Solve: $\frac{x^2-x-6}{x^2}=\frac{x-6}{2 X}+\frac{2 X+12}{x}$After Multiplying Each Side Of The Equation By The LCD And Simplifying, The Resulting Equation Is:

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Introduction

In this article, we will be solving a complex equation involving fractions. The equation is x2βˆ’xβˆ’6x2=xβˆ’62x+2x+12x\frac{x^2-x-6}{x^2}=\frac{x-6}{2 x}+\frac{2 x+12}{x}. To solve this equation, we will first need to find the least common denominator (LCD) of the fractions on both sides of the equation. The LCD is the smallest multiple that all the denominators have in common. Once we have found the LCD, we will multiply each side of the equation by the LCD to eliminate the fractions. After that, we will simplify the resulting equation and solve for x.

Step 1: Find the Least Common Denominator (LCD)

To find the LCD, we need to find the smallest multiple that all the denominators have in common. The denominators on the left side of the equation are x2x^2 and on the right side are 2x2x and xx. The smallest multiple that x2x^2, 2x2x, and xx have in common is 2x22x^2.

Step 2: Multiply Each Side of the Equation by the LCD

Now that we have found the LCD, we can multiply each side of the equation by 2x22x^2 to eliminate the fractions. This will give us the equation:

2x2β‹…x2βˆ’xβˆ’6x2=2x2β‹…(xβˆ’62x+2x+12x)2x^2 \cdot \frac{x^2-x-6}{x^2} = 2x^2 \cdot \left(\frac{x-6}{2 x}+\frac{2 x+12}{x}\right)

Step 3: Simplify the Equation

After multiplying each side of the equation by the LCD, we can simplify the equation by canceling out the common factors. On the left side of the equation, we can cancel out the x2x^2 in the numerator and denominator, leaving us with 2x2β‹…(x2βˆ’xβˆ’6)2x^2 \cdot (x^2-x-6). On the right side of the equation, we can cancel out the xx in the numerator and denominator of the second fraction, leaving us with 2x2β‹…(xβˆ’62x+2x+12x)=2x2β‹…(xβˆ’62x+2(2x+12)x)2x^2 \cdot \left(\frac{x-6}{2 x}+\frac{2 x+12}{x}\right) = 2x^2 \cdot \left(\frac{x-6}{2 x}+\frac{2(2x+12)}{x}\right).

Step 4: Distribute the 2x22x^2 to the Terms Inside the Parentheses

Now that we have simplified the equation, we can distribute the 2x22x^2 to the terms inside the parentheses on the right side of the equation. This will give us:

2x2β‹…(x2βˆ’xβˆ’6)=2x2β‹…(xβˆ’62x+4x+24x)2x^2 \cdot (x^2-x-6) = 2x^2 \cdot \left(\frac{x-6}{2 x}+\frac{4x+24}{x}\right)

Step 5: Simplify the Right Side of the Equation

After distributing the 2x22x^2 to the terms inside the parentheses, we can simplify the right side of the equation by combining the two fractions. This will give us:

2x2β‹…(x2βˆ’xβˆ’6)=2x2β‹…(xβˆ’6+4x+242x)2x^2 \cdot (x^2-x-6) = 2x^2 \cdot \left(\frac{x-6+4x+24}{2x}\right)

Step 6: Simplify the Fraction Inside the Parentheses

Now that we have simplified the right side of the equation, we can simplify the fraction inside the parentheses by combining the terms in the numerator. This will give us:

2x2β‹…(x2βˆ’xβˆ’6)=2x2β‹…(5x+182x)2x^2 \cdot (x^2-x-6) = 2x^2 \cdot \left(\frac{5x+18}{2x}\right)

Step 7: Simplify the Right Side of the Equation

After simplifying the fraction inside the parentheses, we can simplify the right side of the equation by canceling out the common factors. This will give us:

2x2β‹…(x2βˆ’xβˆ’6)=2x2β‹…(5x+182x)=x(5x+18)2x^2 \cdot (x^2-x-6) = 2x^2 \cdot \left(\frac{5x+18}{2x}\right) = x(5x+18)

Step 8: Simplify the Left Side of the Equation

Now that we have simplified the right side of the equation, we can simplify the left side of the equation by multiplying the terms inside the parentheses. This will give us:

2x2β‹…(x2βˆ’xβˆ’6)=2x2β‹…x2βˆ’2x2β‹…xβˆ’2x2β‹…62x^2 \cdot (x^2-x-6) = 2x^2 \cdot x^2 - 2x^2 \cdot x - 2x^2 \cdot 6

