Solving The Equation X²/2x-6 = 9/6x-18 Unveiling True Solutions

Hey everyone! Let's dive into a super interesting math problem today. We're going to break down the equation x²/(2x-6) = 9/(6x-18) and figure out what's really going on with its solutions. It's like we're math detectives, and this equation is our mystery to solve! So, buckle up, and let's get started!

Cracking the Code: Solving the Equation

To solve the mystery, the first step is to manipulate the equation to make it easier to handle. Our main goal here is to get rid of those pesky fractions. We can do this by cross-multiplying. This means we multiply the numerator (the top part) of the first fraction by the denominator (the bottom part) of the second fraction, and vice versa. When we cross multiply in the equation, x²/(2x-6) = 9/(6x-18), we get: x² * (6x - 18) = 9 * (2x - 6).

Now, let's simplify both sides of the equation. We'll distribute the terms, which means multiplying the terms outside the parentheses with the terms inside. So, on the left side, x² * (6x - 18) becomes 6x³ - 18x². On the right side, 9 * (2x - 6) becomes 18x - 54. So our new equation is 6x³ - 18x² = 18x - 54. It looks a bit intimidating, right? But don't worry, we're going to tame it! Our next move is to bring all the terms to one side of the equation. This makes it equal to zero, which is a standard form for solving polynomial equations. We subtract 18x and add 54 to both sides of the equation, this will give us: 6x³ - 18x² - 18x + 54 = 0. Now, we have a cubic equation – an equation where the highest power of x is 3. These can be tricky, but we have some tricks up our sleeves.

The next step is simplifying the equation by dividing both sides by a common factor. Looking at our equation, 6x³ - 18x² - 18x + 54 = 0, we can see that all the coefficients (the numbers in front of the x terms) are divisible by 6. So, let's divide the entire equation by 6. This will make our numbers smaller and the equation easier to work with. Dividing by 6, our equation becomes x³ - 3x² - 3x + 9 = 0. See? Much friendlier! Now, we need to factor this cubic equation. Factoring is like reverse-distributing. We're trying to find expressions that, when multiplied together, give us our equation. One common technique for factoring cubic equations is factoring by grouping. We group the first two terms together and the last two terms together: (x³ - 3x²) + (-3x + 9) = 0. Now, we look for common factors in each group. In the first group, (x³ - 3x²), the common factor is x². We can factor out x² to get: x²(x - 3). In the second group, (-3x + 9), the common factor is -3. Factoring out -3, we get: -3(x - 3). So, our equation now looks like this: x²(x - 3) - 3(x - 3) = 0. Notice something cool? Both terms now have a common factor of (x - 3). This means we can factor out (x - 3) from the entire equation. When we do that, we get: (x - 3)(x² - 3) = 0. We've successfully factored our cubic equation! Now we have a product of two factors that equals zero. This is great because it means that at least one of these factors must be zero.

Now that we've factored the equation, we can use the Zero Product Property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In our case, we have (x - 3)(x² - 3) = 0, so either (x - 3) = 0 or (x² - 3) = 0. Let's solve each of these equations separately. First, we solve x - 3 = 0. To isolate x, we add 3 to both sides of the equation, which gives us x = 3. So, one potential solution is x = 3. Now, let's solve the second equation, x² - 3 = 0. To isolate x², we add 3 to both sides of the equation, which gives us x² = 3. To solve for x, we take the square root of both sides. Remember that when we take the square root, we need to consider both the positive and negative roots. So, x = ±√3. This gives us two more potential solutions: x = √3 and x = -√3. So, based on our calculations, we have three potential solutions: x = 3, x = √3, and x = -√3. But here's where our detective work isn't quite over! We need to check these solutions to make sure they actually work in the original equation.

