Solve for x: $(6 - x)^{4} + (8 - x)^{4} = 16$

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zero

${8, 6}$

${- 8, 6, 7 \pm i \sqrt{7}}$

${8, 6, 7 \pm i \sqrt{7}}$

answer is B.

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Detailed Solution

Given equation is $(6 - x)^{4} + (8 - x)^{4} = 16$ ... (i)

Put $y = \frac{(6 - x) + (8 - x)}{2} = \frac{14 - 2 x}{2} = 7 - x$ .

$\Rightarrow x = 7 - y$

Equation (i) reduces to.

$\begin{array}{l} (6 - 7 + y)^{4} + (8 - 7 + y)^{4} = 16 \\ \Rightarrow (y - 1)^{4} + (y + 1)^{4} = 16 \\ \Rightarrow (y^{4} - 4 y^{3} + 6 y^{2} - 4 y + 1) \\ + (y^{4} + 4 y^{3} + 6 y^{2} + 4 y + 1) = 16 \\ \Rightarrow 2 (y^{4} + 6 y^{2} + 1) = 16 \\ \Rightarrow (y^{4} + 6 y^{2} + 1) = 8 \\ \Rightarrow (y^{4} + 6 y^{2} - 7) = 0 \\ \Rightarrow (y^{2} + 7) (y^{2} - 1) = 0 \\ \Rightarrow (y^{2} + 7) = 0, (y^{2} - 1) = 0 \\ \Rightarrow y^{2} = 1, - 7 \\ \Rightarrow y = \pm 1, \pm i \sqrt{7} \\ \Rightarrow 7 - x = \pm 1, \pm i \sqrt{7} \\ \Rightarrow x = 7 \pm 1, 7 \pm i \sqrt{7} \end{array}$