Question

What is one possible solution to the following equation: \[ \frac{x+1}{x} - \frac{3}{2x^2} = \frac{-5}{2x} \]

• A. -1
• B. 0
• C. \( \frac{-7 + \sqrt{73}}{4} \)
• D. \( \frac{-7 - \sqrt{25}}{4} \)
• E. \( \frac{-1}{2} \)

Hint:

Use the quadratic formula to solve quadratic equations, and always check for multiple solutions.

Answer 1

The Correct Option is C

Multiply the entire equation by \( 2x^2 \) to clear the fractions: \[ 2x^2 \times \left( \frac{x+1}{x} - \frac{3}{2x^2} \right) = 2x^2 \times \frac{-5}{2x} \] Simplifying the terms: \[ 2x(x + 1) - 3 = -5x \] \[ 2x^2 + 2x - 3 = -5x \] \[ 2x^2 + 7x - 3 = 0 \] Now, solve this quadratic equation using the quadratic formula: \[ x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-3)}}{2(2)} \] \[ x = \frac{-7 \pm \sqrt{49 + 24}}{4} \] \[ x = \frac{-7 \pm \sqrt{73}}{4} \] Thus, one possible solution is \( \frac{-7 + \sqrt{73}}{4} \).
Final Answer: \[ \boxed{\frac{-7 + \sqrt{73}}{4}} \]