Spotting the Fakes: Extraneous Solutions

Alright, detectives, now comes the crucial part! We've found our potential solutions: x = 3, x = √3, and x = -√3. But not all that glitters is gold, and in math, not all potential solutions are actual solutions. We need to check for extraneous solutions. These are values that we get when solving the equation, but when we plug them back into the original equation, they cause problems, like division by zero or taking the square root of a negative number. Remember our original equation? x²/(2x - 6) = 9/(6x - 18). The potential trouble spot here is the denominators: (2x - 6) and (6x - 18). If either of these equals zero, we're in trouble because division by zero is a big no-no in math. Let's start by checking x = 3. If we plug x = 3 into the first denominator (2x - 6), we get 2(3) - 6 = 6 - 6 = 0. Uh oh! That's a problem. This means x = 3 makes the denominator zero, which makes the fraction undefined. So, x = 3 is an extraneous solution. It's a fake! Now, let's check x = √3. Plugging this into the denominators: 2(√3) - 6 ≈ -2.53, and 6(√3) - 18 ≈ -7.61. These are not zero, so x = √3 is potentially a valid solution. Finally, let's check x = -√3. Plugging this into the denominators: 2(-√3) - 6 ≈ -9.46, and 6(-√3) - 18 ≈ -28.39. Again, these are not zero, so x = -√3 is also potentially valid. So, we've determined that x = 3 is definitely an extraneous solution, but x = √3 and x = -√3 seem okay so far. To be absolutely sure, we need to plug x = √3 and x = -√3 back into the entire original equation and see if both sides are equal.

Let's verify x = √3 in the original equation x²/(2x - 6) = 9/(6x - 18). Substituting x = √3, we have: (√3)² / (2√3 - 6) = 9 / (6√3 - 18). Simplifying the left side: 3 / (2√3 - 6). Simplifying the right side: 9 / (6√3 - 18). To see if these are equal, we can cross-multiply: 3(6√3 - 18) = 9(2√3 - 6). Expanding both sides: 18√3 - 54 = 18√3 - 54. Hey, look at that! Both sides are equal. This confirms that x = √3 is indeed a valid solution. Now, let's verify x = -√3 in the original equation. Substituting x = -√3, we have: (-√3)² / (2(-√3) - 6) = 9 / (6(-√3) - 18). Simplifying the left side: 3 / (-2√3 - 6). Simplifying the right side: 9 / (-6√3 - 18). Again, let's cross-multiply: 3(-6√3 - 18) = 9(-2√3 - 6). Expanding both sides: -18√3 - 54 = -18√3 - 54. Awesome! Both sides are equal again. This confirms that x = -√3 is also a valid solution. So, after all our detective work, we've found that x = √3 and x = -√3 are the actual solutions to the equation, and x = 3 is an extraneous solution.

The Verdict: Actual Solutions vs. Extraneous Solutions

So, what have we learned today, guys? We've journeyed through the world of equations, cross-multiplied, factored, and even hunted down some sneaky extraneous solutions. It's like we're math detectives, solving mysteries one equation at a time! We started with a seemingly complex equation, x²/(2x - 6) = 9/(6x - 18), and we broke it down step by step. We found potential solutions, but we didn't stop there. We knew that some solutions might be imposters, so we put on our detective hats and checked each one. We discovered that x = 3 was an extraneous solution – a solution that looks good on paper but doesn't actually work in the original equation because it leads to division by zero. We also found two real solutions: x = √3 and x = -√3. These solutions passed our rigorous checks and proved their worth. Remember, when solving equations, especially those with fractions or square roots, it's crucial to check your answers. Extraneous solutions can sneak in and lead you astray if you're not careful.

In conclusion, the solutions to the equation x²/(2x - 6) = 9/(6x - 18) are x = √3 and x = -√3. These are the true solutions, the ones that make the equation balance perfectly. The potential solution x = 3, however, turned out to be an extraneous solution, a mathematical red herring. So, next time you're faced with an equation, remember to solve carefully, check your answers, and watch out for those sneaky extraneous solutions! Keep your math detective hats on, and you'll crack any equation that comes your way!

Therefore, the correct answer is B. x = ±√3, but they are extraneous solutions.