Equation of radicals

$\begin{aligned} \frac {\sqrt{1+x^2}}x & = \sqrt{1+x^2} - 1 & \small \color{#3D99F6} \text{Dividing both sides by }\sqrt{1+x^2} \\ \frac 1x & = 1 - \frac 1{\sqrt{1+x^2}} \\ \frac 1{\sqrt{1+x^2}} & = 1 - \frac 1x & \small \color{#3D99F6} \text{Squaring both sides} \\ \frac 1{1+x^2} & = \left(\frac {x-1}x\right)^2 \\ x^2 & = (1+x^2)(x^2-2x+1) \\ \implies x^4 - 2x^3 + x^2 - 2x + 1 & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }x^2 \\ x^2 - 2x + 1 - \frac 2x + \frac 1{x^2} & = 0 \\ \left(x + \frac 1x\right)^2 - 2\left(x + \frac 1x\right) - 1 & = 0 \\ \implies x + \frac 1x & = 1 \pm \sqrt 2 \\ x^2 - (1 \pm \sqrt 2) x + 1 & = 0 & \small \color{#3D99F6} \text{For positive root }x_0 \\ \implies x_0^2 - (1 + \sqrt 2) x_0 + 1 & = 0 \\ \implies x_0^2 & = (\sqrt 2+1) x_0 - 1 \ \ ...(1) \end{aligned}$

Now we have:

$\begin{aligned} y & = \left((2\sqrt 2 - 2)x_0 - 1\right)^2 \\ & = \left(2(\sqrt 2 - 1)x_0 - 1\right)^2 \\ & = 4(\sqrt 2 - 1)^2{\color{#3D99F6} x_0^2} - 4(\sqrt 2 - 1)x_0 + 1 & \small \color{#3D99F6} (1): \ \ x_0^2 = (\sqrt 2+1) x_0 - 1 \\ & = 4(\sqrt 2 - 1)^2{\color{#3D99F6} \left[(\sqrt 2+1) x_0 - 1\right]} - 4(\sqrt 2 - 1)x_0 + 1 \\ & = 4(\sqrt 2 - 1)x_0 - 4(\sqrt 2 - 1)^2 - 4(\sqrt 2 - 1)x_0 + 1 \\ & = - 4(3-2\sqrt 2) + 1 \\ & = -11 + 8\sqrt 2 \end{aligned}$

$\implies a+b+c = -11+8+2 = \boxed{-1}$

This is the full solution to the equation using trigonometric functions. Since there is an expression $\sqrt{x^2+1}$ , so it might be possible to substitute by $x=\tan\theta$ .

Because we want a positive solution, suppose that $x=\tan\theta$ , where $0 < \theta < \frac{\pi}{2}$ . The equation is transformed into $\frac{\sqrt{\sec^2\theta}}{\tan\theta} = \sqrt{\sec^2\theta}-1.$ Since we choose the range $0 < \theta < \frac{\pi}{2}$ , we have $\sec\theta>0$ . Hence, $\begin{aligned} \frac{\sec\theta}{\tan\theta} &= \sec\theta-1,\\[10pt] \frac{1}{\sin\theta} &= \frac{1}{\cos\theta}-1,\\[10pt] \frac{1}{\cos\theta}-\frac{1}{\sin\theta} &= 1. \end{aligned}$ Squaring equation gives $\begin{aligned} \frac{1}{\cos^2\theta} - \frac{2}{\sin\theta\cos\theta} + \frac{1}{\sin^2\theta} &= 1,\\[10pt] \frac{1}{\sin^2\theta\cos^2\theta}-\frac{2}{\sin\theta\cos\theta} &= 1,\\[10pt] \frac{4}{\sin^2 2\theta} - \frac{4}{\sin\theta} &= 1,\\[10pt] 4\csc^2 2\theta-4\csc 2\theta-1 &= 0,\\[10pt] \csc 2\theta &= \frac{1 \pm \sqrt{2}}{2},\\[10pt] \sin 2\theta &= \frac{2}{1 \pm \sqrt{2}}. \end{aligned}$ Because $0 < 2\theta < \pi$ , we have that $\sin 2\theta>0$ , that is, $\sin 2\theta = \frac{2}{1+\sqrt{2}}=2\sqrt{2}-2$ . As a result, $\cos 2\theta = \pm \sqrt{8\sqrt{2}-11}$ . Therefore, $\tan\theta = \frac{1-\cos 2\theta}{\sin 2\theta} = \frac{1 \pm \sqrt{8\sqrt{2}-11}}{2\sqrt{2}-2}.$ However, the solution must be bigger than $1$ (reason lies below), so that the only positive solution is $x_0 = \frac{1 + \sqrt{8\sqrt{2}-11}}{2\sqrt{2}-2},$ which gives $[(2\sqrt{2}-2)x_0-1]^2 = 8\sqrt{2}-11$ . The solution to this question is $-11+8+2=\boxed{-1}$ , as desired.

In general, if it doesn't specify that the solution is positive, this equation still has unique solution. Obviously, zero isn't solution, so we are looking for nonzero solution. From the equation, we can express it as $x = \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}-1} = \frac{(\sqrt{1+x^2})(\sqrt{1+x^2}+1)}{x^2} > 0.$ This tells us that solution must be positive. Moreover, we obtain from the above equation that $x^3 = 1+x^2+\sqrt{1+x^2} > 1,$ together with the fact that $x>0$ , we finally get $x>1$ . Therefore, this equation has unique solution as derived earlier.