Okay, so I need to solve this equation: √(x + √(3x + 6)) + √(x - √(3x + 6)) = 6. Hmm, that looks a bit complicated with nested square roots. Let me try to break it down step by step.

First, I should probably check the domain of the equation because there are square roots involved. The expressions inside the square roots must be non-negative. Let's start with the innermost square root: √(3x + 6). For this to be real, 3x + 6 must be ≥ 0. So, 3x + 6 ≥ 0 → x ≥ -2. That's the first condition.

Next, looking at the expressions inside the outer square roots: x + √(3x + 6) and x - √(3x + 6). Both of these must be non-negative as well. Let's handle them one by one.

Starting with x + √(3x + 6) ≥ 0. Since √(3x + 6) is non-negative (as x ≥ -2), adding x to it must still be non-negative. So, x + √(3x + 6) ≥ 0. Since x ≥ -2, the smallest x can be is -2. Let's check when x = -2: √(3*(-2) + 6) = √(0) = 0. So, x + 0 = -2 + 0 = -2, which is negative. That's a problem. Wait, but the original equation has √(x + ...), which would require x + ... to be non-negative. So, if x = -2, then x + √(3x + 6) = -2 + 0 = -2 < 0, which is invalid. Therefore, x must be greater than -2. Let's check for x > -2.

Similarly, the second expression is x - √(3x + 6) ≥ 0. So, x ≥ √(3x + 6). Let's solve this inequality. Let me square both sides to get rid of the square root, but I have to be careful because squaring inequalities can sometimes introduce extraneous solutions. So, assuming x ≥ √(3x + 6), then squaring both sides gives x² ≥ 3x + 6. Rearranging, x² - 3x - 6 ≥ 0. Let's solve the quadratic inequality x² - 3x - 6 ≥ 0. The roots of x² - 3x - 6 = 0 are [3 ± √(9 + 24)] / 2 = [3 ± √33]/2. So, the quadratic is positive outside the roots. Since [3 - √33]/2 is approximately (3 - 5.744)/2 ≈ -1.372, and [3 + √33]/2 ≈ (3 + 5.744)/2 ≈ 4.372. Therefore, x ≤ [3 - √33]/2 or x ≥ [3 + √33]/2. But since from the previous condition, x > -2, and [3 - √33]/2 is about -1.372, so x must be ≤ -1.372 or ≥ 4.372. However, since x > -2 from the inner square root, and x must satisfy x ≥ √(3x + 6), which after squaring gives x ≥ [3 + √33]/2 ≈ 4.372 or x ≤ [3 - √33]/2 ≈ -1.372. But since x > -2, the valid intervals are x ∈ (-2, [3 - √33]/2] ∪ [ [3 + √33]/2, ∞). Wait, but [3 - √33]/2 is approximately -1.372, which is greater than -2, so the domain for x is (-2, [3 - √33]/2] ∪ [ [3 + √33]/2, ∞). However, we need to check if x in (-2, [3 - √33]/2] satisfies x ≥ √(3x + 6). Let me test x = -1. Let's see: √(3*(-1) + 6) = √(3) ≈ 1.732. So, x = -1, √(3x + 6) ≈ 1.732. Then x - √(3x + 6) ≈ -1 - 1.732 ≈ -2.732 < 0, which is invalid. Therefore, even though algebraically x ≤ [3 - √33]/2 ≈ -1.372, but in reality, x - √(3x + 6) must be non-negative. So perhaps my previous steps were incorrect. Maybe squaring both sides when x and √(3x +6) can be negative?

Wait, actually, the inequality x ≥ √(3x +6) is tricky because both sides can be positive or negative. But since x is in (-2, [3 - √33]/2] ∪ [ [3 + √33]/2, ∞), and in the interval (-2, [3 - √33]/2], x is negative or up to -1.372. But since √(3x +6) is always non-negative, the inequality x ≥ √(3x +6) can only be true if x is non-negative because the right side is non-negative. Therefore, x must be ≥ √(3x +6) ≥ 0. Therefore, x must be ≥ 0. So, x must be in [0, ∞). But also, x must be ≥ [3 + √33]/2 ≈ 4.372 or x ≤ [3 - √33]/2 ≈ -1.372, but since x ≥ 0, the only valid interval is x ≥ [3 + √33]/2 ≈ 4.372. Wait, that seems conflicting with the previous domain. Let me verify.

