What Is The Sum Of All The Solutions Of The Equation $\frac{4}{x}+\frac{4}{2x+3}=9$?A. $-\frac{5}{6}$ B. $\frac{1}{3}$ C. $\frac{1}{2}$ D. $\frac{11}{6}$

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Introduction

In this article, we will delve into the world of algebra and explore the solution to a complex equation involving fractions. The equation in question is 4x+42x+3=9\frac{4}{x}+\frac{4}{2x+3}=9, and we are tasked with finding the sum of all its solutions. This problem requires a combination of algebraic manipulation and careful analysis to arrive at the correct solution.

Understanding the Equation

The given equation is 4x+42x+3=9\frac{4}{x}+\frac{4}{2x+3}=9. To begin solving this equation, we need to clear the fractions by multiplying both sides by the least common multiple (LCM) of the denominators. In this case, the LCM is x(2x+3)x(2x+3).

Multiplying Both Sides by the LCM

By multiplying both sides of the equation by x(2x+3)x(2x+3), we get:

4(2x+3)+4x=9x(2x+3)4(2x+3)+4x=9x(2x+3)

Expanding and Simplifying

Expanding the left-hand side of the equation, we get:

8x+12+4x=18x2+27x8x+12+4x=18x^2+27x

Simplifying the equation further, we get:

12x+12=18x2+27x12x+12=18x^2+27x

Rearranging the Equation

Rearranging the equation to form a quadratic equation, we get:

18x2+15x+12=018x^2+15x+12=0

Factoring the Quadratic Equation

Unfortunately, this quadratic equation does not factor easily. Therefore, we will use the quadratic formula to find the solutions.

Using the Quadratic Formula

The quadratic formula is given by:

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

In this case, a=18a=18, b=15b=15, and c=12c=12. Plugging these values into the quadratic formula, we get:

x=−15±152−4(18)(12)2(18)x=\frac{-15\pm\sqrt{15^2-4(18)(12)}}{2(18)}

Simplifying the Quadratic Formula

Simplifying the expression under the square root, we get:

x=−15±225−86436x=\frac{-15\pm\sqrt{225-864}}{36}

x=−15±−63936x=\frac{-15\pm\sqrt{-639}}{36}

Complex Solutions

Since the expression under the square root is negative, the solutions will be complex numbers. The solutions are given by:

x=−15±i63936x=\frac{-15\pm i\sqrt{639}}{36}

Sum of the Solutions

To find the sum of the solutions, we add the two complex solutions together:

Sum=−15+i63936+−15−i63936\text{Sum}=\frac{-15+i\sqrt{639}}{36}+\frac{-15-i\sqrt{639}}{36}

Simplifying the Sum

Simplifying the sum, we get:

Sum=−3036\text{Sum}=\frac{-30}{36}

Sum=−56\text{Sum}=-\frac{5}{6}

Conclusion

In this article, we have solved the equation 4x+42x+3=9\frac{4}{x}+\frac{4}{2x+3}=9 and found the sum of its solutions. The sum of the solutions is −56-\frac{5}{6}.

Final Answer

The final answer is −56-\frac{5}{6}.

Introduction

In our previous article, we explored the solution to the complex equation 4x+42x+3=9\frac{4}{x}+\frac{4}{2x+3}=9 and found the sum of its solutions to be −56-\frac{5}{6}. In this article, we will address some of the most frequently asked questions related to this problem.

Q: What is the least common multiple (LCM) of the denominators in the equation?

A: The LCM of the denominators xx and 2x+32x+3 is x(2x+3)x(2x+3).

Q: Why did we multiply both sides of the equation by the LCM?

A: We multiplied both sides of the equation by the LCM to clear the fractions and simplify the equation.

Q: Can you explain the quadratic formula and how it is used to find the solutions?

A: The quadratic formula is given by:

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

In this case, a=18a=18, b=15b=15, and c=12c=12. We used the quadratic formula to find the solutions to the quadratic equation 18x2+15x+12=018x^2+15x+12=0.

Q: Why did the solutions turn out to be complex numbers?

A: The solutions turned out to be complex numbers because the expression under the square root in the quadratic formula was negative.

Q: Can you simplify the sum of the solutions?

A: Yes, the sum of the solutions can be simplified as follows:

Sum=−15+i63936+−15−i63936\text{Sum}=\frac{-15+i\sqrt{639}}{36}+\frac{-15-i\sqrt{639}}{36}

Sum=−3036\text{Sum}=\frac{-30}{36}

Sum=−56\text{Sum}=-\frac{5}{6}

Q: What is the final answer to the problem?

A: The final answer to the problem is −56-\frac{5}{6}.

Q: Can you provide a step-by-step solution to the problem?

A: Yes, here is a step-by-step solution to the problem:

  1. Multiply both sides of the equation by the LCM x(2x+3)x(2x+3).
  2. Expand and simplify the equation.
  3. Rearrange the equation to form a quadratic equation.
  4. Use the quadratic formula to find the solutions.
  5. Simplify the sum of the solutions.

Q: What is the significance of the sum of the solutions?

A: The sum of the solutions is an important concept in algebra and is used to find the total value of all the solutions to an equation.

Q: Can you provide more examples of problems that involve finding the sum of solutions?

A: Yes, here are a few examples of problems that involve finding the sum of solutions:

  • Find the sum of the solutions to the equation x2+5x+6=0x^2+5x+6=0.
  • Find the sum of the solutions to the equation x2−7x+12=0x^2-7x+12=0.
  • Find the sum of the solutions to the equation x2+2x−15=0x^2+2x-15=0.

Conclusion

In this article, we have addressed some of the most frequently asked questions related to the problem of finding the sum of all solutions of the equation 4x+42x+3=9\frac{4}{x}+\frac{4}{2x+3}=9. We hope that this article has provided a clear and concise explanation of the problem and its solution.