<Question>
Find all values of $x$ that satisfy the equation $x = \!\sqrt{11-2x} + 4$.
</Question>

<think>
<Plan_1>
Okay, so I need to solve the equation x equals the square root of (11 minus 2x) plus 4. Let me write that down again to make sure I have it right: x = √(11 - 2x) + 4. Alright, the goal is to find all real values of x that satisfy this equation. Hmm, let's think about how to approach this.

First, since there's a square root involved, I should consider the domain of the equation. The expression inside the square root, which is 11 - 2x, must be greater than or equal to zero. So, 11 - 2x ≥ 0. Solving for x, I subtract 11 from both sides: -2x ≥ -11. Then divide both sides by -2, but remember that dividing by a negative number reverses the inequality sign.
So, x ≤ 11/2. That means x has to be less than or equal to 5.5. Got that. So any solution must satisfy x ≤ 5.5.

Also, the square root itself is always non-negative, so √(11 - 2x) is ≥ 0. Then, adding 4 to that, the right-hand side of the equation is ≥ 4. Therefore, x must be ≥ 4.

<prompt_1.1> Determine the range of possible values for x based on the non-negativity of the square root. </prompt_1.1>

So combining these two domain restrictions, x must be in the interval [4, 5.5].

Now, to solve the equation, let's try to isolate the square root term. So, subtract 4 from both sides:

<prompt_1.2> Isolate the square root term by subtracting 4 from both sides of the equation. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, because if the square root is non-negative, then √(11 - 2x) + 4 is at least 4. So x has to be greater than or equal to 4. Therefore, x is in [4, 5.5]. That's an important restriction.
</execution_1.1>

<execution_1.2>
x - 4 = √(11 - 2x)

Now, to eliminate the square root, I can square both sides of the equation. That should help me get rid of the radical. Let's do that:

(x - 4)² = (√(11 - 2x))²

Simplifying both sides:

Left side: (x - 4)² = x² - 8x + 16

Right side: (√(11 - 2x))² = 11 - 2x

So now the equation is:

x² - 8x + 16 = 11 - 2x
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.2:
Let's bring all terms to one side to form a quadratic equation. Subtract 11 and add 2x to both sides:

<prompt_2.1> Rearrange the equation to form a quadratic equation by moving all terms to one side. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
x² - 8x + 16 - 11 + 2x = 0

Simplify the terms:

x² - 6x + 5 = 0
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So, quadratic equation x² - 6x + 5 = 0. Let's factor this. Looking for two numbers that multiply to 5 and add up to -6.
Hmm, factors of 5 are 1 and 5. Since the middle term is -6x, and the constant term is +5, both numbers should be negative. So, -1 and -5. Yes, (-1) * (-5) = 5 and (-1) + (-5) = -6. Therefore, the quadratic factors as:

(x - 1)(x - 5) = 0

Wait, let's check that.

<prompt_3.1> Verify the factorization of the quadratic equation by expanding (x - 1)(x - 5). </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
If we expand (x - 1)(x - 5), we get x² -5x -x +5 = x² -6x +5. Yes, that's correct. So the solutions are x = 1 and x = 5.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, earlier we established that x must be in [4, 5.5]. So x = 1 is not in this interval. Therefore, x = 1 is an extraneous solution and must be rejected. So the only possible solution is x = 5.

But let's verify this solution in the original equation to make sure.

<prompt_4.1> Verify the solution x = 5 by substituting it back into the original equation. </prompt_4.1>
-----

But let me think again. When we squared both sides of the equation, we might have introduced extraneous solutions, which is why we have to check. But is there another way this could go wrong? Let's see.

Alternatively, suppose there was another solution in [4,5.5]. Let's test x=5 and x=1.

<prompt_4.2> Test alternative solutions within the domain [4, 5.5] to ensure no other solutions exist. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Original equation: x = √(11 - 2x) + 4

Substitute x = 5:

Left side: 5

Right side: √(11 - 2*5) + 4 = √(11 - 10) + 4 = √1 + 4 = 1 + 4 = 5

So, 5 = 5. That checks out.
</execution_4.1>

<execution_4.2>
But x=1 is outside, so it's rejected. What about another number in [4,5.5], say x=4. Let's plug into the original equation:

Left side: x = 4

Right side: √(11 - 2*4) +4 = √(11 -8) +4 = √3 +4 ≈ 1.732 +4 ≈5.732 ≠4. So not a solution.

How about x=5.5?

Left side: x=5.5

Right side: √(11 -2*5.5) +4 = √(11 -11) +4 = √0 +4 =0 +4=4 ≠5.5. So not a solution.

So seems like only x=5 works.

But let's check x=5 again.

Original equation: 5 = √(11 -2*5) +4.

Compute inside the square root: 11 -10 =1. So √1=1. Then 1 +4=5. Correct. So that works.

Therefore, the only solution is x=5.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, but let me make sure there are no other solutions. The quadratic equation gave us two roots, but one was extraneous. Is there a possibility that squaring both sides introduced another one? But since we only got two roots and one was rejected, that's all.

But just to be thorough, let's consider if there could be another solution. Suppose there was another x in [4,5.5] that satisfies the equation. But since we solved algebraically and only got x=5 as a valid solution, and x=1 is outside the domain, there shouldn't be another one.

Wait, but let's test x=5 and maybe another value in the interval. Let's pick x=4.5.

<prompt_5.1> Test additional values within the domain [4, 5.5] to ensure no other solutions exist. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Let's see what the right-hand side gives: √(11 -2*4.5) +4 = √(11 -9) +4 = √2 +4 ≈1.414 +4 ≈5.414, which is not equal to 4.5. So not a solution.

How about x=5. Let's try x=5. Let's see, as before, that works. How about x=4. Let's check x=4 again: RHS is √(11 -8) +4 = √3 +4 ≈5.732, which is not 4. So nope.

Alternatively, maybe graphing both sides would help visualize. The left side is the line y = x, and the right side is y = √(11 -2x) +4. The points where they intersect are the solutions. Since the square root function is a decreasing function (as x increases, 11 -2x decreases, so sqrt decreases), and the line y = x is increasing. So they might intersect at one point, which is x=5.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the final answer is $\boxed{5}$.
</Final_answer>
</think>