Original domain from inner square root: x ≥ -2.

From x + √(3x +6) ≥ 0: For x ≥ -2, but when x is negative, x + √(3x +6) could be negative or positive. For example, x = -1: √(3*(-1) +6) = √3 ≈1.732, so x + √(...) ≈ -1 +1.732 ≈ 0.732 ≥0. So, x can be as low as -2, but x + √(3x +6) must be ≥0. Let's solve x + √(3x +6) ≥0. Let me set t = √(3x +6). Then t ≥0. Then x = (t² -6)/3. The inequality becomes (t² -6)/3 + t ≥0 → (t² -6 +3t)/3 ≥0 → t² +3t -6 ≥0. Solving t² +3t -6 ≥0. The roots are [-3 ± √(9 +24)]/2 = [-3 ± √33]/2. The positive root is [-3 + √33]/2 ≈ (-3 +5.744)/2 ≈1.372. So, t ≤ [-3 -√33]/2 (which is negative, so irrelevant) or t ≥ [-3 +√33]/2 ≈1.372. Since t = √(3x +6) ≥0, we have √(3x +6) ≥ [-3 +√33]/2 ≈1.372. Then squaring both sides: 3x +6 ≥ ([-3 +√33]/2)² = (9 -6√33 +33)/4 = (42 -6√33)/4 = (21 -3√33)/2. Therefore, 3x ≥ (21 -3√33)/2 -6 = (21 -3√33 -12)/2 = (9 -3√33)/2. So, x ≥ (9 -3√33)/6 = (3 -√33)/2 ≈ (3 -5.744)/2 ≈ -1.372. So, combining with the original domain x ≥ -2, the valid interval is x ≥ (3 -√33)/2 ≈-1.372. But wait, x must also satisfy x ≥ [3 +√33]/2 ≈4.372? Wait, this is conflicting. Let me check.

Wait, perhaps I made a mistake here. Let's retrace. The equation x + √(3x +6) ≥0. We set t = √(3x +6) ≥0. Then x = (t² -6)/3. So, substituting into x + t ≥0: (t² -6)/3 + t ≥0 → (t² -6 +3t)/3 ≥0 → t² +3t -6 ≥0. The roots are t = [-3 ±√(9 +24)]/2 = [-3 ±√33]/2. Since t ≥0, the relevant root is t ≥ [-3 +√33]/2 ≈1.372. Therefore, √(3x +6) ≥1.372. Squaring both sides: 3x +6 ≥ ( [-3 +√33]/2 )² = (9 -6√33 +33)/4 = (42 -6√33)/4 = (21 -3√33)/2. Then 3x ≥ (21 -3√33)/2 -6 = (21 -3√33 -12)/2 = (9 -3√33)/2. So, x ≥ (9 -3√33)/6 = (3 -√33)/2 ≈ (3 -5.744)/2 ≈-1.372. Therefore, x must be ≥ (3 -√33)/2 ≈-1.372. So, the domain for x from x + √(3x +6) ≥0 is x ≥ (3 -√33)/2 ≈-1.372. But also, the original inner square root requires x ≥ -2. So, combining these, the domain is x ≥ (3 -√33)/2 ≈-1.372. But we also need x - √(3x +6) ≥0. Let's check that.

From x - √(3x +6) ≥0. Let me solve this inequality. Let me set t = √(3x +6) ≥0. Then x = (t² -6)/3. So, the inequality becomes (t² -6)/3 - t ≥0 → (t² -6 -3t)/3 ≥0 → t² -3t -6 ≥0. Solving t² -3t -6 ≥0. The roots are [3 ±√(9 +24)]/2 = [3 ±√33]/2. Since t ≥0, the relevant root is t ≥ [3 +√33]/2 ≈4.372. Therefore, √(3x +6) ≥ [3 +√33]/2. Squaring both sides: 3x +6 ≥ ([3 +√33]/2)² = (9 +6√33 +33)/4 = (42 +6√33)/4 = (21 +3√33)/2. Therefore, 3x ≥ (21 +3√33)/2 -6 = (21 +3√33 -12)/2 = (9 +3√33)/2. So, x ≥ (9 +3√33)/6 = (3 +√33)/2 ≈ (3 +5.744)/2 ≈4.372. Therefore, x must be ≥ (3 +√33)/2 ≈4.372. So, combining all the domain restrictions:

From x + √(3x +6) ≥0: x ≥ (3 -√33)/2 ≈-1.372.

