`sqrt(x+14-8sqrt(x-2))+ sqrt(x+23-10sqrt(x-2))=3` Find number of real root

Answer

To solve the equation

\[ \sqrt{x + 14 - 8\sqrt{x - 2}} + \sqrt{x + 23 - 10\sqrt{x - 2}} = 3, \]

we will follow these steps:

### Step 1: Simplify the square roots

We can rewrite the expressions under the square roots. Let \( y = \sqrt{x - 2} \). Then, \( x = y^2 + 2 \).

Substituting \( x \) in the equation:

\[ \sqrt{(y^2 + 2) + 14 - 8y} + \sqrt{(y^2 + 2) + 23 - 10y} = 3 \]

This simplifies to:

\[ \sqrt{y^2 - 8y + 16} + \sqrt{y^2 - 10y + 25} = 3 \]

### Step 2: Further simplify the square roots

The expressions inside the square roots can be factored:

\[ \sqrt{(y - 4)^2} + \sqrt{(y - 5)^2} = 3 \]

This gives us:

\[ |y - 4| + |y - 5| = 3 \]

### Step 3: Analyze cases based on the absolute values

We will consider two cases based on the values of \( y \).

**Case 1:** \( y < 4 \)

In this case, both absolute values will be negative:

\[ -(y - 4) - (y - 5) = 3 \implies -2y + 9 = 3 \implies 2y = 6 \implies y = 3 \]

Since \( y < 4 \), this solution is valid.

**Case 2:** \( 4 \leq y < 5 \)

Here, we have:

\[ (y - 4) - (y - 5) = 3 \implies 1 = 3 \]

This case does not yield a valid solution.

**Case 3:** \( y \geq 5 \)

In this case, both absolute values will be positive:

\[ (y - 4) + (y - 5) = 3 \implies 2y - 9 = 3 \implies 2y = 12 \implies y = 6 \]

Since \( y \geq 5 \), this solution is valid.

### Step 4: Convert back to \( x \)

Now we convert the valid \( y \) values back to \( x \):

1. For \( y = 3 \):
\[ \sqrt{x - 2} = 3 \implies x - 2 = 9 \implies x = 11 \]

2. For \( y = 6 \):
\[ \sqrt{x - 2} = 6 \implies x - 2 = 36 \implies x = 38 \]

### Conclusion: Number of real roots

The valid \( x \) values we found are \( 11 \) and \( 38 \). Therefore, the number of real roots is:

\[ \boxed{2} \]