Mastering Algebraic Equations: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving deep into the world of algebra to tackle a juicy equation that might look a bit intimidating at first glance: rac{x^2-x-6}{x^2}=rac{x-6}{2 x}+rac{2 x+12}{x}. Don't sweat it, guys! We're going to break this down piece by piece, making sure you understand every single step. This isn't just about getting the right answer; it's about building your confidence and sharpening your problem-solving skills. So, grab your notebooks, get comfortable, and let's unravel this algebraic puzzle together. We'll explore the strategies needed to simplify complex fractions, find common denominators, and ultimately isolate the variable 'x' to find its true value. This process is fundamental to many areas of mathematics and science, so mastering it will set you up for success in future studies and even in tackling real-world problems that often present themselves in disguised mathematical forms. We'll also touch upon the importance of checking our solutions to ensure accuracy, a crucial habit for any aspiring mathematician or scientist. Get ready to flex those brain muscles and feel that amazing sense of accomplishment when we finally crack this equation!

Understanding the Equation and Initial Steps

Alright, let's start by staring this equation down: rac{x^2-x-6}{x^2}=rac{x-6}{2 x}+rac{2 x+12}{x}. The first thing you'll notice is that we're dealing with fractions. Whenever you see fractions in an equation, especially with variables in the denominator, your Spidey senses should tingle. Our primary goal here is to get rid of those pesky denominators. Why? Because working with whole numbers (or expressions without fractions) is generally much, much easier. To do this, we need to find the least common multiple (LCM) of all the denominators. Our denominators are $x^2$, $2x$, and $x$. Let's break them down:

• $x^2$ is simply $x imes x$
• $2x$ is $2 imes x$
• $x$ is just $x$

To find the LCM, we need to include all the unique factors raised to their highest power. We have a factor of '2' and a factor of 'x'. The highest power of 'x' we see is $x^2$. So, our LCM is $2x^2$. This is the magic number that will help us clear out all the fractions.

Before we go further, a quick but super important note: we need to identify any values of 'x' that would make our denominators zero, as division by zero is undefined. In this equation, the denominators are $x^2$, $2x$, and $x$. Therefore, $x$ cannot be equal to 0. Keep this in mind; it's our restriction!

Now, let's multiply every single term in the equation by our LCM, which is $2x^2$. This is the golden step that will transform our fractional nightmare into a simpler algebraic dream. We'll do this carefully, term by term, to ensure no mistakes are made. This technique is incredibly powerful for solving rational equations, and it's a skill you'll use again and again. Think of it as a universal key that unlocks the solution to many complex algebraic problems. By systematically eliminating the denominators, we simplify the problem significantly, paving the way for straightforward manipulation of the terms. Remember that restriction we found? We'll need to check our final answer against it later to make sure it's valid.

Clearing the Denominators

Okay, team, let's get down to business and multiply each term of our equation rac{x^2-x-6}{x^2}=rac{x-6}{2 x}+rac{2 x+12}{x} by the LCM, which we determined to be $2x^2$. This is where the magic happens, and those fractions start to disappear!

1. Multiplying the first term: (2x^2) imes rac{x^2-x-6}{x^2} Here, the $x^2$ in the denominator cancels out with one of the $x^2$ factors from $2x^2$. What's left is $2 imes (x^2-x-6)$. So, we get $2(x^2-x-6)$. Expanding this gives us $2x^2 - 2x - 12$.

2. Multiplying the second term: (2x^2) imes rac{x-6}{2x} In this case, the '2' in the denominator cancels with the '2' in $2x^2$, and one of the 'x' factors cancels with the 'x' in the denominator. We're left with $x imes (x-6)$. Expanding this gives us $x^2 - 6x$.

3. Multiplying the third term: (2x^2) imes rac{2x+12}{x} Here, one of the 'x' factors from $2x^2$ cancels with the 'x' in the denominator. We are left with $2x imes (2x+12)$. Expanding this gives us $4x^2 + 24x$.

Now, let's put it all together. Our original equation has been transformed into a much simpler form without any denominators:

$2x^2 - 2x - 12 = x^2 - 6x + 4x^2 + 24x$

See? No more fractions! This is a huge step towards solving for 'x'. The next stage involves simplifying this new equation by combining like terms and rearranging it into a standard form, which is typically a quadratic equation ($ax^2 + bx + c = 0$). This process of clearing denominators is a cornerstone of solving rational equations, and it's crucial to perform it meticulously. Each multiplication step needs to be handled with care, ensuring that cancellations are done correctly and that the distributive property is applied accurately when expanding the resulting expressions. The goal is to transition from a complex fractional equation to a more manageable polynomial equation, setting the stage for the final steps of finding the roots.

