Solve (x+10)/(x^2-2) = 4/x: A Step-by-Step Guide

Hey guys! Today, we're diving deep into a fascinating mathematical problem: solving the equation ${\frac{x+10}{x^2-2}=\frac{4}{x}}$. This isn't just about crunching numbers; it’s about understanding the underlying concepts and techniques that make algebra so powerful. We'll break down each step, explain the logic behind it, and ensure you're not just getting the answer, but truly grasping the method. So, grab your pencils, and let's get started!

Understanding the Equation

Before we jump into the nitty-gritty, let’s take a step back and understand what we're dealing with. The equation ${\frac{x+10}{x^2-2}=\frac{4}{x}}$ involves rational expressions, which are essentially fractions where the numerator and the denominator are polynomials. Solving such equations means finding the values of ${x}$ that make the equation true. But here's the catch: we need to be mindful of values that would make the denominators zero, as division by zero is undefined. This is a crucial point to remember throughout the process. Identifying these restrictions upfront helps us avoid extraneous solutions later on.

In this specific equation, we have two denominators: ${x^2 - 2}$ and ${x}$. Setting each to zero gives us potential restrictions. For ${x^2 - 2 = 0}$, we find ${x = \pm\sqrt{2}}$. And for ${x = 0}$, we have a straightforward restriction. So, ${x}$ cannot be ${0}$, ${\sqrt{2}}$, or ${-\sqrt{2}}$. Keeping these values in mind is like setting up guardrails on a highway; they keep us from driving off the road! Understanding the domain of the equation is super important, guys. It's not just about finding any answer; it's about finding valid answers. The domain restrictions are our first checkpoint in this mathematical journey.

Now that we know what we can't do, let's focus on what we can do. Our main goal is to isolate ${x}$ and find its value(s). This usually involves a series of algebraic manipulations, each carefully chosen to simplify the equation while preserving its balance. Think of it like a delicate dance – each step must be precise and purposeful. The initial assessment of the equation is key; it sets the stage for everything that follows. Ignoring the restrictions can lead to wrong turns, so always keep them in the back of your mind. With our guardrails in place, let's move on to the next phase: clearing those fractions!

Clearing the Fractions

The next step in solving the equation ${\frac{x+10}{x^2-2}=\frac{4}{x}}$ is to clear the fractions. This makes the equation much easier to work with. We do this by multiplying both sides of the equation by the least common denominator (LCD) of the fractions involved. In our case, the denominators are ${x^2 - 2}$ and ${x}$, so the LCD is simply their product, ${x(x^2 - 2)}$. Multiplying both sides by the LCD is like using a universal translator – it converts the equation into a language we can more easily understand.

So, we multiply both sides of the equation by ${x(x^2 - 2)}$:

${ x(x^2 - 2) \cdot \frac{x+10}{x^2-2} = x(x^2 - 2) \cdot \frac{4}{x} }$

On the left side, the ${(x^2 - 2)}$ terms cancel out, leaving us with ${x(x + 10)}$. On the right side, the ${x}$ terms cancel out, leaving us with ${4(x^2 - 2)}$. This simplifies the equation to:

${ x(x + 10) = 4(x^2 - 2) }$

Now, we have a much cleaner equation to work with. This step is crucial because it eliminates the fractions, making the equation easier to manipulate and solve. It's like removing obstacles from a path, making the journey smoother and more direct. The elimination of fractions is a pivotal technique in solving rational equations. It transforms a complex problem into a more manageable one. But remember, guys, clearing fractions is not just about making things look simpler; it’s about maintaining the equality of the equation. Whatever we do to one side, we must do to the other.

Clearing fractions is a bit like setting the stage for the main act. With the fractions gone, we can now focus on the core algebraic manipulations needed to solve for ${x}$. This involves expanding, simplifying, and rearranging terms – all techniques that are fundamental to algebra. By clearing the fractions, we've paved the way for a straightforward solution. Next, we'll expand and simplify to get closer to the final answer. Let’s keep the momentum going!

