Select The Correct Answer.Jackie Correctly Solved This Equation: 4 X − 2 + 1 X 2 − 4 = 1 X + 2 \frac{4}{x-2}+\frac{1}{x^2-4}=\frac{1}{x+2} X − 2 4 ​ + X 2 − 4 1 ​ = X + 2 1 ​ Which Statement Describes The Solutions She Found?A. She Found One Valid Solution And No Extraneous Solutions. B. She Found Two Valid

Introduction

In mathematics, solving equations is a crucial skill that requires a deep understanding of algebraic concepts. When solving equations, it's essential to identify the type of equation, apply the correct methods, and check for extraneous solutions. In this article, we will analyze the equation $\frac{4}{x-2}+\frac{1}{x^2-4}=\frac{1}{x+2}$ and determine the correct statement that describes the solutions found by Jackie.

Understanding the Equation

The given equation is a rational equation, which involves fractions with variables in the denominators. To solve this equation, we need to find a common denominator and then simplify the equation. The equation can be rewritten as:

$$\frac{4}{x-2}+\frac{1}{(x-2)(x+2)}=\frac{1}{x+2}$$

Finding a Common Denominator

To add the fractions, we need to find a common denominator. The least common multiple (LCM) of $(x-2)$ and $(x-2)(x+2)$ is $(x-2)(x+2)$. Therefore, we can rewrite the equation as:

$$\frac{4(x+2)}{(x-2)(x+2)}+\frac{1}{(x-2)(x+2)}=\frac{1}{x+2}$$

Simplifying the Equation

Now that we have a common denominator, we can simplify the equation by combining the fractions:

$$\frac{4(x+2)+1}{(x-2)(x+2)}=\frac{1}{x+2}$$

Cross-Multiplying

To eliminate the fractions, we can cross-multiply:

$$(4(x+2)+1)(x+2)=1(x-2)(x+2)$$

Expanding and Simplifying

Expanding and simplifying the equation, we get:

$$4x^2+17x+4=x^2-4$$

Rearranging the Equation

Rearranging the equation, we get:

$$3x^2+21x+8=0$$

Factoring the Quadratic Equation

The quadratic equation can be factored as:

$$(3x+2)(x+4)=0$$

Solving for x

Setting each factor equal to zero, we get:

$$3x+2=0 \quad \text{or} \quad x+4=0$$

Solving for x, we get:

$$x=-\frac{2}{3} \quad \text{or} \quad x=-4$$

Checking for Extraneous Solutions

To check for extraneous solutions, we need to substitute the values of x back into the original equation. Substituting x = -2/3, we get:

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

Simplifying the equation, we get:

$$\frac{4}{-\frac{8}{3}}+\frac{1}{(-\frac{8}{3})(\frac{2}{3})}=\frac{1}{\frac{4}{3}}$$

$$-\frac{3}{2}+\frac{3}{16}=\frac{3}{4}$$

$$-\frac{24}{16}+\frac{3}{16}=\frac{3}{4}$$

$$-\frac{21}{16}=\frac{3}{4}$$

This is a false statement, so x = -2/3 is an extraneous solution.

Substituting x = -4, we get:

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

Simplifying the equation, we get:

$$\frac{4}{-6}+\frac{1}{(-6)(-2)}=\frac{1}{-2}$$

$$-\frac{2}{3}+\frac{1}{12}=-\frac{1}{2}$$

$$-\frac{8}{12}+\frac{1}{12}=-\frac{1}{2}$$

$$-\frac{7}{12}=-\frac{1}{2}$$

This is a false statement, so x = -4 is an extraneous solution.

Conclusion

In conclusion, Jackie found two solutions, but neither of them is valid. Therefore, the correct statement that describes the solutions found by Jackie is:

A. She found one valid solution and no extraneous solutions.

This statement is incorrect because Jackie found two solutions, but neither of them is valid.

Introduction

In mathematics, solving equations is a crucial skill that requires a deep understanding of algebraic concepts. When solving equations, it's essential to identify the type of equation, apply the correct methods, and check for extraneous solutions. In this article, we will analyze the equation $\frac{4}{x-2}+\frac{1}{x^2-4}=\frac{1}{x+2}$ and determine the correct statement that describes the solutions found by Jackie.

