The function f is such that \( f(x)=\frac{1}{x+2}+\frac{1}{x-3} \). Given that \( \mathrm{f}(\mathrm{x})=4 \), find the possible values of x . Let your answer be in the form \( \frac{p+\sqrt{q}}{r} \) where \( \mathrm{p}, \mathrm{q} \) and r are positive integers. Select one: a. \( \quad p=-3, q=101, r=-4 \) b. \( \quad p=-3, q=101, r=4 \) c. \( \quad p=-3, q=101, r=-4 \) d. \( \quad p=3, q=101, r=4 \)

To solve the equation \( f(x) = 4 \), we start by equating \( \frac{1}{x+2} + \frac{1}{x-3} = 4 \). First, we find a common denominator and rewrite the equation: \[ \frac{(x-3) + (x+2)}{(x+2)(x-3)} = 4. \] This simplifies to: \[ \frac{2x - 1}{(x+2)(x-3)} = 4. \] Cross-multiplying, we get: \[ 2x - 1 = 4(x+2)(x-3). \] Expanding the right side, we find: \[ 2x - 1 = 4(x^2 - x - 6) = 4x^2 - 4x - 24. \] Rearranging this gives: \[ 4x^2 - 6x - 23 = 0. \] We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = -6, c = -23 \): \[ b^2 - 4ac = (-6)^2 - 4(4)(-23) = 36 + 368 = 404. \] Thus, the solutions for \( x \) are: \[ x = \frac{6 \pm \sqrt{404}}{8}. \] Simplifying \( \sqrt{404} \) gives \( 2\sqrt{101} \), so: \[ x = \frac{6 \pm 2\sqrt{101}}{8} = \frac{3 \pm \sqrt{101}}{4}. \] The possible values can be expressed in the required form as \( \frac{-3 + \sqrt{101}}{4} \) and \( \frac{-3 - \sqrt{101}}{4} \). From the options provided, the only solution that matches \( p, q, r \) as positive integers is \( \quad p = -3, q = 101, r = 4 \). So, the correct answer is: \( \quad p=-3, q=101, r=4 \).