In the real number system the equation
`sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x -1))`=1 has

Options

A. No solution
B. Exactly two distinct solution
C. Exactly four distinct solution
D. Infinitely many solution

Answer

To solve the equation

\[ \sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1, \]

we will follow these steps:

### Step 1: Substitute \( t = \sqrt{x - 1} \)
We start by letting \( t = \sqrt{x - 1} \). Then we have:

\[ x = t^2 + 1. \]

### Step 2: Rewrite the equation in terms of \( t \)
Substituting \( x \) into the equation gives:

\[ \sqrt{(t^2 + 1) + 3 - 4t} + \sqrt{(t^2 + 1) + 8 - 6t} = 1. \]

This simplifies to:

\[ \sqrt{t^2 - 4t + 4} + \sqrt{t^2 - 6t + 9} = 1. \]

### Step 3: Simplify the square roots
The expressions inside the square roots can be factored:

\[ \sqrt{(t - 2)^2} + \sqrt{(t - 3)^2} = 1. \]

This simplifies to:

\[ |t - 2| + |t - 3| = 1. \]

### Step 4: Analyze the absolute value equation
We need to consider different cases based on the values of \( t \):

1. **Case 1: \( t < 2 \)**
\[ -(t - 2) - (t - 3) = 1 \implies -2t + 5 = 1 \implies -2t = -4 \implies t = 2. \]

Since \( t < 2 \) is not satisfied, this case gives no solutions.

2. **Case 2: \( 2 \leq t < 3 \)**
\[ (t - 2) - (t - 3) = 1 \implies 1 = 1. \]

This is always true for \( 2 \leq t < 3 \).

3. **Case 3: \( t \geq 3 \)**
\[ (t - 2) + (t - 3) = 1 \implies 2t - 5 = 1 \implies 2t = 6 \implies t = 3. \]

This is valid since \( t \geq 3 \).

### Step 5: Determine the range of \( t \)
From the analysis, we find:

- \( t \) can take any value in the interval \( [2, 3] \).

### Step 6: Convert back to \( x \)
Recall that \( t = \sqrt{x - 1} \). Therefore,

\[ 2 \leq \sqrt{x - 1} \leq 3. \]

Squaring the inequalities gives:

\[ 4 \leq x - 1 \leq 9 \implies 5 \leq x \leq 10. \]

### Conclusion
The solution set for \( x \) is the interval \( [5, 10] \). Since this interval contains infinitely many values, we conclude that the equation has infinitely many solutions.