Solving (h+x)(S+x) = (x+x)(9+x) A Comprehensive Mathematical Guide

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Hey guys! Ever stumbled upon an equation that looks like a cryptic puzzle? Today, we're diving deep into solving an equation that might seem intimidating at first glance: (h+x)(S+x) = (x+x)(9+x). Don't worry, we'll break it down step-by-step, making it super easy to understand. We're not just going to give you the answer; we're going to equip you with the knowledge to tackle similar problems on your own. So, buckle up and let's get started!

Understanding the Equation

Before we jump into the solution, let's take a good look at the equation (h+x)(S+x) = (x+x)(9+x). It's crucial to understand what each part represents and how they interact. This equation involves variables, constants, and algebraic operations. Variables, in this case, are represented by letters like 'h', 'S', and 'x', which can take on different values. Constants are fixed numbers, such as 9 in our equation. The operations involved are multiplication and addition. Understanding these basic elements is the first step to unraveling the mystery of the equation.

When we see an equation like this, it's often helpful to think of it as a balance. The left side (h+x)(S+x) must equal the right side (x+x)(9+x). Our goal is to find the value(s) of 'x' that make this balance true. It's like a detective trying to solve a case; we need to gather clues and piece them together to find the solution. In mathematical terms, we are essentially looking for the roots or solutions of the equation. These are the values of 'x' that, when plugged into the equation, will satisfy the equality. Understanding the structure and components of the equation sets the stage for employing various algebraic techniques to isolate 'x' and ultimately find the solution. Let's move forward and explore the steps we can take to solve this intriguing equation.

Simplifying the Equation

Alright, let's get our hands dirty and start simplifying the equation (h+x)(S+x) = (x+x)(9+x). The first thing we can do is expand both sides of the equation. This means we'll be multiplying out the terms within the parentheses. Remember the FOIL method (First, Outer, Inner, Last)? It's a handy tool for multiplying binomials (expressions with two terms). On the left side, we have (h+x)(S+x). Let's apply FOIL:

  • First: h * S = hS
  • Outer: h * x = hx
  • Inner: x * S = Sx
  • Last: x * x = x²

So, the left side expands to hS + hx + Sx + x². Now, let's tackle the right side, which is (x+x)(9+x). We can simplify (x+x) to 2x. So the right side becomes 2x(9+x). Expanding this, we get:

  • 2x * 9 = 18x
  • 2x * x = 2x²

Thus, the right side expands to 18x + 2x². Now our equation looks like this: hS + hx + Sx + x² = 18x + 2x². We've successfully expanded both sides, and the equation is starting to take a clearer shape. By expanding, we've removed the parentheses, which allows us to combine like terms and move closer to isolating 'x'. The next step involves rearranging the terms and potentially simplifying further by combining similar terms. This simplification process is a crucial step in solving any algebraic equation, as it reduces complexity and makes the equation easier to manipulate. So, let's move on to the next phase and see how we can further simplify and rearrange the terms to bring us closer to the solution.

Rearranging Terms

Now that we've expanded the equation to hS + hx + Sx + x² = 18x + 2x², it's time to rearrange the terms. Our goal here is to get all the terms on one side of the equation, ideally leaving zero on the other side. This is a common strategy when dealing with equations, especially quadratic equations (equations where the highest power of the variable is 2). Let's move everything to the right side of the equation. We'll do this by subtracting the terms on the left side from both sides. So, we subtract hS, hx, Sx, and x² from both sides:

  • 0 = 18x + 2x² - hS - hx - Sx - x²

Now, let's simplify by combining like terms. We have 2x² and -x², which combine to x². Our equation now looks like this:

  • 0 = x² + 18x - hx - Sx - hS

We can further group the terms involving 'x' to make the equation look cleaner:

  • 0 = x² + (18 - h - S)x - hS

This form of the equation is quite interesting. It's a quadratic equation in the form of ax² + bx + c = 0, where:

  • a = 1
  • b = (18 - h - S)
  • c = -hS

Rearranging the equation into this standard quadratic form is a significant step forward. It allows us to leverage well-established methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. By rearranging the terms and identifying the coefficients a, b, and c, we've set the stage for applying these powerful techniques. Understanding the structure of the equation and manipulating it into a recognizable form is a key skill in algebra. Now, let's explore how we can use this quadratic form to find the values of 'x' that satisfy the equation.