Understanding the Equation

The given equation is a rational equation, which involves fractions with variables in the denominators. To solve this equation, we need to find a common denominator and then simplify the equation. The equation can be rewritten as:

$$\frac{4}{x-2}+\frac{1}{(x-2)(x+2)}=\frac{1}{x+2}$$

Finding a Common Denominator

To add the fractions, we need to find a common denominator. The least common multiple (LCM) of $(x-2)$ and $(x-2)(x+2)$ is $(x-2)(x+2)$. Therefore, we can rewrite the equation as:

$$\frac{4(x+2)}{(x-2)(x+2)}+\frac{1}{(x-2)(x+2)}=\frac{1}{x+2}$$

Simplifying the Equation

Now that we have a common denominator, we can simplify the equation by combining the fractions:

$$\frac{4(x+2)+1}{(x-2)(x+2)}=\frac{1}{x+2}$$

Cross-Multiplying

To eliminate the fractions, we can cross-multiply:

$$(4(x+2)+1)(x+2)=1(x-2)(x+2)$$

Expanding and Simplifying

Expanding and simplifying the equation, we get:

$$4x^2+17x+4=x^2-4$$

Rearranging the Equation

Rearranging the equation, we get:

$$3x^2+21x+8=0$$

Factoring the Quadratic Equation

The quadratic equation can be factored as:

$$(3x+2)(x+4)=0$$

Solving for x

Setting each factor equal to zero, we get:

$$3x+2=0 \quad \text{or} \quad x+4=0$$

Solving for x, we get:

$$x=-\frac{2}{3} \quad \text{or} \quad x=-4$$

Checking for Extraneous Solutions

To check for extraneous solutions, we need to substitute the values of x back into the original equation. Substituting x = -2/3, we get:

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

Simplifying the equation, we get:

$$\frac{4}{-\frac{8}{3}}+\frac{1}{(-\frac{8}{3})(\frac{2}{3})}=\frac{1}{\frac{4}{3}}$$

$$-\frac{3}{2}+\frac{3}{16}=\frac{3}{4}$$

$$-\frac{24}{16}+\frac{3}{16}=\frac{3}{4}$$

$$-\frac{21}{16}=\frac{3}{4}$$

This is a false statement, so x = -2/3 is an extraneous solution.

Substituting x = -4, we get:

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

Simplifying the equation, we get:

$$\frac{4}{-6}+\frac{1}{(-6)(-2)}=\frac{1}{-2}$$

$$-\frac{2}{3}+\frac{1}{12}=-\frac{1}{2}$$

$$-\frac{8}{12}+\frac{1}{12}=-\frac{1}{2}$$

$$-\frac{7}{12}=-\frac{1}{2}$$

This is a false statement, so x = -4 is an extraneous solution.

Conclusion

In conclusion, Jackie found two solutions, but neither of them is valid. Therefore, the correct statement that describes the solutions found by Jackie is:

A. She found one valid solution and no extraneous solutions.

This statement is incorrect because Jackie found two solutions, but neither of them is valid.

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Q: What type of equation is the given equation?

A: The given equation is a rational equation, which involves fractions with variables in the denominators.

Q: How do we find a common denominator for the fractions?

A: To find a common denominator, we need to find the least common multiple (LCM) of the denominators. In this case, the LCM of $(x-2)$ and $(x-2)(x+2)$ is $(x-2)(x+2)$.

Q: How do we simplify the equation after finding a common denominator?

A: After finding a common denominator, we can simplify the equation by combining the fractions.

Q: What is the first step in solving the equation?

A: The first step in solving the equation is to find a common denominator and then simplify the equation.

Q: How do we check for extraneous solutions?

A: To check for extraneous solutions, we need to substitute the values of x back into the original equation.

Q: What is the final answer to the equation?

A: The final answer to the equation is that Jackie found two solutions, but neither of them is valid.

Q: What is the correct statement that describes the solutions found by Jackie?

A: The correct statement that describes the solutions found by Jackie is:

A. She found one valid solution and no extraneous solutions.

This statement is incorrect because Jackie found two solutions, but neither of them is valid.

Q: What is the main concept of the article?

A: The main concept of the article is solving rational equations and checking for extraneous solutions.

Q: What is the importance of checking for extraneous solutions?

A: Checking for extraneous solutions is important because it ensures that the solutions found are valid and not false.

Q: What is the final conclusion of the article?

A: The final conclusion of the article is that Jackie found two solutions, but neither of them is valid.