Solving Quadratic Equations: A Step-by-Step Guide

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Solving Quadratic Equations: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into the world of quadratic equations, and we're going to solve one together. Specifically, we'll be tackling the equation: x2+2xβˆ’35=0x^2 + 2x - 35 = 0. Don't worry if this looks a bit intimidating at first – we'll break it down step-by-step, making it super easy to understand. Quadratic equations are fundamental in algebra, popping up in all sorts of applications, from physics and engineering to finance and even video game design. Understanding how to solve them is a crucial skill for anyone looking to level up their math game. So, grab your pencils, and let's get started. We'll be using a common and effective method called factoring. Factoring is like reverse distribution; it's the process of breaking down a quadratic expression into a product of two simpler expressions (usually binomials). This method is particularly useful when the quadratic equation is easily factorable, as it directly leads to the solutions (also known as roots or zeros) of the equation. We'll also quickly touch on the quadratic formula, just in case factoring isn't straightforward. Remember, practice is key! The more you work through these problems, the more comfortable and confident you'll become. By the end of this guide, you'll be able to solve quadratic equations with ease. Let's start with a general overview of the forms of quadratic equations and the common methods used to solve them.

Understanding Quadratic Equations

First things first, what exactly is a quadratic equation? Simply put, it's an equation that can be written in the form ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. The 'x' is our variable, and the equation's solutions are the values of 'x' that make the equation true. The highest power of the variable is 2, hence the name "quadratic" (from the Latin "quadratus," meaning square). The graph of a quadratic equation is a parabola, a U-shaped curve that can open upwards or downwards. The points where the parabola intersects the x-axis are the solutions to the equation – these are also known as the roots or zeros. There are several ways to find these roots, each with its own advantages. The most common methods are factoring, completing the square, using the quadratic formula, and graphing. Factoring is usually the quickest method if the equation is easily factorable. Completing the square is a more general method and always works, but it can be a bit more involved. The quadratic formula is a universal tool – it works for any quadratic equation, regardless of how complex it looks. Graphing provides a visual representation of the solutions, but it might not always give you the exact values, especially if the roots are not integers. Understanding the different forms of quadratic equations, such as standard form (ax2+bx+c=0ax^2 + bx + c = 0) and vertex form (a(xβˆ’h)2+k=0a(x - h)^2 + k = 0), can also help you solve them. Being familiar with these different forms can make the solving process much easier, since you can quickly determine which methods might be most appropriate for each. Remember, the goal is always to find the values of x that satisfy the equation, and choosing the right method depends on the specific equation you're dealing with.

The Quadratic Formula

Just a quick note about the quadratic formula. This is a lifesaver when factoring gets tough. The quadratic formula is: x = rac{-b obreak{\pm} obreak{\sqrt{b^2 - 4ac}}}{2a}. You simply plug in the values of a, b, and c from your equation, and boom – you've got your solutions. We will cover this later. But for now, let's stick with factoring, since our equation is a good candidate for this method.

Solving the Equation by Factoring

Okay, guys, let's get down to business and solve our equation, x2+2xβˆ’35=0x^2 + 2x - 35 = 0. The first step in factoring is to look for two numbers that multiply to give you c (-35 in this case) and add up to give you b (2 in this case). This may seem like a little puzzle. Think of it as finding the right combination of numbers that fit the equation. Sometimes, it takes a little trial and error, but with practice, you'll get the hang of it. Let's list the factors of -35: (1, -35), (-1, 35), (5, -7), and (-5, 7). Now, let's see which of these pairs adds up to 2. Aha! We see that 7 and -5 is the pair we are looking for because 7 + (-5) = 2. These numbers will be used to rewrite the middle term of the quadratic expression. Our next step is to rewrite the middle term (2x2x) using the two numbers we just found. This means we rewrite the equation as x2βˆ’5x+7xβˆ’35=0x^2 - 5x + 7x - 35 = 0. Notice that we have not changed the equation; we've just rewritten it in a slightly different form. This step is critical because it sets us up for the next step: factoring by grouping. We group the first two terms and the last two terms together. This gives us (x2βˆ’5x)+(7xβˆ’35)=0(x^2 - 5x) + (7x - 35) = 0. Now, we factor out the greatest common factor (GCF) from each group. For the first group, the GCF is x, so we have x(xβˆ’5)x(x - 5). For the second group, the GCF is 7, so we have 7(xβˆ’5)7(x - 5). So now our equation looks like this: x(xβˆ’5)+7(xβˆ’5)=0x(x - 5) + 7(x - 5) = 0. Notice that both terms now have a common factor of (xβˆ’5)(x - 5). Now, we factor out (xβˆ’5)(x - 5), which gives us (xβˆ’5)(x+7)=0(x - 5)(x + 7) = 0. We've successfully factored the quadratic expression! The hard part is over.

Finding the Solutions

We're in the home stretch, guys! Now that we have our factored form, (xβˆ’5)(x+7)=0(x - 5)(x + 7) = 0, we can find the solutions. The key here is the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This means that either (xβˆ’5)=0(x - 5) = 0 or (x+7)=0(x + 7) = 0. So we set each factor equal to zero and solve for x. For (xβˆ’5)=0(x - 5) = 0, we add 5 to both sides to get x=5x = 5. For (x+7)=0(x + 7) = 0, we subtract 7 from both sides to get x=βˆ’7x = -7. Therefore, the solutions to the quadratic equation x2+2xβˆ’35=0x^2 + 2x - 35 = 0 are x=5x = 5 and x=βˆ’7x = -7. These are the values of x that make the original equation true. To check your answer, you can plug these values back into the original equation and verify that both sides are equal. Congratulations, you've successfully solved a quadratic equation! You have learned an incredibly valuable skill. Now it’s time to practice, practice, practice! Make sure to work through similar examples, so that you get the hang of this. It will become a whole lot easier.

Conclusion: Mastering Quadratic Equations

Awesome work, everyone! You've successfully navigated the world of quadratic equations and emerged victorious. You've not only solved x2+2xβˆ’35=0x^2 + 2x - 35 = 0 but also gained a solid understanding of factoring and the zero-product property. Remember, solving quadratic equations is like building a puzzle, and each method provides you with a different set of tools to solve the puzzle. Always start by trying to factor. If that doesn’t work, then you should consider other strategies. The quadratic formula is a universal tool, but factoring is often quicker and simpler. Remember to check your answers by plugging them back into the original equation to ensure they are correct. Now that you've mastered the basics, consider exploring more complex quadratic equations and different methods like completing the square or using the quadratic formula. These advanced techniques will broaden your problem-solving toolkit and boost your mathematical confidence. Keep practicing, stay curious, and you'll become a quadratic equation whiz in no time. If you got through this guide, then you can solve quadratic equations.