What Is True About The Solution Of X 2 2 X − 6 = 9 6 X − 18 \frac{x^2}{2x-6}=\frac{9}{6x-18} 2 X − 6 X 2 ​ = 6 X − 18 9 ​ ?A. X = ± 3 X = \pm \sqrt{3} X = ± 3 ​ , And They Are Actual Solutions.B. X = ± 3 X = \pm \sqrt{3} X = ± 3 ​ , But They Are Extraneous Solutions.C. X = 3 X = 3 X = 3 , And It Is An Actual

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Introduction

When solving equations involving fractions, it's essential to consider the restrictions on the variables and the potential for extraneous solutions. In this case, we're given the equation $\frac{x^2}{2x-6}=\frac{9}{6x-18}$, and we need to determine the validity of the solutions $x = \pm \sqrt{3}$.

Step 1: Multiply both sides by the LCD

To eliminate the fractions, we multiply both sides of the equation by the least common denominator (LCD), which is $(2x-6)(6x-18)$. This gives us:

$$x^2(6x-18) = 9(2x-6)$$

Step 2: Expand and simplify

Expanding and simplifying the equation, we get:

$$6x^3 - 18x^2 = 18x - 54$$

Step 3: Move all terms to one side

Moving all terms to one side of the equation, we get:

$$6x^3 - 18x^2 - 18x + 54 = 0$$

Step 4: Factor the equation

Factoring the equation, we get:

$$(2x-3)(3x^2-9x+18) = 0$$

Step 5: Solve for x

Setting each factor equal to zero, we get:

$$2x-3 = 0 \quad \text{or} \quad 3x^2-9x+18 = 0$$

Solving for x, we get:

$$x = \frac{3}{2} \quad \text{or} \quad x = \frac{3 \pm \sqrt{3}}{2}$$

Step 6: Check for extraneous solutions

We need to check if the solutions $x = \pm \sqrt{3}$ are extraneous. Plugging these values back into the original equation, we get:

$$\frac{(\pm \sqrt{3})^2}{2(\pm \sqrt{3})-6} = \frac{9}{6(\pm \sqrt{3})-18}$$

Simplifying, we get:

$$\frac{3}{\pm 2\sqrt{3}-6} = \frac{9}{\pm 6\sqrt{3}-18}$$

Step 7: Determine the validity of the solutions

Since the denominators are zero when $x = \pm \sqrt{3}$, these values are not valid solutions. Therefore, the correct answer is:

Conclusion

The solutions $x = \pm \sqrt{3}$ are extraneous solutions, and the only valid solution is $x = 3$. Therefore, the correct answer is:

B. $x = \pm \sqrt{3}$, but they are extraneous solutions.

This answer is supported by the fact that the denominators are zero when $x = \pm \sqrt{3}$, making these values invalid solutions.

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Introduction

In our previous article, we explored the solution of the equation $\frac{x^2}{2x-6}=\frac{9}{6x-18}$. We determined that the solutions $x = \pm \sqrt{3}$ are extraneous, and the only valid solution is $x = 3$. In this article, we'll answer some frequently asked questions about the solution of this equation.

Q: What is the least common denominator (LCD) of the fractions in the equation?

A: The LCD of the fractions in the equation is $(2x-6)(6x-18)$.

Q: Why do we need to multiply both sides of the equation by the LCD?

A: We need to multiply both sides of the equation by the LCD to eliminate the fractions and simplify the equation.

Q: How do we determine if a solution is extraneous?

A: A solution is extraneous if it makes the denominator of the original equation equal to zero. In this case, the solutions $x = \pm \sqrt{3}$ make the denominator zero, so they are extraneous.

Q: What is the difference between an actual solution and an extraneous solution?

A: An actual solution is a value that satisfies the original equation, while an extraneous solution is a value that makes the equation true but is not a valid solution.

Q: Why is it important to check for extraneous solutions?

A: It's essential to check for extraneous solutions because they can arise from the process of solving the equation. If we don't check for extraneous solutions, we may end up with incorrect answers.

Q: Can you provide an example of how to check for extraneous solutions?

A: Let's say we have the equation $\frac{x}{x-1} = 2$. If we solve for x, we get $x = 2$. However, if we plug this value back into the original equation, we get $\frac{2}{2-1} = 2$, which is true. But if we plug $x = 2$ into the denominator, we get $2-1 = 1$, which is not zero. Therefore, $x = 2$ is an actual solution.

Q: What is the final answer to the equation $\frac{x^2}{2x-6}=\frac{9}{6x-18}$?

A: The final answer to the equation is $x = 3$, which is the only valid solution.

Q: Why is $x = 3$ the only valid solution?

A: $x = 3$ is the only valid solution because it satisfies the original equation and does not make the denominator zero.

Q: Can you provide a summary of the steps to solve the equation?

A: Here's a summary of the steps to solve the equation:

1. Multiply both sides of the equation by the LCD.
2. Expand and simplify the equation.
3. Move all terms to one side of the equation.
4. Factor the equation.
5. Solve for x.
6. Check for extraneous solutions.

Conclusion

In this article, we answered some frequently asked questions about the solution of the equation $\frac{x^2}{2x-6}=\frac{9}{6x-18}$. We that the solutions $x = \pm \sqrt{3}$ are extraneous, and the only valid solution is $x = 3$. We also provided a summary of the steps to solve the equation and discussed the importance of checking for extraneous solutions.