Solve: 2 + X 2 − X + 2 − X 2 + X = 4 1 4 \frac{2+x}{2-x}+\frac{2-x}{2+x}=4 \frac{1}{4} 2 − X 2 + X ​ + 2 + X 2 − X ​ = 4 4 1 ​

by ADMIN 127 views

Introduction

Mathematical equations involving fractions can be challenging to solve, especially when they involve complex expressions. In this article, we will focus on solving the equation 2+x2−x+2−x2+x=414\frac{2+x}{2-x}+\frac{2-x}{2+x}=4 \frac{1}{4}, which involves two fractions with different denominators. We will use algebraic techniques to simplify the equation and find the solution.

Understanding the Equation

The given equation is 2+x2−x+2−x2+x=414\frac{2+x}{2-x}+\frac{2-x}{2+x}=4 \frac{1}{4}. To begin solving this equation, we need to understand the concept of equivalent fractions. Equivalent fractions are fractions that have the same value, but with different numerators and denominators. In this case, we can rewrite the fractions as follows:

2+x2−x=(2+x)(2+x)(2−x)(2+x)=4+x2+2x4−x2\frac{2+x}{2-x} = \frac{(2+x)(2+x)}{(2-x)(2+x)} = \frac{4+x^2+2x}{4-x^2}

2−x2+x=(2−x)(2+x)(2+x)(2−x)=4−x24+x2\frac{2-x}{2+x} = \frac{(2-x)(2+x)}{(2+x)(2-x)} = \frac{4-x^2}{4+x^2}

Simplifying the Equation

Now that we have rewritten the fractions, we can substitute them back into the original equation:

4+x2+2x4−x2+4−x24+x2=414\frac{4+x^2+2x}{4-x^2} + \frac{4-x^2}{4+x^2} = 4 \frac{1}{4}

To simplify the equation further, we can multiply both sides by the least common multiple (LCM) of the denominators, which is (4−x2)(4+x2)(4-x^2)(4+x^2). This will eliminate the fractions and make it easier to solve the equation.

(4+x2+2x)(4+x2)+(4−x2)(4−x2)=(414)(4−x2)(4+x2)(4+x^2+2x)(4+x^2) + (4-x^2)(4-x^2) = (4 \frac{1}{4})(4-x^2)(4+x^2)

Expanding and Simplifying

Now that we have multiplied both sides by the LCM, we can expand and simplify the equation:

16+8x2+4x4+8x+16+x4−8x2+16−x4=(414)(4−x2)(4+x2)16 + 8x^2 + 4x^4 + 8x + 16 + x^4 - 8x^2 + 16 - x^4 = (4 \frac{1}{4})(4-x^2)(4+x^2)

Combine like terms:

48+4x4+8x=(414)(4−x2)(4+x2)48 + 4x^4 + 8x = (4 \frac{1}{4})(4-x^2)(4+x^2)

Solving for x

Now that we have simplified the equation, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by subtracting 48 from both sides:

4x4+8x=(414)(4−x2)(4+x2)−484x^4 + 8x = (4 \frac{1}{4})(4-x^2)(4+x^2) - 48

Next, we can simplify the right-hand side of the equation:

4x4+8x=(1714)(4−x2)(4+x2)−484x^4 + 8x = (17 \frac{1}{4})(4-x^2)(4+x^2) - 48

Factoring the Quadratic

Now that we have simplified the equation, we can factor the quadratic expression on the right-hand side:

(4−x2)(4+x2)=16−x4(4-x^2)(4+x^2) = 16 - x^4

Substitute this expression back into the equation:

4x4+8x=(1714)(16−x4)−484x^4 + 8x = (17 \frac{1}{4})(16 - x^4) - 48

Simplifying the Equation

Now that we have factored the quadratic expression, we can simplify the equation further:

4x4+8x=(1714)(16)−(1714)x4−484x^4 + 8x = (17 \frac{1}{4})(16) - (17 \frac{1}{4})x^4 - 48

Combine like terms:

4x4+8x=272−(1714)x4−484x^4 + 8x = 272 - (17 \frac{1}{4})x^4 - 48

Solving for x

Now that we have simplified the equation, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by adding (1714)x4(17 \frac{1}{4})x^4 to both sides:

4x4+(1714)x4+8x=272−484x^4 + (17 \frac{1}{4})x^4 + 8x = 272 - 48

Combine like terms:

(1714)x4+4x4+8x=224(17 \frac{1}{4})x^4 + 4x^4 + 8x = 224

Factoring the Quadratic

Now that we have simplified the equation, we can factor the quadratic expression:

(1714)x4+4x4=(2112)x4(17 \frac{1}{4})x^4 + 4x^4 = (21 \frac{1}{2})x^4

Substitute this expression back into the equation:

(2112)x4+8x=224(21 \frac{1}{2})x^4 + 8x = 224

Solving for x

Now that we have factored the quadratic expression, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by subtracting 8x from both sides:

(2112)x4=224−8x(21 \frac{1}{2})x^4 = 224 - 8x

Next, we can divide both sides by (2112)(21 \frac{1}{2}):

x4=224−8x2112x^4 = \frac{224 - 8x}{21 \frac{1}{2}}

Simplifying the Equation

Now that we have simplified the equation, we can simplify the right-hand side:

x4=2242112−8x2112x^4 = \frac{224}{21 \frac{1}{2}} - \frac{8x}{21 \frac{1}{2}}

x4=44843−8x2112x^4 = \frac{448}{43} - \frac{8x}{21 \frac{1}{2}}

Solving for x

Now that we have simplified the equation, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by adding 8x2112\frac{8x}{21 \frac{1}{2}} to both sides:

x4+8x2112=44843x^4 + \frac{8x}{21 \frac{1}{2}} = \frac{448}{43}

Next, we can multiply both sides by 43:

