The equation `(x+3-4(x-1)^(1//2))^(1//2)+(x+8-6(x-1)^(1//2))^(1//2)=1` has (A) no solution (B) only `1` solution (C) only `2` solutions (D) more than `2` solutions

Options

A. no solution
B. only `1` solution
C. only `2` solutions
D. more than `2` solutions

Explanation

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 \( t = \sqrt{x - 1} \)

This gives us \( x = t^2 + 1 \). Now, we can rewrite the equation in terms of \( t \):

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

### Step 2: Simplify the equation

Substituting \( x \) into the equation:

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

This simplifies to:

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

### Step 3: Rewrite using absolute values

The equation becomes:

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

### Step 4: Analyze the cases for \( t \)

We need to consider the different cases based on the values of \( t \):

1. **Case 1:** \( t < 2 \)
- Here, \( |t - 2| = 2 - t \) and \( |t - 3| = 3 - t \).
- The equation becomes:
\[ (2 - t) + (3 - t) = 1 \implies 5 - 2t = 1 \implies 2t = 4 \implies t = 2. \]
- This case gives us no valid solution since \( t < 2 \) is not satisfied.

2. **Case 2:** \( 2 \leq t < 3 \)
- Here, \( |t - 2| = t - 2 \) and \( |t - 3| = 3 - t \).
- The equation becomes:
\[ (t - 2) + (3 - t) = 1 \implies 1 = 1. \]
- This is true for all \( t \) in the interval \( [2, 3) \).

3. **Case 3:** \( t \geq 3 \)
- Here, \( |t - 2| = t - 2 \) and \( |t - 3| = t - 3 \).
- The equation becomes:
\[ (t - 2) + (t - 3) = 1 \implies 2t - 5 = 1 \implies 2t = 6 \implies t = 3. \]
- This case gives us a valid solution at \( t = 3 \).

### Step 5: Find the corresponding \( x \) values

From the intervals we found for \( t \):

- For \( 2 \leq t < 3 \):
- \( t = \sqrt{x - 1} \) implies \( 2 \leq \sqrt{x - 1} < 3 \).
- Squaring gives \( 4 \leq x - 1 < 9 \) or \( 5 \leq x < 10 \).

- For \( t = 3 \):
- \( t = 3 \) gives \( \sqrt{x - 1} = 3 \) or \( x - 1 = 9 \) which gives \( x = 10 \).

### Conclusion

Thus, the solutions for \( x \) are in the interval \( [5, 10] \), which includes all values from 5 to just below 10, and also includes the endpoint \( x = 10 \). Therefore, there are infinitely many solutions in this interval.

The answer is:

**(D) more than 2 solutions.**