From x - √(3x +6) ≥0: x ≥ (3 +√33)/2 ≈4.372.

Therefore, the overall domain is x ≥ (3 +√33)/2 ≈4.372. So, x must be at least approximately 4.372. Okay, that's a key point. So, the solution must be x ≥ (3 +√33)/2. But let's also check if there's an upper limit. Since the original equation has √(x + ...) and √(x - ...), as x increases, the expressions inside the square roots increase and decrease respectively. But since we have x ≥4.372, there's no upper limit here; x can be as large as needed. So, the domain is x ≥ (3 +√33)/2.

Now, moving on to solving the equation: √(x + √(3x +6)) + √(x - √(3x +6)) =6. Let me denote A = √(x + √(3x +6)) and B = √(x - √(3x +6)). Then, the equation is A + B =6. Let me try squaring both sides to eliminate the square roots. So, (A + B)^2 = 36. Expanding, A² + 2AB + B² =36. But A² is x + √(3x +6) and B² is x - √(3x +6). Therefore, A² + B² = [x + √(3x +6)] + [x - √(3x +6)] = 2x. So, substituting back: 2x + 2AB =36. Therefore, 2AB =36 -2x → AB =18 -x. So, now we have AB =18 -x. Let's compute AB. AB = √(x + √(3x +6)) * √(x - √(3x +6)). This is equal to √[ (x + √(3x +6))(x - √(3x +6)) ] = √[x² - (√(3x +6))²] = √[x² - (3x +6)] = √(x² -3x -6). Therefore, AB = √(x² -3x -6). So, from earlier, AB =18 -x. Therefore, √(x² -3x -6) =18 -x. Let's square both sides again: x² -3x -6 = (18 -x)² = 324 -36x +x². Simplify: x² -3x -6 =324 -36x +x². Subtract x² from both sides: -3x -6 =324 -36x. Bring all terms to one side: -3x -6 -324 +36x =0 →33x -330=0 →33x=330 →x=10. So, x=10. But we need to check if this solution is valid in the original equation and within the domain.

First, check if x=10 is in the domain x ≥(3 +√33)/2 ≈4.372. Yes, 10>4.372. Now, check the original equation: √(10 + √(3*10 +6)) + √(10 - √(3*10 +6)).

Compute √(3*10 +6) =√(36)=6. Then, 10 +6=16, √16=4. 10 -6=4, √4=2. So, 4 +2=6. Which matches the equation. Therefore, x=10 is a valid solution.

Is there another solution? Let's check. When we squared the equations, sometimes extraneous solutions can appear. So, let's verify if there are other solutions. Let me see. When we squared the equation √(x² -3x -6) =18 -x, we must ensure that both sides are non-negative. The left side √(x² -3x -6) is non-negative by definition. The right side 18 -x must also be non-negative, so 18 -x ≥0 →x ≤18. So, any solution must satisfy x ≤18. But our domain is x ≥(3 +√33)/2 ≈4.372 and x ≤18. So, x ∈ [4.372,18]. However, when we solved, we got x=10. Let me check if there's another solution in this interval.

Suppose there's another solution. Let me assume that when we squared √(x² -3x -6) =18 -x, we might get another solution. Let's solve the equation again: √(x² -3x -6) =18 -x. Square both sides: x² -3x -6 = (18 -x)² =324 -36x +x². Subtract x²: -3x -6 =324 -36x. Then, 33x =330 →x=10. So, only one solution here. Therefore, x=10 is the only solution in the domain. So, the answer is x=10.

But let me check if there's another possible solution when we squared the equation. For example, sometimes when you square both sides, you can get solutions that don't satisfy the original equation. But in this case, we only got x=10, which works. Also, when we considered the domain, x must be ≥ (3 +√33)/2 ≈4.372 and ≤18. So, x=10 is within this interval and satisfies the original equation. Therefore, x=10 is the only solution.

Therefore, the solution to the equation is x=10.
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