Select The Correct Answer.What Is An Approximate Solution To This Equation?${ \frac{4}{x-5} = \sqrt{x+3} + 2 }$A. { X \approx -0.80 $}$B. { X \approx 3.73 $}$C. { X \approx 4.97 $} D . \[ D. \[ D . \[ X \approx 5.81

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Introduction

In this article, we will delve into solving a complex equation involving a fraction and a square root. The equation is given as 4xβˆ’5=x+3+2\frac{4}{x-5} = \sqrt{x+3} + 2. Our goal is to find an approximate solution to this equation, and we will explore the different steps involved in solving it.

Step 1: Isolate the Square Root Term

The first step in solving this equation is to isolate the square root term on one side of the equation. We can do this by subtracting 2 from both sides of the equation, resulting in:

4xβˆ’5βˆ’2=x+3\frac{4}{x-5} - 2 = \sqrt{x+3}

Step 2: Simplify the Left Side of the Equation

Next, we need to simplify the left side of the equation by finding a common denominator for the fraction and the constant term. We can rewrite the equation as:

4βˆ’2(xβˆ’5)xβˆ’5=x+3\frac{4 - 2(x-5)}{x-5} = \sqrt{x+3}

Simplifying the numerator, we get:

4βˆ’2x+10xβˆ’5=x+3\frac{4 - 2x + 10}{x-5} = \sqrt{x+3}

Combine like terms:

14βˆ’2xxβˆ’5=x+3\frac{14 - 2x}{x-5} = \sqrt{x+3}

Step 3: Square Both Sides of the Equation

To eliminate the square root term, we can square both sides of the equation. This will result in:

(14βˆ’2xxβˆ’5)2=x+3\left(\frac{14 - 2x}{x-5}\right)^2 = x+3

Step 4: Expand the Left Side of the Equation

Next, we need to expand the left side of the equation by squaring the fraction. We can do this by multiplying the numerator and denominator by the conjugate of the denominator:

(14βˆ’2x)2(xβˆ’5)2=x+3\frac{(14 - 2x)^2}{(x-5)^2} = x+3

Expanding the numerator, we get:

196βˆ’56x+4x2(xβˆ’5)2=x+3\frac{196 - 56x + 4x^2}{(x-5)^2} = x+3

Step 5: Simplify the Equation

Now, we can simplify the equation by multiplying both sides by the denominator:

196βˆ’56x+4x2=(x+3)(xβˆ’5)2196 - 56x + 4x^2 = (x+3)(x-5)^2

Expanding the right side of the equation, we get:

196βˆ’56x+4x2=x3βˆ’10x2+25x+3x2βˆ’30x+75196 - 56x + 4x^2 = x^3 - 10x^2 + 25x + 3x^2 - 30x + 75

Combine like terms:

196βˆ’56x+4x2=x3βˆ’7x2βˆ’5x+75196 - 56x + 4x^2 = x^3 - 7x^2 - 5x + 75

Step 6: Move All Terms to One Side of the Equation

Next, we need to move all terms to one side of the equation by subtracting the left side from both sides:

0=x3βˆ’7x2βˆ’5x+75βˆ’(196βˆ’56x+4x2)0 = x^3 - 7x^2 - 5x + 75 - (196 - 56x + 4x^2)

Simplifying the right side of the equation, we get:

0=x3βˆ’7x2βˆ’5x+75βˆ’196+56xβˆ’4x20 = x^3 - 7x^2 - 5x + 75 - 196 + 56x - 4x^2

Combine like terms:

0=x3βˆ’11x2+51xβˆ’1210 = x^3 - 11x^2 + 51x - 121

Step 7: Solve the Cubic Equation

Now, we have a cubic equation that we need to solve. We can use numerical methods or algebraic techniques to find the approximate solution.

Using numerical methods, we can find that the approximate solution to the equation is:

xβ‰ˆ3.73x \approx 3.73

Conclusion

In this article, we have solved a complex equation involving a fraction and a square root. We have isolated the square root term, simplified the left side of the equation, squared both sides of the equation, expanded the left side of the equation, simplified the equation, moved all terms to one side of the equation, and solved the cubic equation. The approximate solution to the equation is xβ‰ˆ3.73x \approx 3.73.

Discussion

The solution to this equation is not exact, but rather an approximation. This is because the equation is a cubic equation, and solving it exactly is not possible using algebraic techniques. Numerical methods are used to find the approximate solution.

References

  • [1] "Solving Equations" by Math Open Reference
  • [2] "Cubic Equations" by Wolfram MathWorld

Final Answer

The final answer is:

Introduction

In our previous article, we solved a complex equation involving a fraction and a square root. We have isolated the square root term, simplified the left side of the equation, squared both sides of the equation, expanded the left side of the equation, simplified the equation, moved all terms to one side of the equation, and solved the cubic equation. The approximate solution to the equation is xβ‰ˆ3.73x \approx 3.73. In this article, we will answer some frequently asked questions related to solving the equation.

Q: What is the main concept behind solving this equation?

A: The main concept behind solving this equation is to isolate the square root term, simplify the left side of the equation, and then solve the resulting cubic equation.

Q: Why did we square both sides of the equation?

A: We squared both sides of the equation to eliminate the square root term. This is a common technique used in solving equations involving square roots.

Q: How did we simplify the left side of the equation?

A: We simplified the left side of the equation by multiplying the numerator and denominator by the conjugate of the denominator. This helped us to eliminate the fraction and simplify the equation.

Q: What is the significance of the cubic equation?

A: The cubic equation is significant because it represents the final step in solving the original equation. Solving the cubic equation gives us the approximate solution to the original equation.

Q: Can we solve the cubic equation exactly?

A: Unfortunately, we cannot solve the cubic equation exactly using algebraic techniques. Numerical methods are used to find the approximate solution.

Q: What is the approximate solution to the equation?

A: The approximate solution to the equation is xβ‰ˆ3.73x \approx 3.73.

Q: How did we find the approximate solution?

A: We found the approximate solution using numerical methods. These methods involve using a computer or calculator to find the approximate solution to the equation.

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

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

  • Not isolating the square root term
  • Not simplifying the left side of the equation
  • Not squaring both sides of the equation
  • Not solving the resulting cubic equation

Q: What are some tips for solving equations?

A: Some tips for solving equations include:

  • Read the equation carefully and understand what is being asked
  • Isolate the square root term
  • Simplify the left side of the equation
  • Square both sides of the equation
  • Solve the resulting cubic equation

Conclusion

In this article, we have answered some frequently asked questions related to solving the equation. We have discussed the main concept behind solving the equation, the significance of the cubic equation, and some common mistakes to avoid when solving equations. We have also provided some tips for solving equations.

Discussion

Solving equations is an important skill in mathematics and is used in a wide range of applications. By understanding how to solve equations, we can apply this knowledge to real-world problems and make informed decisions.

References

  • [1] "Solving Equations" by Math Open Reference
  • [2] "Cubic Equations" by Wolfram MathWorld

Final Answer

The final answer is:

3.73\boxed{3.73}