Identify Extraneous Solutions In Rational Equations

When we're solving equations, especially those involving fractions like rational expressions, it's super important to keep an eye out for extraneous solutions. These are solutions that pop up during our solving process but don't actually work when plugged back into the original equation. Think of them as little mathematical imposters! Today, we're going to dive deep into how to spot these tricky values in the equation $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$. Understanding this concept is key to mastering algebra and ensuring your answers are always correct.

Understanding Extraneous Solutions in Rational Equations

Let's really get a handle on what extraneous solutions are in the context of solving equations, particularly those with fractions (rational equations). When you're faced with an equation that has variables in the denominator, like the one we're about to tackle, there's a crucial rule: the denominator can never be zero. If a potential solution makes any denominator in the original equation equal to zero, then that solution is not valid. It's extraneous. So, the first and most vital step in solving any rational equation is to identify the values of the variable that would make any denominator zero. These are your 'forbidden' values, the ones you must exclude from your possible solutions. For our equation, $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$, we have two denominators: $(x-1)$ and $(2x^2-2)$. We need to find the values of $x$ that make either of these zero. Setting the first denominator to zero, we get $x-1=0$, which means $x=1$. If $x=1$, the left side of the equation becomes $\frac{1}{1-1} = \frac{1}{0}$, which is undefined. Now, let's look at the second denominator: $2x^2-2$. We set this to zero: $2x^2-2=0$. We can factor out a 2, giving us $2(x^2-1)=0$. The term $(x^2-1)$ is a difference of squares, which factors into $(x-1)(x+1)$. So, the equation becomes $2(x-1)(x+1)=0$. This equation is true if either $(x-1)=0$ or $(x+1)=0$. This gives us two more forbidden values: $x=1$ (which we already found) and $x=-1$. Therefore, for the equation $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$, any solution we find that is $x=1$ or $x=-1$ will be an extraneous solution because it would lead to division by zero in the original equation. This pre-analysis of potential problem values is a game-changer for efficiently solving rational equations and avoiding those pesky extraneous solutions.

Solving the Rational Equation Step-by-Step

Now that we've identified the values that could cause problems ($x=1$ and $x=-1$), let's proceed with solving the equation $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$. Our goal is to eliminate the denominators and get a simpler equation. A great way to do this is to multiply both sides of the equation by the least common denominator (LCD). First, let's factor the denominators completely. We already factored the second denominator: $2x^2-2 = 2(x^2-1) = 2(x-1)(x+1)$. The first denominator is just $(x-1)$. So, the LCD for this equation is $2(x-1)(x+1)$. Now, we multiply both sides of the equation by this LCD:

$$2(x-1)(x+1) \cdot \frac{1}{x-1} = 2(x-1)(x+1) \cdot \frac{x-2}{2(x-1)(x+1)}$$

On the left side, the $(x-1)$ terms cancel out, leaving us with $2(x+1) \cdot 1$, which simplifies to $2(x+1)$.

On the right side, the entire denominator $2(x-1)(x+1)$ cancels out with the multiplier, leaving us with just $(x-2)$.

So, our equation transforms into a much simpler linear equation:

$$2(x+1) = x-2$$

Now, we just need to solve for $x$. First, distribute the 2 on the left side:

$$2x + 2 = x-2$$

Next, we want to gather all the $x$ terms on one side and the constant terms on the other. Subtract $x$ from both sides:

$$2x - x + 2 = x - x - 2$$

$$x + 2 = -2$$

Finally, subtract 2 from both sides to isolate $x$:

$$x + 2 - 2 = -2 - 2$$

$$x = -4$$

So, our algebraic process yields a single potential solution: $x=-4$. This step-by-step approach, starting with identifying potential issues and then systematically clearing denominators, is fundamental to confidently solving these types of problems.

Verifying the Solution and Identifying Extraneous Values

We've arrived at a potential solution, $x=-4$, after carefully solving the equation $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$. However, the job isn't quite done yet! We absolutely must verify this solution by plugging it back into the original equation. This is where we confirm if our solution is valid or if it's actually an extraneous one. Remember, we previously identified that $x=1$ and $x=-1$ would make the denominators zero, and thus are values to watch out for. Our potential solution, $x=-4$, is not one of those values, which is a good sign, but verification is still essential.

Let's substitute $x=-4$ into the original equation:

$$LHS = \frac{1}{x-1} = \frac{1}{-4-1} = \frac{1}{-5} = -\frac{1}{5}$$

$$RHS = \frac{x-2}{2 x^2-2} = \frac{-4-2}{2(-4)^2-2} = \frac{-6}{2(16)-2} = \frac{-6}{32-2} = \frac{-6}{30} = -\frac{1}{5}$$

Since the Left Hand Side (LHS) equals the Right Hand Side (RHS) when $x=-4$, this solution is valid. It works perfectly in the original equation.

Now, let's consider the options given in the question: A. $x=1$ and $x=-4$, B. neither $x=1$ or $x=-4$, C. $x=1$, D. $x=-4$. The question asks which solution is extraneous. We found that $x=-4$ is a valid solution. What about $x=1$? We determined early on that if $x=1$, the denominator $(x-1)$ in the original equation becomes $1-1=0$, which is undefined. Therefore, $x=1$ is an extraneous solution. It arose from the algebraic manipulation but does not satisfy the original equation due to division by zero.

So, to directly answer the question, the extraneous solution is $x=1$. This means option C is the correct choice. It's crucial to remember that even if you find a value during the solving process, you must always check if it leads to division by zero in the original equation. If it does, it's extraneous, no matter how neatly it came out of your algebra.

Conclusion: Mastering Extraneous Solutions

We've successfully navigated the complexities of solving a rational equation and, more importantly, identified an extraneous solution. The equation $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$ presented us with an opportunity to practice a vital skill in algebra: recognizing and discarding extraneous solutions. We learned that these 'fake' solutions appear when our algebraic steps, while sound in isolating the variable, produce a result that violates the fundamental rules of the original equation. In the case of rational equations, this violation most often involves making a denominator equal to zero, leading to an undefined expression.

Our process involved two key stages. First, we proactively identified the values of $x$ that would make any denominator in the original equation zero. For $\frac{1}{x-1}=\frac{x-2}{2 x^2-2}$, these values were $x=1$ (from $x-1=0$) and $x=-1$ (from $2x^2-2=0$). These are our 'danger zones'. Second, we solved the equation using algebraic methods, which led us to a potential solution of $x=-4$. Finally, and critically, we verified this solution by plugging it back into the original equation. Since $x=-4$ did not cause any denominators to be zero and resulted in a true statement ($-1/5 = -1/5$), it is a valid solution. However, the value $x=1$, which we flagged as a potential problem from the start, is indeed an extraneous solution because it would force a division by zero in the original equation.

Therefore, when asked which solution is extraneous, the answer is x=1. Understanding this concept is not just about passing a test; it's about developing a rigorous and accurate approach to problem-solving in mathematics. Always remember to check your solutions in the original equation, especially when dealing with rational expressions, radical equations, or logarithmic equations, as these types of equations are prone to generating extraneous results. Keep practicing, and you'll become a pro at spotting these tricky mathematical imposters!

For further exploration into solving rational equations and understanding extraneous solutions, you can refer to resources from Khan Academy and Paul's Online Math Notes. These sites offer comprehensive explanations and practice problems that can solidify your understanding.