If the product of the roots of the equation `x^(2) - 2sqrt(2) kx + 2e^(2 log k) -1 = 0` is 31, then the roots of the equation are real for k equal to

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the value of \( k \) such that the product of the roots of the quadratic equation

\[ x^2 - 2\sqrt{2}kx + (2e^{2 \log k} - 1) = 0 \]

is equal to 31.

### Step 1: Identify the coefficients
In the general form of a quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = -2\sqrt{2}k \)
- \( c = 2e^{2 \log k} - 1 \)

### Step 2: Use the product of roots formula
The product of the roots of a quadratic equation is given by the formula:

\[ \text{Product of roots} = \frac{c}{a} \]

Substituting the values of \( c \) and \( a \):

\[ \text{Product of roots} = 2e^{2 \log k} - 1 \]

### Step 3: Set the product equal to 31
According to the problem, the product of the roots is equal to 31. Therefore, we can set up the equation:

\[ 2e^{2 \log k} - 1 = 31 \]

### Step 4: Solve for \( e^{2 \log k} \)
Adding 1 to both sides:

\[ 2e^{2 \log k} = 32 \]

Dividing both sides by 2:

\[ e^{2 \log k} = 16 \]

### Step 5: Simplify \( e^{2 \log k} \)
Using the property of logarithms, we know that \( e^{\log a} = a \). Therefore:

\[ e^{2 \log k} = (e^{\log k})^2 = k^2 \]

So we can rewrite the equation as:

\[ k^2 = 16 \]

### Step 6: Solve for \( k \)
Taking the square root of both sides gives us:

\[ k = 4 \quad \text{or} \quad k = -4 \]

### Step 7: Determine the valid value of \( k \)
Since \( k \) must be greater than 0 for \( \log k \) to be defined, we discard \( k = -4 \). Thus, the only valid solution is:

\[ k = 4 \]

### Conclusion
The value of \( k \) for which the roots of the equation are real is:

\[ \boxed{4} \]