Solve the equation $\frac{1}{4x} = \frac{2}{x^2} + \frac{3}{x-3}$.

1. Problem Description

Solve the equation $\frac{1}{4x} = \frac{2}{x^2} + \frac{3}{x-3}$.

2. Solution Steps

First, we note that $x$ cannot be $0$ or $3$ because those values would make the denominators zero.

Multiply both sides of the equation by $4x^2(x-3)$ to clear the fractions.

$$4x^2(x-3) \cdot \frac{1}{4x} = 4x^2(x-3) \cdot \frac{2}{x^2} + 4x^2(x-3) \cdot \frac{3}{x-3}$$

$$x(x-3) = 8(x-3) + 12x^2$$

$$x^2 - 3x = 8x - 24 + 12x^2$$

$$0 = 11x^2 + 11x - 24$$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a=11$, $b=11$, and $c=-24$. We can solve for $x$ using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

$$x = \frac{-11 \pm \sqrt{11^2 - 4(11)(-24)}}{2(11)}$$

$$x = \frac{-11 \pm \sqrt{121 + 1056}}{22}$$

$$x = \frac{-11 \pm \sqrt{1177}}{22}$$

$$x = \frac{-11 \pm \sqrt{1177}}{22}$$

Thus, $x = \frac{-11 + \sqrt{1177}}{22}$ and $x = \frac{-11 - \sqrt{1177}}{22}$.

Since $\sqrt{1177} \approx 34.307$, we have:

$x_1 = \frac{-11 + 34.307}{22} \approx \frac{23.307}{22} \approx 1.059$

$x_2 = \frac{-11 - 34.307}{22} \approx \frac{-45.307}{22} \approx -2.059$

Since neither of these solutions is equal to $0$ or $3$, they are both valid.

3. Final Answer

$x = \frac{-11 \pm \sqrt{1177}}{22}$

A. $x = \frac{-11 + \sqrt{1177}}{22}, \frac{-11 - \sqrt{1177}}{22}$