The Golden Equation

We have $\begin{aligned}\sqrt{1+\dfrac{1}{1+x}}+\dfrac{1}{\sqrt{1+x}} = & \sqrt{x}+\dfrac{1}{\sqrt{x}} \\ \dfrac{\sqrt{x+2}+1}{\sqrt{x+1}}= & \dfrac{x+1}{\sqrt{x}} \quad \quad & [\text{simplifying}]\\ \sqrt{x}(\sqrt{x+2}+1)= & (x+1)\sqrt{x+1} \\ x(x+3+2\sqrt{x+2})= & (x+1)^3 \quad \quad & [\because (x+1)\sqrt{x+1}=(x+1)^{\frac{3}{2}}] \\ x^2+3x+2x\sqrt{x+2}= & x^3+3x^2+3x+1 \\ x^3+2x^2-2x \sqrt{x+2}+1= & 0 \\ (x\sqrt{x+2})^2-2x\sqrt{x+2}+1= & 0 \\ (x\sqrt{x+2}-1)^2= & 0 \\ x\sqrt{x+2}= & 1 \\ x^2(x+2)= & 1 \\ x^3+2x^2-1= & 0 \\ (x+1)(x^2+x-1)= & 0 \\ x= & -1,\dfrac{-1\pm\sqrt{1+4}}{2} \\ = & -1,\dfrac{-1\pm\sqrt5}{2} \\ \end{aligned}$ But notice that $-1,\dfrac{-1-\sqrt5}{2}$ do not satisfy the above equation. $\therefore \begin{aligned} x= & \boxed{\dfrac{\sqrt5-1}{2}} \\ = & \phi-1 \end{aligned}$