Explanation

(d)
Given, $\left( 3 + 2\sqrt{2} \right)^{x^{2} - 8} + \left( 3 + 2\sqrt{2} \right)^{8 - x^{2}} = 6$
Let $\left( 3 + 2\sqrt{2} \right)^{x^{2} - 8} = y$
$\therefore\ \ y + y^{- 1} = 6$
$\Rightarrow y^{2} - 6y + 1 = 0$
$\Rightarrow \ \ y = \frac{6 \pm \sqrt{36 - 4}}{2 \times 1}$
$= \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}$
For positive sign
$\left( 3 + 2\sqrt{2} \right)^{x^{2} - 8} = 3 + 2\sqrt{2}$
$\Rightarrow x^{2} - 8 = 1\ \ \Rightarrow \ \ \ x = \pm 3$
For negative sign
$\left\lbrack \left( 3 + 2\sqrt{2} \right)^{- 1} \right\rbrack^{8 - x^{2}} = 3 - 2\sqrt{2}$
$\Rightarrow \left( 3 - 2\sqrt{2} \right)^{8 - x^{2}} = 3 - 2\sqrt{2}$
$\Rightarrow 8 - x^{2} = 1\ \ \ \Rightarrow \ \ \ \ x^{2} = 7$
$\Rightarrow \ \ \ x = \pm \sqrt{7}$