Expanding and Simplifying

Now that we've cleared the fractions in the equation ${x(x + 10) = 4(x^2 - 2)}$, it’s time to expand and simplify. This step is all about getting the equation into a form that's easier to solve, usually a standard polynomial equation. Expanding means removing the parentheses by applying the distributive property. Simplifying means combining like terms to make the equation as concise as possible. Think of this as decluttering a room – you're organizing and tidying up to make the space more functional.

Let's start by expanding both sides of the equation. On the left side, we distribute ${x}$ across ${(x + 10)}$, which gives us ${x^2 + 10x}$. On the right side, we distribute ${4}$ across ${(x^2 - 2)}$, which gives us ${4x^2 - 8}$. So, the equation becomes:

${ x^2 + 10x = 4x^2 - 8 }$

Now, we need to simplify further. The goal is to get all the terms on one side of the equation, setting the other side to zero. This will allow us to solve the equation more easily, especially if it turns out to be a quadratic equation. Subtracting ${x^2}$ and ${10x}$ from both sides, we get:

${ 0 = 3x^2 - 10x - 8 }$

We now have a quadratic equation in the standard form ${ax^2 + bx + c = 0}$, where ${a = 3}$, ${b = -10}$, and ${c = -8}$. This form is perfect for applying various methods to solve for ${x}$, such as factoring, completing the square, or using the quadratic formula. The transformation into a standard quadratic form is a key achievement in this process. It’s like translating a problem into a language you already know how to solve. Expanding and simplifying is not just about making the equation look neater; it's about revealing its underlying structure and making it solvable.

Remember, guys, precision is key in these steps. A small error in expansion or simplification can lead to a completely wrong solution. So, always double-check your work! With our equation now in standard quadratic form, we’re ready to tackle the next challenge: finding the values of ${x}$ that satisfy this equation. Let’s move on to the solution phase!

Solving the Quadratic Equation

With our equation now simplified to the standard quadratic form ${3x^2 - 10x - 8 = 0}$, it's time to solve for ${x}$. There are several methods we can use, but two of the most common are factoring and the quadratic formula. Let's explore both, starting with factoring, as it can sometimes be the quickest route.

Factoring the Quadratic

Factoring involves expressing the quadratic expression as a product of two binomials. If we can factor the quadratic, we can then set each factor equal to zero and solve for ${x}$. For the equation ${3x^2 - 10x - 8 = 0}$, we look for two binomials that multiply to give us this quadratic. This can sometimes be a bit of a puzzle, requiring some trial and error. After some thought, we find that:

${ 3x^2 - 10x - 8 = (3x + 2)(x - 4) }$

Setting each factor equal to zero gives us:

${ 3x + 2 = 0 \quad \text{or} \quad x - 4 = 0 }$

Solving these linear equations, we get:

${ x = -\frac{2}{3} \quad \text{or} \quad x = 4 }$

So, factoring has given us two potential solutions: ${x = -\frac{2}{3}}$ and ${x = 4}$. But before we declare victory, we need to check these solutions against our initial restrictions. Factoring is like finding the right ingredients for a recipe; if you mix them correctly, you get a delicious result!

Using the Quadratic Formula

If factoring proves difficult, the quadratic formula is a reliable alternative. The quadratic formula provides a direct way to find the solutions of any quadratic equation in the form ${ax^2 + bx + c = 0}$. The formula is:

${ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }$

For our equation ${3x^2 - 10x - 8 = 0}$, we have ${a = 3}$, ${b = -10}$, and ${c = -8}$. Plugging these values into the quadratic formula, we get:

${ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(-8)}}{2(3)} }$

${ x = \frac{10 \pm \sqrt{100 + 96}}{6} }$

${ x = \frac{10 \pm \sqrt{196}}{6} }$

${ x = \frac{10 \pm 14}{6} }$

This gives us two solutions:

${ x = \frac{10 + 14}{6} = \frac{24}{6} = 4 }$

${ x = \frac{10 - 14}{6} = \frac{-4}{6} = -\frac{2}{3} }$

As you can see, the quadratic formula gives us the same solutions as factoring: ${x = -\frac{2}{3}}$ and ${x = 4}$. The quadratic formula is like a universal key; it unlocks the solutions to any quadratic equation, no matter how complex. Now that we have our potential solutions, it's time for the final step: checking for extraneous solutions.

