## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$x=\frac{3-\sqrt{5}}{2},\:x=\frac{3+\sqrt{5}}{2}$$
Solving by setting the expressions equal to each other and then using the quadratic formula, we find: $$\frac{3-x}{4}=\frac{1}{4x}\\ \left(3-x\right)\cdot \:4x=4\cdot \:1\\ -4x^2+12x-4=0\\ x_{1,\:2}=\frac{-12\pm \sqrt{12^2-4\left(-4\right)\left(-4\right)}}{2\left(-4\right)}\\ x=\frac{3-\sqrt{5}}{2},\:x=\frac{3+\sqrt{5}}{2}$$