Okay, so I need to solve the equation x = sqrt(11 - 2x) + 4. Hmm, let's see. First, I remember that when dealing with square roots, I have to make sure that the expression inside the square root is non-negative because you can't take the square root of a negative number in real numbers. So, maybe I should start by figuring out the domain of this equation. 

Let me write down the equation again: x = sqrt(11 - 2x) + 4. The term inside the square root is 11 - 2x. For this to be valid, 11 - 2x must be greater than or equal to zero. So, 11 - 2x ≥ 0. Solving that inequality: subtract 11 from both sides, -2x ≥ -11. Then divide both sides by -2, but remember that dividing by a negative number flips the inequality sign. So, x ≤ 11/2. Which is x ≤ 5.5. So, x has to be less than or equal to 5.5. That's the domain I need to consider.

Now, the equation itself. To solve for x, maybe I can isolate the square root term first. Let me subtract 4 from both sides: x - 4 = sqrt(11 - 2x). Now, to eliminate the square root, I can square both sides of the equation. But before I do that, I need to remember that squaring both sides can sometimes introduce extraneous solutions, so I have to check any solutions I get at the end.

So, squaring both sides: (x - 4)^2 = (sqrt(11 - 2x))^2. That simplifies to (x - 4)^2 = 11 - 2x. Let's expand the left side: (x - 4)(x - 4) = x^2 - 8x + 16. So, the equation becomes x^2 - 8x + 16 = 11 - 2x.

Now, let's bring all terms to one side to form a quadratic equation. Subtract 11 and add 2x to both sides: x^2 - 8x + 16 - 11 + 2x = 0. Simplify that: x^2 - 6x + 5 = 0. So, the quadratic equation is x^2 - 6x + 5 = 0. Let's factor this quadratic. Looking for two numbers that multiply to 5 and add to -6. Hmm, factors of 5 are 1 and 5. Since the middle term is -6x, both numbers should be negative. So, -1 and -5. Therefore, (x - 1)(x - 5) = 0. So, the solutions are x = 1 and x = 5.

But wait, I need to check these solutions in the original equation because we squared both sides, which can introduce extraneous solutions. Let's check x = 1 first. Plugging into the original equation: left side is 1. Right side is sqrt(11 - 2*1) + 4 = sqrt(9) + 4 = 3 + 4 = 7. So, 1 ≠ 7. That's not valid. So, x = 1 is an extraneous solution and should be discarded.

Now, check x = 5. Left side is 5. Right side is sqrt(11 - 2*5) + 4 = sqrt(11 - 10) + 4 = sqrt(1) + 4 = 1 + 4 = 5. So, 5 = 5. That works. So, x = 5 is a valid solution.

But wait, earlier we found the domain was x ≤ 5.5, so 5 is within that domain. So, that's good. So, the only solution is x = 5.

Wait, but let me think again. When we squared both sides, we might have lost some information. Let me verify again. The original equation is x = sqrt(11 - 2x) + 4. When we subtract 4, we get x - 4 = sqrt(11 - 2x). Then, squaring gives (x - 4)^2 = 11 - 2x. Then expanding to quadratic, which gives x^2 - 8x + 16 = 11 - 2x. Then moving terms: x^2 - 6x + 5 = 0. Factored as (x - 1)(x - 5) = 0. So, solutions 1 and 5. Then check in original equation.

x = 1: sqrt(11 - 2*1) + 4 = sqrt(9) +4 = 3 +4=7 ≠1. So invalid.

x=5: sqrt(11 - 10) +4=1 +4=5. Valid.

So, only x=5 is the solution. Therefore, the answer is 5.

But let me just make sure I didn't make a mistake in the algebra. Let me retrace:

Original equation: x = sqrt(11 - 2x) +4.

Isolate sqrt: x -4 = sqrt(11 -2x).

Square both sides: (x -4)^2 = 11 -2x.

Expand left: x² -8x +16 = 11 -2x.

Bring all terms to left: x² -8x +16 -11 +2x = x² -6x +5=0.

Factor: x² -6x +5=(x-5)(x-1)=0.

Solutions x=5, x=1. Check in original equation.

x=5: valid. x=1: invalid.

So, yes, only x=5. Therefore, the answer is 5.

But wait, let me check if there's another possible solution. Suppose I consider the function f(x) = sqrt(11 -2x) +4. Its domain is x ≤5.5. The left side is x. So, when x is 5, f(x)=5. When x approaches negative infinity, f(x) approaches sqrt(11 -2x) +4, which would be sqrt(something very large) +4, which goes to infinity. But the left side x is going to negative infinity. So, the functions x and f(x) might intersect only once. Since at x=5, they are equal. Let's check another point. Let's pick x=0. Then f(0)=sqrt(11) +4≈3.316+4≈7.316>0. So, f(0)=7.316>0. But x=0, so 0 <7.316. So, at x=0, f(x) >x. At x=5, f(x)=5. What about x=6? Wait, x=6 is not in the domain. But x=5.5, which is the upper limit. f(5.5)=sqrt(11 -11) +4=0 +4=4. But x=5.5, so 5.5>4. So, at x=5.5, x=5.5>4. So, the left side is greater than the right side at x=5.5. At x=5, they are equal. Let's check between x=5 and x=5.5. For example, x=5.25. Then f(x)=sqrt(11 -10.5)+4=sqrt(0.5)+4≈0.707+4≈4.707. But x=5.25>4.707. So, left side > right side. So, between x=5 and x=5.5, x>f(x). So, the only crossing point is at x=5. So, that's the only solution.

Therefore, the answer is 5.

**Final Answer**
The value of \( x \) that satisfies the equation is \boxed{5}.
Therefore, the solution to the equation $x = \!\sqrt{11-2x} + 4$ is $\boxed{5}$.
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