Solving the Quadratic Equation

Okay, we've arrived at the quadratic equation: 0 = x² + (18 - h - S)x - hS. Now comes the exciting part – solving for 'x'! As we discussed, there are a few methods we can use to tackle this. Let's consider two common approaches: factoring and the quadratic formula.

Factoring

Factoring involves expressing the quadratic equation as a product of two binomials. If we can find two numbers that multiply to -hS (the constant term) and add up to (18 - h - S) (the coefficient of the x term), then we can factor the equation. This method is often the quickest, but it's not always straightforward, especially if the numbers are complex or non-integer. Let's assume, for a moment, that we can factor the equation into the form (x + A)(x + B) = 0. Then, A * B would have to equal -hS, and A + B would have to equal (18 - h - S). Finding such A and B might require some trial and error, or a clever insight into the relationship between h and S. If we successfully factor the equation, we can then set each factor equal to zero and solve for x. This gives us two potential solutions for x.

Quadratic Formula

When factoring isn't immediately obvious or seems too challenging, the quadratic formula is our trusty backup. The quadratic formula provides a direct way to find the solutions for any quadratic equation in the form ax² + bx + c = 0. The formula is:

  • x = (-b ± √(b² - 4ac)) / (2a)

In our case, a = 1, b = (18 - h - S), and c = -hS. Plugging these values into the formula gives us:

  • x = (-(18 - h - S) ± √((18 - h - S)² - 4(1)(-hS))) / (2(1))

This might look a bit intimidating, but it's just a matter of carefully substituting the values and simplifying. The expression inside the square root, (b² - 4ac), is called the discriminant. The discriminant tells us about the nature of the solutions:

  • If (b² - 4ac) > 0, there are two distinct real solutions.
  • If (b² - 4ac) = 0, there is exactly one real solution (a repeated root).
  • If (b² - 4ac) < 0, there are two complex solutions.

By using the quadratic formula, we can find the values of 'x' that satisfy our equation, regardless of how messy the coefficients are. It's a powerful tool in our algebraic arsenal. Whether we choose to factor or use the quadratic formula, the goal is the same: to find the values of 'x' that make the equation true. Let's move on to the next section, where we'll discuss potential solutions and interpretations.

Interpreting the Solutions

After applying either factoring or the quadratic formula, we'll arrive at one or two possible values for 'x'. But what do these solutions actually mean? It's crucial to interpret the solutions in the context of the original equation and any real-world scenarios it might represent. Let's explore some possibilities.

Real Solutions

If we obtain real number solutions for 'x', it means there are actual values that, when substituted back into the original equation (h+x)(S+x) = (x+x)(9+x), will make the equation true. These solutions can be positive, negative, or zero. Each solution represents a point where the balance in the equation is achieved. For example, if we find that x = 5 is a solution, it means that plugging in 5 for 'x' on both sides of the equation will result in the same value. In real-world applications, these solutions might represent physical quantities, such as distances, times, or quantities of a substance. The interpretation depends heavily on the specific context of the problem.

Complex Solutions

Sometimes, when using the quadratic formula, we might encounter a negative number inside the square root. This indicates that the equation has complex solutions, which involve the imaginary unit 'i' (where i² = -1). Complex solutions don't have a direct representation on the real number line. While they might seem abstract, complex numbers are essential in many areas of mathematics, physics, and engineering. They often arise in situations involving oscillations, waves, and electrical circuits. In the context of our equation, complex solutions might suggest that there are no real-world values of 'x' that satisfy the equation, but there are solutions within the complex number system.