Step 9: Simplify the Left Side of the Equation

After multiplying the terms inside the parentheses, we can simplify the left side of the equation by combining the like terms. This will give us:

2x2β‹…(x2βˆ’xβˆ’6)=2x4βˆ’2x3βˆ’12x22x^2 \cdot (x^2-x-6) = 2x^4 - 2x^3 - 12x^2

Step 10: Set the Two Sides of the Equation Equal to Each Other

Now that we have simplified both sides of the equation, we can set the two sides equal to each other. This will give us:

2x4βˆ’2x3βˆ’12x2=x(5x+18)2x^4 - 2x^3 - 12x^2 = x(5x+18)

Step 11: Distribute the xx to the Terms Inside the Parentheses

Now that we have set the two sides of the equation equal to each other, we can distribute the xx to the terms inside the parentheses on the right side of the equation. This will give us:

2x4βˆ’2x3βˆ’12x2=5x2+18x2x^4 - 2x^3 - 12x^2 = 5x^2 + 18x

Step 12: Subtract 5x25x^2 from Both Sides of the Equation

Now that we have distributed the xx to the terms inside the parentheses, we can subtract 5x25x^2 from both sides of the equation to get rid of the 5x25x^2 term on the right side of the equation. This will give us:

2x4βˆ’2x3βˆ’17x2=18x2x^4 - 2x^3 - 17x^2 = 18x

Step 13: Subtract 18x18x from Both Sides of the Equation

Now that we have subtracted 5x25x^2 from both sides of the equation, we can subtract 18x18x from both sides of the equation to get rid of the 18x18x term on the right side of the equation. This will give us:

2x4βˆ’2x3βˆ’17x2βˆ’18x=02x^4 - 2x^3 - 17x^2 - 18x = 0

Step 14: Factor the Left Side of the Equation

Now that we have subtracted 18x18x from both sides of the equation, we can factor the left side of the equation by grouping the terms. This will give us:

(2x4βˆ’2x3)βˆ’(17x2+18x)=0(2x^4 - 2x^3) - (17x^2 + 18x) = 0

Step 15: Factor Out the GCF from Each Group of Terms

Now that we have factored the left side of the equation, we can factor out the greatest common factor (GCF) from each group of terms. This will give us:

2x3(xβˆ’1)βˆ’17x(x+1)=02x^3(x - 1) - 17x(x + 1) = 0

Step 16: Factor Out the GCF from Each Group of Terms

After factoring out the GCF from each group of terms, we can factor out the GCF from each group of terms again. This will give us:

x(2x2(xβˆ’1)βˆ’17(x+1))=0x(2x^2(x - 1) - 17(x + 1)) = 0

Step 17: Factor the Quadratic Expression Inside the Parentheses

Now that we have factored out the GCF from each group of terms, we can factor the quadratic expression inside the parentheses. This will give us:

x(2x2(xβˆ’1)βˆ’17(x+1))=x(2x3βˆ’2x2βˆ’17xβˆ’17)x(2x^2(x - 1) - 17(x + 1)) = x(2x^3 - 2x^2 - 17x - 17)

Step 18: Factor the Quadratic Expression Inside the Parentheses

After factoring the quadratic expression inside the parentheses, we can factor the quadratic expression inside the parentheses again. This will give us:

x(2x3βˆ’2x2βˆ’17xβˆ’17)=x(2x2(xβˆ’1)βˆ’17(x+1))x(2x^3 - 2x^2 - 17x - 17) = x(2x^2(x - 1) - 17(x + 1))

Step 19: Factor the Quadratic Expression Inside the Parentheses

Now that we have factored the quadratic expression inside the parentheses, we can factor the quadratic expression inside the parentheses again. This will give us:

x(2x2(xβˆ’1)βˆ’17(x+1))=x(2x2(xβˆ’1)βˆ’17(x+1))x(2x^2(x - 1) - 17(x + 1)) = x(2x^2(x - 1) - 17(x + 1))

Step 20: Solve for x

Now that we have factored the quadratic expression inside the parentheses, we can solve for x by setting each factor equal to zero. This will give us:

x=0Β orΒ 2x2(xβˆ’1)βˆ’17(x+1)=0x = 0 \text{ or } 2x^2(x - 1) - 17(x + 1) = 0

Step 21: Solve the Quadratic Equation

Now that we have solved for x, we can solve the quadratic equation by factoring. This will give us:

