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

Feb 26, 2025 by ADMIN 214 views

$What is true about the solution of $\frac{x^2}{2x-6} = \frac{9}{6x-18}$?$

Introduction

When solving equations involving fractions, it's essential to be cautious and methodical in our approach. The given equation, $\frac{x^2}{2x-6} = \frac{9}{6x-18}$ , is a rational equation that requires careful manipulation to isolate the variable. In this article, we will delve into the solution of this equation and determine the validity of the given options.

Step 1: Simplify the Equation

The first step in solving the equation is to simplify it by eliminating the fractions. We can do this by multiplying both sides of the equation by the least common multiple (LCM) of the denominators, which is $(2x-6)(6x-18)$ .

\frac{x^2}{2x-6} = \frac{9}{6x-18}

Multiplying both sides by $(2x-6)(6x-18)$ , we get:

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

Solving for x, we get:

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

Solving the first equation, we get:

x = \frac{3}{2}

Solving the second equation, we get:

x = \frac{3 \pm \sqrt{3^2-4(3)(18)}}{2(3)}

Simplifying further, we get:

x = \frac{3 \pm \sqrt{-171}}{6}

Conclusion

The solutions to the equation are $x = \frac{3}{2}$ and $x = \frac{3 \pm \sqrt{-171}}{6}$ . However, we need to check if these solutions are valid by plugging them back into the original equation.

Checking the Solutions

Plugging $x = \frac{3}{2}$ back into the original equation, we get:

\frac{\left(\frac{3}{2}\right)^2}{2\left(\frac{3}{2}\right)-6} = \frac{9}{6\left(\frac{3}{2}\right)-18}

Simplifying, we get:

\frac{\frac{9}{4}}{0} = \frac{9}{0}

This is undefined, so $x = \frac{3}{2}$ is not a valid solution.

Plugging $x = \frac{3 \pm \sqrt{-171}}{6}$ back into the original equation, we get:

\frac{\left(\frac{3 \pm \sqrt{-171}}{6}\right)^2}{2\left(\frac{3 \pm \sqrt{-171}}{6}\right)-6} = \frac{9}{6\left(\frac{3 \pm \sqrt{-171}}{6}\right)-18}

Simplifying, we get:

\frac{\frac{9 \pm 9\sqrt{-1}}{36}}{0} = \frac{9}{0}

This is undefined, so $x = \frac{3 \pm \sqrt{-171}}{6}$ is not a valid solution.

Final Conclusion

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

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

Discussion

In this article, we solved the equation $\frac{x^2}{2x-6} = \frac{9}{6x-18}$ and determined that the only valid solution is $x = \pm \sqrt{3}$ , but they are extraneous solutions. This highlights the importance of checking the solutions by plugging them back into the original equation to ensure their validity.

References

[1] "Algebra" by Michael Artin
[2] "Calculus" by Michael Spivak

Keywords

Rational equation
Simplifying equations
Factoring equations
Solving equations
Checking solutions
Validity of solutions
Extraneous solutions

Introduction

In our previous article, we solved the rational equation $\frac{x^2}{2x-6} = \frac{9}{6x-18}$ and determined that the only valid solution is $x = \pm \sqrt{3}$ , but they are extraneous solutions. In this article, we will answer some frequently asked questions related to rational equations and solutions.

Q1: What is a rational equation?

A rational equation is an equation that contains one or more fractions, where the numerator and denominator are polynomials.

Q2: How do I simplify a rational equation?

To simplify a rational equation, you can multiply both sides of the equation by the least common multiple (LCM) of the denominators.

Q3: What is the least common multiple (LCM) of the denominators?

The LCM of the denominators is the smallest polynomial that is divisible by all the denominators.

Q4: How do I factor a rational equation?

To factor a rational equation, you can look for common factors in the numerator and denominator, and then factor out the greatest common factor (GCF).

Q5: What is an extraneous solution?

An extraneous solution is a solution that is not valid or is not a solution to the original equation.

Q6: How do I check if a solution is valid?

To check if a solution is valid, you can plug it back into the original equation and see if it satisfies the equation.

Q7: What is the difference between a rational equation and a polynomial equation?

A rational equation is an equation that contains one or more fractions, while a polynomial equation is an equation that contains only polynomials.

Q8: Can I use the quadratic formula to solve a rational equation?

No, the quadratic formula is used to solve quadratic equations, not rational equations.

Q9: How do I solve a rational equation with a variable in the denominator?

To solve a rational equation with a variable in the denominator, you can multiply both sides of the equation by the conjugate of the denominator.

Q10: What is the conjugate of a denominator?

The conjugate of a denominator is the same expression with the opposite sign.

Q&A Examples

Example 1

Solve the rational equation $\frac{x}{x-1} = \frac{2}{x+1}$ .

Solution

To solve the equation, we can multiply both sides by the LCM of the denominators, which is $(x-1)(x+1)$ .

\frac{x}{x-1} = \frac{2}{x+1}

Multiplying both sides by $(x-1)(x+1)$ , we get:

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

Expanding and simplifying, we get:

x^2 + x = 2x - 2

Rearranging the equation, we get:

x^2 - x + 2 = 0

Factoring the equation, we get:

(x - 1)^2 + 2 = 0

This equation has no real solutions.

Example 2

Solve the rational equation $\frac{x^2}{x-1} = \frac{4}{x+1}$ .