Extraneous Solutions

It's also important to be aware of extraneous solutions. These are solutions that we obtain through the algebraic process but do not actually satisfy the original equation. Extraneous solutions can arise when we perform operations that are not reversible, such as squaring both sides of an equation. To check for extraneous solutions, we must always substitute our solutions back into the original equation and verify that they make the equation true. If a solution doesn't work in the original equation, we discard it. Interpreting solutions correctly is a critical step in problem-solving. It ensures that we not only find the mathematical answers but also understand their significance and validity. By considering the nature of the solutions and checking for extraneous ones, we can confidently arrive at the correct conclusion.

Practical Applications

So, we've cracked the code of the equation (h+x)(S+x) = (x+x)(9+x). But where does this kind of equation actually show up in the real world? You might be surprised to learn that equations of this form, particularly quadratic equations, have a wide range of practical applications across various fields.

Physics

In physics, quadratic equations are used to describe projectile motion. For example, if you throw a ball in the air, the path it follows can be modeled by a quadratic equation. The variable 'x' might represent time, and the solutions to the equation could tell you when the ball will hit the ground. Similarly, quadratic equations are used in optics to describe the focusing of light by lenses and mirrors. They also appear in the study of energy and motion, such as in calculating kinetic energy or potential energy.

Engineering

Engineers use quadratic equations in structural design. For instance, when designing bridges or buildings, engineers need to calculate the forces and stresses acting on the structure. Quadratic equations can help determine the optimal dimensions and materials to ensure the structure is stable and safe. In electrical engineering, quadratic equations are used to analyze circuits and calculate currents, voltages, and power. They are also crucial in control systems, where engineers design systems to automatically regulate processes, such as temperature control or robotic movements.

Economics and Finance

Even in economics and finance, quadratic equations have their place. They can be used to model cost and revenue functions, helping businesses determine the optimal production levels to maximize profit. Quadratic equations also appear in financial modeling, such as in calculating compound interest or the return on investment. They can help investors make informed decisions about their portfolios and manage risk. These are just a few examples of the many practical applications of equations like the one we've been solving. The ability to manipulate and solve these equations is a valuable skill in many fields. Understanding the underlying principles and techniques allows us to tackle real-world problems and make informed decisions. So, the next time you encounter a quadratic equation, remember that you're not just solving an abstract problem; you're equipping yourself to address challenges in a wide range of disciplines.

Conclusion

Alright, guys, we've reached the end of our journey into solving the equation (h+x)(S+x) = (x+x)(9+x). We've taken it apart, simplified it, and explored different methods for finding the solutions. From expanding and rearranging terms to using the quadratic formula, we've covered a lot of ground. But more importantly, we've learned that solving equations isn't just about finding numbers; it's about understanding the relationships between variables and applying logical steps to uncover the hidden values.

We started by understanding the equation's structure and recognizing it as a quadratic equation in disguise. We then used algebraic techniques to simplify and rearrange the terms, putting it into a standard form that we could work with. We explored two powerful methods for solving quadratic equations: factoring and the quadratic formula. Each method has its strengths, and choosing the right one can make the process much smoother. We also discussed the importance of interpreting the solutions, considering real-world scenarios, and checking for extraneous solutions. This ensures that we not only find the mathematical answers but also understand their significance and validity.

Finally, we touched on the practical applications of quadratic equations in various fields, from physics and engineering to economics and finance. This highlights the relevance of the skills we've developed and shows how they can be applied to solve real-world problems. Solving equations is a fundamental skill in mathematics and a valuable tool in many disciplines. By mastering these techniques, you're not just learning how to manipulate numbers; you're developing critical thinking skills that will serve you well in any field you pursue. So, keep practicing, keep exploring, and keep challenging yourself with new equations. The more you practice, the more confident and proficient you'll become. And remember, every equation is a puzzle waiting to be solved, a challenge to be overcome, and an opportunity to expand your knowledge and skills.