Let S = {x ∈ R: x ≥ 0} and 2|√x - 3| + √x(√x - 3) + 6 = 0. Then S contains exactly two elements.

x ≥ 0 (given) ——— (1)
2|√x - 3| + √x(√x - 3) + 6 = 0 ———(2)
Consider x < 9, √x - 3 < 0. For x < 9, eq(2) becomes
-2(√x - 3) + √x(√x - 3) + 6 = 0
-2√x + 6 + x - 3√x + 6 = 0
x - 5√x + 12 = 0
(√x - 3)(√x - 4) = 0
⇒ √x = 3, 4
x = 9, 16
Since x < 9, so x = 9 is invalid. x = 16 is also invalid because we assumed x < 9.
For x ≥ 9
2(√x - 3) + √x(√x - 3) + 6 = 0
2√x - 6 + x - 3√x + 6 = 0
x - √x = 0
√x(√x - 1) = 0
⇒ √x = 0 or √x = 1
x = 0, 1
Since x ≥ 9, x = 0 and x = 1 are invalid.
Let's reconsider the equation:
2|√x - 3| + √x(√x - 3) + 6 = 0
Let u = √x - 3
Then √x = u + 3
x = (u+3)²
2|u| + (u+3)u + 6 = 0
2|u| + u² + 3u + 6 = 0
If u ≥ 0, 2u + u² + 3u + 6 = 0
u² + 5u + 6 = 0
(u + 2)(u + 3) = 0
u = -2, -3
If u < 0, -2u + u² + 3u + 6 = 0
u² + u + 6 = 0
The discriminant is 1 - 4(6) = -23 < 0, so there are no real solutions.
If u = -2, √x - 3 = -2 ⇒ √x = 1 ⇒ x = 1
If u = -3, √x - 3 = -3 ⇒ √x = 0 ⇒ x = 0
Since x ≥ 0, both x = 0 and x = 1 are possible solutions. But these solutions don't satisfy the condition x ≥ 9.
Let's try x = 4. Then 2|2-3| + 2(2-3) + 6 = 2(1) - 2 + 6 = 6 ≠ 0.
Let's try x = 16. Then 2|4-3| + 4(4-3) + 6 = 2 + 4 + 6 = 12 ≠ 0.
If x = 4, 2|2 - 3| + 2(2 - 3) + 6 = 2(1) - 2 + 6 = 6 ≠ 0.
If x = 16, 2|4 - 3| + 4(4 - 3) + 6 = 2 + 4 + 6 = 12 ≠ 0.
There must be a mistake in the problem statement or my calculations.