Checking for Extraneous Solutions

We've arrived at a crucial stage in solving the equation ${\frac{x+10}{x^2-2}=\frac{4}{x}}$: checking for extraneous solutions. Remember those restrictions we identified at the very beginning? This is where they come into play. Extraneous solutions are values that we obtain through the algebraic process that do not actually satisfy the original equation. They typically arise when we perform operations that are not reversible, such as squaring both sides or, in our case, clearing denominators.

Our potential solutions are ${x = -\frac{2}{3}}$ and ${x = 4}$. Our restrictions, as we determined earlier, are ${x \neq 0}$, ${x \neq \sqrt{2}}$, and ${x \neq -\sqrt{2}}$. Both ${-\frac{2}{3}}$ and ${4}$ satisfy these restrictions, so they are likely valid solutions. But we can't be absolutely sure until we plug them back into the original equation. This step is like the final quality check in a manufacturing process; it ensures that our product—the solution—meets the required standards.

Let’s first check ${x = -\frac{2}{3}}$. Plugging this value into the original equation, we get:

${ \frac{-\frac{2}{3} + 10}{\left(-\frac{2}{3}\right)^2 - 2} = \frac{4}{-\frac{2}{3}} }$

Simplifying the left side:

${ \frac{\frac{28}{3}}{\frac{4}{9} - 2} = \frac{\frac{28}{3}}{\frac{4 - 18}{9}} = \frac{\frac{28}{3}}{-\frac{14}{9}} = \frac{28}{3} \cdot \left(-\frac{9}{14}\right) = -6 }$

Simplifying the right side:

${ \frac{4}{-\frac{2}{3}} = 4 \cdot \left(-\frac{3}{2}\right) = -6 }$

Since both sides are equal, ${x = -\frac{2}{3}}$ is indeed a solution. Yay!

Now let’s check ${x = 4}$. Plugging this value into the original equation, we get:

${ \frac{4 + 10}{4^2 - 2} = \frac{4}{4} }$

Simplifying the left side:

${ \frac{14}{16 - 2} = \frac{14}{14} = 1 }$

Simplifying the right side:

${ \frac{4}{4} = 1 }$

Since both sides are equal, ${x = 4}$ is also a solution. Double yay!

Thus, after careful checking, we can confidently conclude that both ${x = -\frac{2}{3}}$ and ${x = 4}$ are valid solutions to the original equation. This final check is not just a formality; it's a critical step in the problem-solving process. It ensures that our solutions are not just mathematically correct but also valid within the context of the original equation.

Conclusion

So, guys, we've successfully solved the equation ${\frac{x+10}{x^2-2}=\frac{4}{x}}$! We navigated through the complexities of rational expressions, cleared fractions, simplified a quadratic equation, and rigorously checked our solutions. The solutions we found are ${x = -\frac{2}{3}}$ and ${x = 4}$. This journey illustrates the power and beauty of algebra in solving real mathematical problems.

Remember, solving equations isn’t just about finding the right numbers; it’s about understanding the process, the logic, and the underlying concepts. Each step we took—from identifying restrictions to checking for extraneous solutions—was crucial in ensuring the accuracy and validity of our results. This problem encapsulates many of the key skills in algebra, including working with rational expressions, solving quadratic equations, and the importance of careful verification.

Keep practicing, keep exploring, and most importantly, keep questioning. The world of mathematics is vast and fascinating, and every problem you solve is a step further on your journey. Until next time, happy solving! You've got this! Solving mathematical equations is like piecing together a puzzle; each step is a piece, and when they all fit, you see the complete picture.