Let S = { x in R : x ge 0 text{ and } 2|sqrt{ | vedclass

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(B) Let $t = \sqrt{x}$,where $t \ge 0$. The equation becomes $2|t - 3| + t(t - 6) + 6 = 0$.Case-$I$: $0 \le t < 3$ (i.e.,$0 \le x < 9$)$2(3 - t) + t^2

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contains exactly two elements.

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(B) Let $t = \sqrt{x}$,where $t \ge 0$. The equation becomes $2|t - 3| + t(t - 6) + 6 = 0$.Case-$I$: $0 \le t $2(3 - t) + t^2 - 6t + 6 = 0$$6 - 2t + t^2 - 6t + 6 = 0$$t^2 - 8t + 12 = 0$$(t - 6)(t - 2) = 0$$t = 6$ or $t = 2$. Since $0 \le t Case-$II$: $t \ge 3$ (i.e.,$x \ge 9$)$2(t - 3) + t^2 - 6t + 6 = 0$$2t - 6 + t^2 - 6t + 6 = 0$$t^2 - 4t = 0$$t(t - 4) = 0$$t = 0$ or $t = 4$. Since $t \ge 3$,we have $t = 4$,which implies $x = 16$.Thus,$S = \{4, 16\}$,which contains exactly two elements.

Let $S = \{ x \in R : x \ge 0 \text{ and } 2|\sqrt{x} - 3| + \sqrt{x}(\sqrt{x} - 6) + 6 = 0 \}$. Then $S$:

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