2
6
18
12
5
Correct answer is D
\((3x + 1)^{2} = (3x - 1)^{2} + x^{2}\) (Pythagoras's theorem) \(9x^{2} + 6x + 1 = 9x^{2} - 6x + 1 + x^{2}\) x2 - 12x = 0 x(x - 12) = 0 x = 0 or 12
The sides cannot be 0 hence x = 12.
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