The number of real solutions the equation `sqrt(x+14-8sqrt(x-2))+sqrt(x+23-10sqrt(x-2))=3` are

Options

A. 2
B. 4
C. 8
D. infinite

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: Substitute for \( \sqrt{x - 2} \)
Let \( \sqrt{x - 2} = t \). Then, we have:
\[ x = t^2 + 2 \]
This substitution simplifies our equation.

### Step 2: Rewrite the equation
Substituting \( x \) in the original equation gives:
\[ \sqrt{(t^2 + 2) + 14 - 8t} + \sqrt{(t^2 + 2) + 23 - 10t} = 3 \]
This simplifies to:
\[ \sqrt{t^2 - 8t + 16} + \sqrt{t^2 - 10t + 25} = 3 \]

### Step 3: Simplify the square roots
Notice that:
\[ \sqrt{t^2 - 8t + 16} = \sqrt{(t - 4)^2} = |t - 4| \]
\[ \sqrt{t^2 - 10t + 25} = \sqrt{(t - 5)^2} = |t - 5| \]
Thus, the equation becomes:
\[ |t - 4| + |t - 5| = 3 \]

### Step 4: Solve the absolute value equation
We will consider different cases based on the values of \( t \).

#### Case 1: \( t < 4 \)
In this case, both \( t - 4 < 0 \) and \( t - 5 < 0 \):
\[ -(t - 4) - (t - 5) = 3 \implies -t + 4 - t + 5 = 3 \implies -2t + 9 = 3 \implies -2t = -6 \implies t = 3 \]
Since \( t = 3 < 4 \), this solution is valid.

#### Case 2: \( 4 \leq t < 5 \)
Here, \( t - 4 \geq 0 \) and \( t - 5 < 0 \):
\[ (t - 4) - (t - 5) = 3 \implies t - 4 - t + 5 = 3 \implies 1 = 3 \]
This is a contradiction, so there are no solutions in this case.

#### Case 3: \( t \geq 5 \)
In this case, both \( t - 4 \geq 0 \) and \( t - 5 \geq 0 \):
\[ (t - 4) + (t - 5) = 3 \implies t - 4 + t - 5 = 3 \implies 2t - 9 = 3 \implies 2t = 12 \implies t = 6 \]
Since \( t = 6 \geq 5 \), this solution is valid.

### Step 5: Find corresponding values of \( x \)
We have two valid values for \( t \):
1. \( t = 3 \)
2. \( t = 6 \)

Using the relation \( x = t^2 + 2 \):
- For \( t = 3 \):
\[ x = 3^2 + 2 = 9 + 2 = 11 \]
- For \( t = 6 \):
\[ x = 6^2 + 2 = 36 + 2 = 38 \]

### Conclusion
The number of real solutions to the equation is 2, corresponding to \( x = 11 \) and \( x = 38 \).