, huh?

Step 3: Verifying the Solutions (Optional but Recommended)

Okay, we've found our solutions using the quadratic formula: x = 1 + √47 and x = 1 - √47. But, just to be absolutely sure we haven't made any sneaky mistakes along the way, it's always a fantastic idea to verify our solutions. This step is optional, but seriously, it's like having a built-in safety net – it can catch any errors before they become a problem. Verifying simply means plugging our solutions back into the original equation and making sure both sides of the equation are still equal. If they are, we know we're golden! If not, it's a sign we need to revisit our steps and hunt down that little gremlin that's causing the trouble.

Let's start with x = 1 + √47. Our original equation was 2x² + 3x - 7 = x² + 5x + 39. Plugging in our value for x, we get: 2(1 + √47)² + 3(1 + √47) - 7 = (1 + √47)² + 5(1 + √47) + 39. This looks a bit messy, I know, but hang in there! Let's carefully expand those squares and simplify. Remember that (a + b)² = a² + 2ab + b². So, (1 + √47)² = 1 + 2√47 + 47 = 48 + 2√47. Substituting this back into our equation gives us: 2(48 + 2√47) + 3 + 3√47 - 7 = 48 + 2√47 + 5 + 5√47 + 39. Now, let's distribute and combine like terms. On the left side, we have 96 + 4√47 + 3 + 3√47 - 7 = 92 + 7√47. On the right side, we have 48 + 2√47 + 5 + 5√47 + 39 = 92 + 7√47. Voila! Both sides are equal! That's a great sign. It means x = 1 + √47 is indeed a valid solution.

Now, let's do the same for x = 1 - √47. Plugging this into our original equation gives us: 2(1 - √47)² + 3(1 - √47) - 7 = (1 - √47)² + 5(1 - √47) + 39. Again, we expand the squares: (1 - √47)² = 1 - 2√47 + 47 = 48 - 2√47. Substituting back into the equation: 2(48 - 2√47) + 3 - 3√47 - 7 = 48 - 2√47 + 5 - 5√47 + 39. Distributing and combining like terms, we get: 96 - 4√47 + 3 - 3√47 - 7 = 92 - 7√47 on the left side and 48 - 2√47 + 5 - 5√47 + 39 = 92 - 7√47 on the right side. Boom! Both sides are equal again! This confirms that x = 1 - √47 is also a valid solution. So, we've successfully verified both of our solutions. Pat yourselves on the back, guys – you've earned it!

Conclusion

Awesome work, everyone! We've successfully solved for x in the quadratic equation 2x² + 3x - 7 = x² + 5x + 39. We started by rearranging the equation into standard form (x² - 2x - 46 = 0), then we skillfully applied the quadratic formula to find our solutions (x = 1 + √47 and x = 1 - √47), and finally, we even verified our answers to make sure everything was shipshape. Remember, solving quadratic equations is a fundamental skill in algebra and has tons of applications in the real world. So, mastering this process is a huge win. Keep practicing, and you'll become a quadratic equation-solving pro in no time! And if you ever get stuck, don't hesitate to revisit these steps – they'll guide you through. Until next time, happy solving!