Hot And Fresh

Let $y =\sqrt{2+x} + \sqrt{2-x}$ ; then we have:

$\begin{aligned} 3\log_3^2 y + 2\log_\frac 13 y\log_3(9x^2) + (1-\log_\frac 13 x)^2 & = 0 \\ 3\log_3^2 y + 2 \cdot \frac{\log_3 y}{\log_3 \frac 13} \cdot 2 \log_3 (3x) + (1- \frac{\log_3 x}{\log_3 \frac 13})^2 & = 0 \\ 3\log_3^2 y - 4\log_3 y \log_3 (3x) + (1+\log_3 x)^2 & = 0 \\ 3\log_3^2 y - 4\log_3 y \log_3 (3x) + \log_3^2 (3x) & = 0 \\ \left(3\log_3 y - \log_3 (3x)\right) \left(\log_3 y - \log_3 (3x)\right) & = 0 \end{aligned}$

$\implies \begin{cases} 3\log_3 y = \log_3 (3x) & \implies \left(\sqrt{2+x} + \sqrt{2-x}\right)^3 = 3x \\ \log_3 y = \log_3 (3x) & \implies \sqrt{2+x} + \sqrt{2-x} = 3x \end{cases}$

Considering $\left(\sqrt{2+x} + \sqrt{2-x}\right)^3 = 3x$ . We note that the minumum value of $\sqrt{2+x} + \sqrt{2-x}$ is 2, when $x=\pm 2$ . Therefore, minimum value of LHS $\min (\sqrt{2+x} + \sqrt{2-x})^3 = 8$ . Since the equation is only valid for $x \in [-2,2]$ , the maximum value of RHS $\max (3x) = 6$ , $\implies$ LHS $>$ RHS for $x \in [-2,2]$ , therefore there is no root.

Solving the following:

$\begin{aligned} \sqrt{2+x} + \sqrt{2-x} & = 3x \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ 2+x + 2\sqrt{4-x^2} +2-x & = 9x^2 \\ 2\sqrt{4-x^2} & = 9x^2 - 4 \quad \quad \small \color{#3D99F6}{\text{Squaring both sides}} \\ 16 - 4x^2 & = 81x^4 -72x^2 + 16 \\ 81x^4 -68x^2 & = 0 \\ x^2(81x^2-68) & = 0 \\ x & = \frac {2\sqrt{17}}9 \quad \quad \small \color{#3D99F6}{\text{Non-zero solution}} \end{aligned}$

$\implies a+b+c = 2+17+9 = \boxed{28}$ .