Solve `sqrt(x-2)+sqrt(4-x)=2.`

Explanation

To solve the equation \( \sqrt{x-2} + \sqrt{4-x} = 2 \), we will follow these steps:

### Step 1: Isolate one of the square root terms
We can rearrange the equation to isolate one of the square roots:
\[ \sqrt{x-2} = 2 - \sqrt{4-x} \]

### Step 2: Square both sides
Next, we square both sides to eliminate the square root:
\[ (\sqrt{x-2})^2 = (2 - \sqrt{4-x})^2 \]
This simplifies to:
\[ x - 2 = 4 - 4\sqrt{4-x} + (4-x) \]
Combining like terms gives:
\[ x - 2 = 8 - x - 4\sqrt{4-x} \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate the square root:
\[ 4\sqrt{4-x} = 8 - x + 2 \]
This simplifies to:
\[ 4\sqrt{4-x} = 10 - x \]

### Step 4: Divide by 4
Next, we divide both sides by 4:
\[ \sqrt{4-x} = \frac{10 - x}{4} \]

### Step 5: Square both sides again
We square both sides again to eliminate the square root:
\[ (4-x) = \left(\frac{10 - x}{4}\right)^2 \]
This expands to:
\[ 4 - x = \frac{(10 - x)^2}{16} \]

### Step 6: Multiply through by 16
To eliminate the fraction, we multiply both sides by 16:
\[ 16(4 - x) = (10 - x)^2 \]
This simplifies to:
\[ 64 - 16x = 100 - 20x + x^2 \]

### Step 7: Rearrange to form a quadratic equation
Rearranging gives:
\[ x^2 - 4x + 36 = 0 \]

### Step 8: Solve the quadratic equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -4, c = 36 \):
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 - 144}}{2} \]
\[ x = \frac{4 \pm \sqrt{-128}}{2} \]
Since the discriminant is negative, there are no real solutions to this equation.

### Step 9: Check for extraneous solutions
However, we should check if any solutions we found are valid in the original equation.

### Final Solution
Since we found that the quadratic has no real solutions, we conclude that the original equation has no solution.