Extraneous Solutions: Is X=-1 Or X=10 Valid?

Hey guys! Let's dive into a fun math problem today where we're going to figure out if some solutions we found are actually valid, or if they're just trying to trick us! We're looking at the equation $\frac{12}{x^2-1}=\frac{4}{3x+3}$, and we've got two potential solutions: $x=10$ and $x=-1$. The big question is: are either of these solutions extraneous? In simpler terms, does plugging them back into the equation cause any problems? We'll break it down step by step so you can become a pro at spotting extraneous solutions. So, grab your thinking caps, and let’s get started!

Understanding Extraneous Solutions

Before we jump into solving the problem, let's make sure we're all on the same page about what extraneous solutions actually are. Extraneous solutions are basically fake answers. They pop up when we're solving equations, especially rational equations (equations with fractions where the variable is in the denominator), but they don't actually work when you plug them back into the original equation. They're like those friends who say they'll help you move but mysteriously disappear on moving day!

So, why do these sneaky solutions appear? Well, often it's because of operations we perform while solving the equation, like squaring both sides or, in our case, dealing with fractions that have variables in the denominator. These operations can sometimes introduce solutions that weren't there in the first place. That's why it's super important to always check your solutions by plugging them back into the original equation. If a solution makes the denominator of any fraction zero, it's an extraneous solution and we have to throw it out.

To spot these mathematical imposters, the golden rule is: Check, check, and double-check! We'll apply this rule rigorously as we analyze our potential solutions for the equation.

Solving the Equation

Okay, let's get our hands dirty and solve the equation $\frac{12}{x^2-1}=\frac{4}{3x+3}$. Our first goal is to simplify things and get rid of those pesky fractions. To do that, we'll try to find a common denominator and then multiply both sides of the equation by it. This will clear out the fractions and make the equation much easier to handle.

First, we need to factor the denominators to see what we're working with. Notice that $x^2 - 1$ is a difference of squares, which factors into $(x - 1)(x + 1)$. And $3x + 3$ can be factored by pulling out a 3, giving us $3(x + 1)$. So, our equation now looks like this: $\frac{12}{(x-1)(x+1)} = \frac{4}{3(x+1)}$.

Now we can identify the least common denominator (LCD). The LCD is the smallest expression that both denominators divide into evenly. In this case, the LCD is $3(x - 1)(x + 1)$. To clear the fractions, we'll multiply both sides of the equation by this LCD:

$3(x - 1)(x + 1) \cdot \frac{12}{(x-1)(x+1)} = 3(x - 1)(x + 1) \cdot \frac{4}{3(x+1)}$

On the left side, the $(x - 1)(x + 1)$ terms cancel out, and on the right side, the $3(x + 1)$ terms cancel out. This leaves us with:

$3 \cdot 12 = 4(x - 1)$

Simplifying this, we get:

$36 = 4x - 4$

Now we can solve for $x$. Add 4 to both sides:

$40 = 4x$

Divide both sides by 4:

$x = 10$

But wait! We're not done yet. We also need to consider the potential for extraneous solutions. Let's not forget about our other candidate, $x = -1$, which was given in the problem. So, we have two potential solutions: $x = 10$ and $x = -1$. Now comes the crucial step: we need to check these solutions in the original equation to see if they are valid.

Checking for Extraneous Solutions

This is the most important part, guys! We've found our potential solutions, but now we need to put on our detective hats and see if they hold up under scrutiny. Remember, an extraneous solution is one that doesn't actually work when you plug it back into the original equation, usually because it makes a denominator zero.

So, let's take each of our solutions, $x = 10$ and $x = -1$, and plug them back into the original equation: $\frac{12}{x^2-1}=\frac{4}{3x+3}$.

Checking x = 10

First, let's try $x = 10$. Plugging it into the equation, we get:

$\frac{12}{10^2-1} = \frac{4}{3(10)+3}$

Simplify the denominators:

$\frac{12}{100-1} = \frac{4}{30+3}$

$\frac{12}{99} = \frac{4}{33}$

Now, we can simplify both fractions. The left side simplifies to $\frac{4}{33}$, and the right side is already $\frac{4}{33}$. So, we have:

$\frac{4}{33} = \frac{4}{33}$

This is a true statement! So, $x = 10$ is a valid solution. It passes the test with flying colors. No problem here!

Checking x = -1

Now, let's check the other solution, $x = -1$. Plugging it into the original equation, we get:

$\frac{12}{(-1)^2-1} = \frac{4}{3(-1)+3}$

Simplify the denominators:

$\frac{12}{1-1} = \frac{4}{-3+3}$

$\frac{12}{0} = \frac{4}{0}$

Uh oh! We've got a problem. We have zero in the denominator on both sides of the equation. Dividing by zero is a big no-no in mathematics – it's undefined! This means that $x = -1$ makes the denominators zero, and therefore, it is an extraneous solution. It's a mathematical imposter!

Conclusion

Alright guys, we've reached the end of our mathematical journey for today. We started with the equation $\frac{12}{x^2-1}=\frac{4}{3x+3}$, found two potential solutions ($x = 10$ and $x = -1$), and then put them to the test. By plugging them back into the original equation, we discovered that:

• $x = 10$ is a valid solution. It works perfectly and makes the equation true.
• $x = -1$ is an extraneous solution. It makes the denominators zero, which is a mathematical crime! So, we have to reject it.

Therefore, the final answer is that the solution $x = -1$ is extraneous. Always remember to check your solutions, especially when dealing with rational equations. You never know when an extraneous solution might be lurking around the corner, trying to trick you! Keep practicing, and you'll become a master at spotting these mathematical imposters. You got this!