1. **Stating the problem:** Given the equation $$\frac{x}{x^2 + 3x - 1} = \frac{1}{4}$$, find the value of $x$. 2. **Rewrite the equation:** Multiply both sides by the denominator to clear the fraction: $$x = \frac{1}{4} (x^2 + 3x - 1)$$ 3. **Multiply both sides by 4:** $$4x = x^2 + 3x - 1$$ 4. **Bring all terms to one side:** $$x^2 + 3x - 1 - 4x = 0$$ $$x^2 - x - 1 = 0$$ 5. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, the solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-1$, $c=-1$. 6. **Calculate the discriminant:** $$\Delta = (-1)^2 - 4(1)(-1) = 1 + 4 = 5$$ 7. **Find the roots:** $$x = \frac{1 \pm \sqrt{5}}{2}$$ **Final answer:** $$x = \frac{1 + \sqrt{5}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{5}}{2}$$