2x^2(x -<br/> # Q&A: Solving the Equation $\frac{x^2-x-6}{x^2}=\frac{x-6}{2 x}+\frac{2 x+12}{x}$ ## Introduction In our previous article, we solved the equation $\frac{x^2-x-6}{x^2}=\frac{x-6}{2 x}+\frac{2 x+12}{x}$ by finding the least common denominator (LCD) and multiplying each side of the equation by the LCD. We then simplified the resulting equation and solved for x. In this article, we will answer some common questions that readers may have about solving this equation. ## Q: What is the least common denominator (LCD) of the fractions on both sides of the equation? A: The LCD of the fractions on both sides of the equation is $2x^2$. ## Q: Why do we need to find the LCD of the fractions? A: We need to find the LCD of the fractions so that we can multiply each side of the equation by the LCD and eliminate the fractions. ## Q: How do we simplify the equation after multiplying each side by the LCD? A: We simplify the equation by canceling out the common factors and combining the like terms. ## Q: What is the final simplified equation after multiplying each side by the LCD? A: The final simplified equation is $2x^4 - 2x^3 - 17x^2 - 18x = 0$. ## Q: How do we solve the equation $2x^4 - 2x^3 - 17x^2 - 18x = 0$? A: We solve the equation by factoring the left side of the equation and setting each factor equal to zero. ## Q: What are the solutions to the equation $2x^4 - 2x^3 - 17x^2 - 18x = 0$? A: The solutions to the equation are $x = 0$ and $2x^2(x - 1) - 17(x + 1) = 0$. ## Q: How do we solve the quadratic equation $2x^2(x - 1) - 17(x + 1) = 0$? A: We solve the quadratic equation by factoring and setting each factor equal to zero. ## Q: What are the solutions to the quadratic equation $2x^2(x - 1) - 17(x + 1) = 0$? A: The solutions to the quadratic equation are $x = 0$ and $x = 1$. ## Q: What is the final solution to the equation $\frac{x^2-x-6}{x^2}=\frac{x-6}{2 x}+\frac{2 x+12}{x}$? A: The final solution to the equation is $x = 0$ and $x = 1$. ## Conclusion In this article, we answered some common questions that readers may have about solving the equation $\frac{x^2-x-6}{x^2}=\frac{x-6}{2 x}+\frac{2 x+12}{x}$. We provided step-by-step solutions to the equation and explained the reasoning behind each step. We also provided the final solution to the equation, which is $x = 0$ and $x = 1$. ## Additional Resources For more information on solving equations and quadratic equations, please refer to the following resources: * [Solving Equations](https://www.mathsisfun.com/algebra/solving-equations.html) * [Quadratic Equations](https://www.mathsisfun.com/algebra/quadratic-equations.html) * [Factoring Quadratic Expressions](https://www.mathsisfun.com/algebra/factoring-quadratic-expressions.html) ## Frequently Asked Questions * Q: What is the least common denominator (LCD) of the fractions on both sides of the equation? A: The LCD of the fractions on both sides of the equation is $2x^2$. * Q: Why do we need to find the LCD of the fractions? A: We need to find the LCD of the fractions so that we can multiply each side of the equation by the LCD and eliminate the fractions. * Q: How do we simplify the equation after multiplying each side by the LCD? A: We simplify the equation by canceling out the common factors and combining the like terms. * Q: What is the final simplified equation after multiplying each side by the LCD? A: The final simplified equation is $2x^4 - 2x^3 - 17x^2 - 18x = 0$. * Q: How do we solve the equation $2x^4 - 2x^3 - 17x^2 - 18x = 0$? A: We solve the equation by factoring the left side of the equation and setting each factor equal to zero. * Q: What are the solutions to the equation $2x^4 - 2x^3 - 17x^2 - 18x = 0$? A: The solutions to the equation are $x = 0$ and $2x^2(x - 1) - 17(x + 1) = 0$. * Q: How do we solve the quadratic equation $2x^2(x - 1) - 17(x + 1) = 0$? A: We solve the quadratic equation by factoring and setting each factor equal to zero. * Q: What are the solutions to the quadratic equation $2x^2(x - 1) - 17(x + 1) = 0$? A: The solutions to the quadratic equation are $x = 0$ and $x = 1$. * Q: What is the final solution to the equation $\frac{x^2-x-6}{x^2}=\frac{x-6}{2 x}+\frac{2 x+12}{x}$? A: The final solution to the equation is $x = 0$ and $x = 1$.