Instructions: Solve The Equation And Check For Extraneous Solutions.$\[ \frac{6}{x^2} = \frac{x-6}{x} - \frac{1}{x^2} \\]Find \[$ X = \square, \square \$\]

Introduction

In this article, we will guide you through the process of solving a given equation and checking for extraneous solutions. The equation we will be working with is $\frac{6}{x^2} = \frac{x-6}{x} - \frac{1}{x^2}$. We will use algebraic techniques to solve for the variable $x$ and then verify our solutions to ensure they are valid.

Step 1: Multiply Both Sides by $x^2$

To begin solving the equation, we need to eliminate the fractions. We can do this by multiplying both sides of the equation by $x^2$. This will give us a new equation without any fractions.

$$\frac{6}{x^2} \cdot x^2 = \left(\frac{x-6}{x} - \frac{1}{x^2}\right) \cdot x^2$$

Simplifying both sides of the equation, we get:

$$6 = (x-6)x - 1$$

Step 2: Expand and Simplify the Equation

Next, we need to expand and simplify the equation. We can do this by multiplying the terms inside the parentheses.

$$6 = x^2 - 6x - 1$$

Now, we can combine like terms and simplify the equation further.

$$6 = x^2 - 6x - 1$$

Step 3: Rearrange the Equation

To make it easier to solve for $x$, we can rearrange the equation by moving all the terms to one side.

$$x^2 - 6x - 1 - 6 = 0$$

Simplifying the equation, we get:

$$x^2 - 6x - 7 = 0$$

Step 4: Factor the Quadratic Equation

Now, we need to factor the quadratic equation. We can do this by finding two numbers whose product is $-7$ and whose sum is $-6$.

$$x^2 - 6x - 7 = (x - a)(x - b)$$

Factoring the quadratic equation, we get:

$$(x - 7)(x + 1) = 0$$

Step 5: Solve for $x$

Now, we can solve for $x$ by setting each factor equal to zero.

$$x - 7 = 0 \quad \text{or} \quad x + 1 = 0$$

Solving for $x$, we get:

$$x = 7 \quad \text{or} \quad x = -1$$

Step 6: Check for Extraneous Solutions

Before we can consider our solutions valid, we need to check for extraneous solutions. We can do this by plugging our solutions back into the original equation and checking if they are true.

Plugging $x = 7$ into the original equation, we get:

$$\frac{6}{7^2} = \frac{7-6}{7} - \frac{1}{7^2}$$

Simplifying the equation, we get:

$$\frac{6}{49} = \frac{1}{7} - \frac{1}{49}$$

This is a true statement, so $x = 7$ is a valid solution.

Plugging $x = -1$ into the original equation, we get:

$$\frac{6}{(-1)^2} = \frac{-1-6}{-1} - \frac{1}{(-1)^2}$$

Simplifying the equation, we get:

$$6 = \frac{7}{-1} + 1$$

This is not a true statement, so $x = -1$ is an extraneous solution.

Conclusion

$$$$$$

Introduction

In our previous article, we guided you through the process of solving a given equation and checking for extraneous solutions. The equation we worked with was $\frac{6}{x^2} = \frac{x-6}{x} - \frac{1}{x^2}$. We used algebraic techniques to solve for the variable $x$ and then verified our solutions to ensure they were valid. In this article, we will answer some frequently asked questions related to solving equations and checking for extraneous solutions.

Q: What is an extraneous solution?

A: An extraneous solution is a solution to an equation that is not valid. It is a solution that makes the equation true, but it is not a real or actual solution to the equation.

Q: How do I check for extraneous solutions?

A: To check for extraneous solutions, you need to plug your solutions back into the original equation and check if they are true. If the solution makes the equation true, then it is a valid solution. If the solution does not make the equation true, then it is an extraneous solution.

Q: Why do I need to check for extraneous solutions?

A: You need to check for extraneous solutions because some solutions may not be valid. For example, in the equation $\frac{6}{x^2} = \frac{x-6}{x} - \frac{1}{x^2}$, the solution $x = -1$ is an extraneous solution. If you do not check for extraneous solutions, you may end up with incorrect answers.

Q: How do I know if a solution is extraneous or not?

A: To determine if a solution is extraneous or not, you need to plug the solution back into the original equation and check if it is true. If the solution makes the equation true, then it is a valid solution. If the solution does not make the equation true, then it is an extraneous solution.

Q: Can I always check for extraneous solutions?

A: No, you cannot always check for extraneous solutions. In some cases, it may be difficult or impossible to check for extraneous solutions. For example, in the equation $x^2 + 2x + 1 = 0$, it is not possible to check for extraneous solutions because the equation has no real solutions.

Q: What are some common mistakes to avoid when checking for extraneous solutions?

A: Some common mistakes to avoid when checking for extraneous solutions include:

• Not plugging the solution back into the original equation
• Not checking if the solution makes the equation true
• Not considering the possibility of extraneous solutions
• Not being careful when simplifying the equation

Conclusion

In this article, we answered some frequently asked questions related to solving equations and checking for extraneous solutions. We hope this article has been helpful in understanding how to solve equations and check for extraneous solutions. Remember to always check for extraneous solutions to ensure that your answers are correct.

Additional Resources

• Solving Equations
• Checking for Extraneous Solutions
• Algebraic Techniques

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