Solving Quadratic Equations: Find The Other Solution

Hey guys! Let's dive into a fun math problem today that involves solving quadratic equations. Quadratic equations might sound intimidating, but they're actually super manageable once you get the hang of them. We're going to break down a specific example step-by-step, so you can see exactly how it's done. Our main goal here is to not just give you the answer, but to help you understand the process of finding it. Understanding the process is key because you'll be able to tackle similar problems with confidence. So, let's jump right into it and unlock the secrets of quadratic equations together! We'll explore the fascinating world of quadratic equations, focusing on a specific problem where we need to find the missing solution. So, buckle up, grab your thinking caps, and let's get started!

Understanding the Problem

The question we're tackling today is a classic example of a quadratic equation problem. We're given the equation $x^2 - 2x - 15 = 0$ and told that one of the solutions is $x = -3$. Our mission, should we choose to accept it (and we do!), is to find the other solution. Before we start crunching numbers, it's important to understand what a solution to a quadratic equation actually means. A solution, or root, of an equation is a value that, when plugged in for the variable (in this case, x), makes the equation true. Think of it like a magic key that unlocks the equation! So, we already have one key ($x = -3$), and we need to find the other one that fits. Now, let's think about quadratic equations in general. They're called "quadratic" because the highest power of the variable is 2 (that x² term). This also tells us that a quadratic equation can have up to two solutions. That's why we're looking for another one! We can visualize a quadratic equation as a parabola (a U-shaped curve) on a graph, and the solutions are where the parabola crosses the x-axis. Knowing this gives us a visual context for what we're trying to find. We're not just blindly manipulating numbers; we're actually finding the points where the parabola intersects the horizontal line. This understanding will help us in choosing the right method to solve the problem and in interpreting our final answer. So, with this foundation in place, let's move on to the methods we can use to crack this mathematical puzzle.

Methods to Solve Quadratic Equations

There are several ways to solve quadratic equations, each with its own strengths and when it's best applied. For this particular problem, we'll focus on two main methods: factoring and using the quadratic formula. Let's dive into each one. First up, factoring. Factoring is like reverse distribution. We want to break down the quadratic expression ($x^2 - 2x - 15$) into two binomials (expressions with two terms) that multiply together to give us the original quadratic. It's a bit like finding the ingredients that make up a dish. When factoring, we look for two numbers that multiply to the constant term (-15) and add up to the coefficient of the x term (-2). This might sound like a puzzle, and it is! But with a little practice, you'll get the hang of it. Factoring is often the quickest method when it works, but it's not always straightforward for every quadratic equation. Sometimes the numbers are tricky, or the expression simply doesn't factor nicely. That's where the next method comes in handy. The quadratic formula is a universally applicable method for solving quadratic equations. It's a formula that directly gives you the solutions for x, no matter how messy the equation looks. The formula is: $x = (-b ± √(b^2 - 4ac)) / (2a)$, where a, b, and c are the coefficients of the quadratic equation in the form $ax^2 + bx + c = 0$. Now, this formula might look a bit intimidating at first glance, but trust me, it's a lifesaver. It's like having a Swiss Army knife for quadratic equations – it can handle anything! The quadratic formula guarantees that you'll find the solutions, even if they are fractions, decimals, or involve square roots. In our case, we already know one solution and want to find the other. We can use either method, but let's see how factoring works for this specific problem. This will give us a good feel for both methods and help us choose the most efficient one. So, let's roll up our sleeves and try factoring first!

Solving by Factoring

Let's tackle the problem using the factoring method. Remember, we want to rewrite the quadratic expression $x^2 - 2x - 15$ as a product of two binomials. To do this, we need to find two numbers that multiply to -15 (the constant term) and add up to -2 (the coefficient of the x term). Think of it as a little number puzzle! What two numbers come to mind? After a bit of thought, you might realize that 3 and -5 fit the bill. Why? Because 3 multiplied by -5 equals -15, and 3 plus -5 equals -2. Bingo! Now we can rewrite our quadratic expression in factored form. It becomes $(x + 3)(x - 5) = 0$. Notice how the numbers 3 and -5 directly appear in the binomials. The x comes from splitting the x² term. The whole idea behind factoring is that if the product of two factors is zero, then at least one of those factors must be zero. This is a crucial principle that allows us to find the solutions. So, we set each factor equal to zero and solve for x:

• $x + 3 = 0$
• $x - 5 = 0$

Solving the first equation, we subtract 3 from both sides to get $x = -3$. This is the solution we were given in the problem, so it checks out! Solving the second equation, we add 5 to both sides to get $x = 5$. And there you have it – the other solution! So, by factoring the quadratic equation, we've successfully found both solutions: $x = -3$ and $x = 5$. Factoring worked nicely in this case because the numbers were relatively straightforward. But what if we had a more complicated equation that didn't factor easily? That's where the quadratic formula comes to the rescue. However, for this problem, factoring was a pretty efficient way to find the answer. We've seen how to break down the quadratic, find the right numbers, and use those numbers to get our solutions. Now, to solidify our understanding, let's verify that this solution is correct in the original problem.

Verifying the Solution

It's always a good idea to double-check your work, especially in math! Verifying our solution ensures that we haven't made any silly mistakes along the way. So, let's take the solution we found, $x = 5$, and plug it back into the original equation: $x^2 - 2x - 15 = 0$. We're substituting x with 5, so the equation becomes: $(5)^2 - 2(5) - 15 = 0$. Now, let's simplify: 25 - 10 - 15 = 0. And further: 15 - 15 = 0. Finally: 0 = 0. Hooray! The equation holds true. This confirms that $x = 5$ is indeed a valid solution to the quadratic equation. We've not only found the solution but also rigorously verified that it's correct. This process of verification is crucial in mathematics. It helps you catch errors, build confidence in your answers, and deepen your understanding of the concepts. Think of it as the final step in a detective story – making sure all the pieces fit together perfectly. We knew one solution was $x = -3$, and now we've confirmed that the other solution is $x = 5$. So, we've successfully solved the problem using factoring and verified our answer. But what if factoring wasn't so straightforward? Let's briefly consider how we could have used the quadratic formula to solve this problem as well. This will give us a more complete picture of how to tackle quadratic equations. Though factoring worked well here, understanding the quadratic formula gives us a powerful backup plan for more complex scenarios.

Conclusion

So, we've successfully navigated the world of quadratic equations! We started with the equation $x^2 - 2x - 15 = 0$, were given one solution ($x = -3$), and set out to find the other. We explored the method of factoring, which proved to be an efficient way to break down the quadratic expression and find the solutions. We identified the numbers 3 and -5 as the key to factoring, rewrote the equation as $(x + 3)(x - 5) = 0$, and ultimately found the other solution: $x = 5$. But we didn't stop there! We also emphasized the importance of verifying our solution. By plugging $x = 5$ back into the original equation, we confirmed that it indeed makes the equation true. This step is a critical part of the problem-solving process, ensuring accuracy and building confidence. While factoring worked well for this particular problem, we also briefly touched upon the quadratic formula as a more universally applicable method. The quadratic formula is like a safety net – it guarantees that you'll find the solutions, even when factoring is difficult. Remember, mastering quadratic equations is a fundamental step in your mathematical journey. They pop up in various fields, from physics to engineering to computer science. By understanding how to solve them, you're equipping yourself with a valuable tool. Keep practicing, keep exploring different methods, and don't be afraid to tackle challenging problems. The more you practice, the more comfortable and confident you'll become. And who knows, maybe you'll even start to enjoy solving quadratic equations! So, keep up the great work, and let's continue to explore the fascinating world of mathematics together!