Simplifying and Rearranging the Equation

We've successfully cleared the denominators, and our equation now looks like this: $2x^2 - 2x - 12 = x^2 - 6x + 4x^2 + 24x$. Our next mission, should we choose to accept it (and we totally should!), is to simplify this beast. This means gathering all the terms onto one side of the equation to set it equal to zero. This is the standard form for a quadratic equation, which we can then solve using various methods.

First, let's combine the like terms on the right-hand side of the equation:

• Combine the $x^2$ terms: $x^2 + 4x^2 = 5x^2$
• Combine the $x$ terms: $-6x + 24x = 18x$

So, the right-hand side simplifies to $5x^2 + 18x$. Now our equation is:

$2x^2 - 2x - 12 = 5x^2 + 18x$

Awesome! Now, let's move all the terms from the right side over to the left side. To do this, we subtract $5x^2$ and $18x$ from both sides of the equation:

$(2x^2 - 5x^2) - 2x - 18x - 12 = 0$

Let's simplify this further:

• Combine the $x^2$ terms: $2x^2 - 5x^2 = -3x^2$
• Combine the $x$ terms: $-2x - 18x = -20x$

So, our equation becomes:

$-3x^2 - 20x - 12 = 0$

This is a quadratic equation! However, it's often easier to work with quadratic equations when the leading coefficient (the coefficient of the $x^2$ term) is positive. To achieve this, we can multiply the entire equation by -1. This doesn't change the solutions, it just makes the numbers look a bit friendlier:

$(-1) imes (-3x^2 - 20x - 12) = (-1) imes 0$

Which gives us:

$3x^2 + 20x + 12 = 0$

There we have it – our simplified quadratic equation in standard form! The process of combining like terms and rearranging is essential for isolating the unknown and setting up the equation for solving. It involves careful attention to signs and coefficients to avoid errors. By transforming the equation into the $ax^2 + bx + c = 0$ format, we make it amenable to standard solving techniques like factoring or the quadratic formula. This simplification step is crucial; a single sign error here can lead to a completely incorrect final answer, so always double-check your arithmetic and algebraic manipulations.

Solving the Quadratic Equation

We've arrived at a beautiful, standard quadratic equation: $3x^2 + 20x + 12 = 0$. Now it's time to find the values of 'x' that satisfy this equation. There are a couple of popular methods for solving quadratic equations: factoring and using the quadratic formula. Let's explore both!

Method 1: Factoring

Factoring works best when the quadratic expression can be easily broken down into two binomials. We are looking for two numbers that multiply to give us $a imes c$ (which is $3 imes 12 = 36$) and add up to give us $b$ (which is 20). Let's list factors of 36:

• 1 and 36 (sum = 37)
• 2 and 18 (sum = 20)
• 3 and 12 (sum = 15)
• 4 and 9 (sum = 13)
• 6 and 6 (sum = 12)

Bingo! The pair 2 and 18 adds up to 20. Now we can rewrite the middle term ($20x$) using these numbers:

$3x^2 + 2x + 18x + 12 = 0$

Next, we group the terms and factor by grouping:

• Group the first two terms: $(3x^2 + 2x)$
• Group the last two terms: $(18x + 12)$

Factor out the greatest common factor (GCF) from each group:

• From $(3x^2 + 2x)$, the GCF is $x$. So we get $x(3x + 2)$.
• From $(18x + 12)$, the GCF is 6. So we get $6(3x + 2)$.

Now our equation looks like this:

$x(3x + 2) + 6(3x + 2) = 0$

Notice that we have a common binomial factor, $(3x + 2)$. We can factor this out:

$(3x + 2)(x + 6) = 0$

For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for 'x':

• 3x + 2 = 0 ewline ightarrow 3x = -2 ewline ightarrow x = -rac{2}{3}
• $x + 6 = 0 ewline ightarrow x = -6$

So, our potential solutions are x = -rac{2}{3} and $x = -6$.

Method 2: The Quadratic Formula

If factoring seems tricky, the quadratic formula is your reliable backup. For any quadratic equation in the form $ax^2 + bx + c = 0$, the solutions are given by:

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