43x4+344x2112=44843x^4 + \frac{344x}{21 \frac{1}{2}} = 448

Simplifying the Equation

Now that we have simplified the equation, we can simplify the right-hand side:

43x4+344x2112=44843x^4 + \frac{344x}{21 \frac{1}{2}} = 448

x4+8x2112=44843x^4 + \frac{8x}{21 \frac{1}{2}} = \frac{448}{43}

Solving for x

Now that we have simplified the equation, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by multiplying both sides by 21 \frac{1}{2}:

2112x4+8x=44843×211221 \frac{1}{2}x^4 + 8x = \frac{448}{43} \times 21 \frac{1}{2}

Next, we can simplify the right-hand side:

2112x4+8x=22421 \frac{1}{2}x^4 + 8x = 224

Factoring the Quadratic

Now that we have simplified the equation, we can factor the quadratic expression:

2112x4+8x=22421 \frac{1}{2}x^4 + 8x = 224

x4+8x2112=2242112x^4 + \frac{8x}{21 \frac{1}{2}} = \frac{224}{21 \frac{1}{2}}

Solving for x

Now that we have factored the quadratic expression, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by multiplying both sides by 21 \frac{1}{2}:

2112x4+8x=22421 \frac{1}{2}x^4 + 8x = 224

x4+8x2112=2242112x^4 + \frac{8x}{21 \frac{1}{2}} = \frac{224}{21 \frac{1}{2}}

Solving the Equation

Now that we have simplified the equation, we can solve for x. To do this, we need to isolate x on one side of the equation. We can start by multiplying both sides by 21 \frac{1}{2}:

2112x4+8x=22421 \frac{1}{2}x^4 + 8x = 224

x4+8x2112=2242112x^4 + \frac{8x}{21 \frac{1}{2}} = \frac{224}{21 \frac{1}{2}}

Final Solution

After simplifying the equation and solving for x, we get:

x=±2242112−821124x = \pm \sqrt[4]{\frac{224}{21 \frac{1}{2}} - \frac{8}{21 \frac{1}{2}}}

x=±224−821124x = \pm \sqrt[4]{\frac{224 - 8}{21 \frac{1}{2}}}

x=±21621124x = \pm \sqrt[4]{\frac{216}{21 \frac{1}{2}}}

x=±216434x = \pm \sqrt[4]{\frac{216}{43}}

x=±216434x = \pm \sqrt[4]{\frac{216}{43}}

Conclusion

In this article, we have solved the equation $\frac{2+x}{2-x}+\frac{2-x}{2+x}=4 \frac

Introduction

In our previous article, we solved the equation 2+x2−x+2−x2+x=414\frac{2+x}{2-x}+\frac{2-x}{2+x}=4 \frac{1}{4}. In this article, we will answer some of the most frequently asked questions about the solution to this equation.

Q: What is the main concept used to solve this equation?

A: The main concept used to solve this equation is algebraic manipulation, specifically the use of equivalent fractions and factoring quadratic expressions.

Q: Why is it necessary to multiply both sides of the equation by the least common multiple (LCM) of the denominators?

A: It is necessary to multiply both sides of the equation by the LCM of the denominators to eliminate the fractions and make it easier to solve the equation.

Q: How do you simplify the equation after multiplying both sides by the LCM?

A: After multiplying both sides by the LCM, we can simplify the equation by combining like terms and factoring quadratic expressions.

Q: What is the final solution to the equation?

A: The final solution to the equation is x=±216434x = \pm \sqrt[4]{\frac{216}{43}}.

Q: What is the significance of the solution to this equation?

A: The solution to this equation is significant because it demonstrates the use of algebraic manipulation to solve a complex equation involving fractions.

Q: Can you provide more examples of equations that can be solved using this method?

A: Yes, there are many examples of equations that can be solved using this method. Some examples include:

  • x+2x−2+x−2x+2=4\frac{x+2}{x-2} + \frac{x-2}{x+2} = 4
  • 2x+1x−1+x−12x+1=3\frac{2x+1}{x-1} + \frac{x-1}{2x+1} = 3
  • x+3x−3+x−3x+3=2\frac{x+3}{x-3} + \frac{x-3}{x+3} = 2

Q: How do you determine the correct solution to an equation involving fractions?

A: To determine the correct solution to an equation involving fractions, you need to follow the order of operations and simplify the equation step by step. You also need to check your solution by plugging it back into the original equation.

Q: What are some common mistakes to avoid when solving equations involving fractions?

A: Some common mistakes to avoid when solving equations involving fractions include:

  • Not multiplying both sides of the equation by the LCM of the denominators
  • Not simplifying the equation after multiplying both sides by the LCM
  • Not checking the solution by plugging it back into the original equation

Q: Can you provide more tips for solving equations involving fractions?

A: Yes, here are some additional tips for solving equations involving fractions:

  • Make sure to simplify the equation as much as possible before solving for the variable
  • Use equivalent fractions to make the equation easier to solve
  • Check your solution by plugging it back into the original equation
  • Use a calculator to check your solution if necessary

Conclusion

In this article, we have answered some of the most frequently asked questions about solving the equation 2+x2−x+2−x2+x=414\frac{2+x}{2-x}+\frac{2-x}{2+x}=4 \frac{1}{4}. We hope that this article has provided you with a better understanding of how to solve equations involving fractions and has given you the confidence to tackle more complex equations.