Number of real roots of the equation `sqrt(x)+sqrt(x-sqrt((1-x)))=1` is

Explanation

To find the number of real roots of the equation \( \sqrt{x} + \sqrt{x - \sqrt{1 - x}} = 1 \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \sqrt{x} + \sqrt{x - \sqrt{1 - x}} = 1 \]
We can isolate one of the square root terms:
\[ \sqrt{x - \sqrt{1 - x}} = 1 - \sqrt{x} \]

### Step 2: Square Both Sides
Next, we square both sides to eliminate the square root:
\[ x - \sqrt{1 - x} = (1 - \sqrt{x})^2 \]
Expanding the right side:
\[ x - \sqrt{1 - x} = 1 - 2\sqrt{x} + x \]
Now, simplify the equation:
\[ -\sqrt{1 - x} = 1 - 2\sqrt{x} \]
Rearranging gives:
\[ \sqrt{1 - x} = 2\sqrt{x} - 1 \]

### Step 3: Square Again
Square both sides again:
\[ 1 - x = (2\sqrt{x} - 1)^2 \]
Expanding the right side:
\[ 1 - x = 4x - 4\sqrt{x} + 1 \]
Now, simplify:
\[ -x = 4x - 4\sqrt{x} \]
Rearranging gives:
\[ 5x = 4\sqrt{x} \]

### Step 4: Isolate the Square Root
Now, isolate the square root:
\[ \sqrt{x} = \frac{5x}{4} \]

### Step 5: Square Again
Square both sides once more:
\[ x = \left(\frac{5x}{4}\right)^2 \]
This simplifies to:
\[ x = \frac{25x^2}{16} \]
Rearranging gives:
\[ 25x^2 - 16x = 0 \]
Factoring out \( x \):
\[ x(25x - 16) = 0 \]

### Step 6: Find the Roots
Setting each factor to zero gives:
1. \( x = 0 \)
2. \( 25x - 16 = 0 \) which leads to \( x = \frac{16}{25} \)

### Step 7: Check the Validity of Roots
We have two potential roots: \( x = 0 \) and \( x = \frac{16}{25} \). We need to check if these values satisfy the original equation.

1. For \( x = 0 \):
\[ \sqrt{0} + \sqrt{0 - \sqrt{1 - 0}} = 0 + \sqrt{-1} \quad \text{(not valid)} \]

2. For \( x = \frac{16}{25} \):
\[ \sqrt{\frac{16}{25}} + \sqrt{\frac{16}{25} - \sqrt{1 - \frac{16}{25}}} \]
Calculate \( \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \):
\[ \sqrt{\frac{16}{25}} + \sqrt{\frac{16}{25} - \frac{3}{5}} = \frac{4}{5} + \sqrt{\frac{16}{25} - \frac{15}{25}} = \frac{4}{5} + \sqrt{\frac{1}{25}} = \frac{4}{5} + \frac{1}{5} = 1 \quad \text{(valid)} \]

### Conclusion
Thus, the only valid root is \( x = \frac{16}{25} \). Therefore, the number of real roots of the equation is **1**.