If the roots of the equation `x^(2) - 4x- log_(3) N =0` are real, then what is the minimum vlaue of N?

Options

A. `1//256`
B. `1//27`
C. `1//64`
D. `1//81`

Explanation

To find the minimum value of \( N \) such that the roots of the equation

\[ x^2 - 4x - \log_3 N = 0 \]

are real, we will use the condition that the discriminant of the quadratic equation must be non-negative.

### Step 1: Identify the coefficients
The given quadratic equation can be compared with the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 1 \)
- \( b = -4 \)
- \( c = -\log_3 N \)

### Step 2: Write the discriminant condition
The discriminant \( D \) of a quadratic equation is given by:

\[ D = b^2 - 4ac \]

For the roots to be real, we need:

\[ D \geq 0 \]

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

\[ (-4)^2 - 4 \cdot 1 \cdot (-\log_3 N) \geq 0 \]

### Step 3: Simplify the discriminant
Calculating the discriminant:

\[ 16 + 4\log_3 N \geq 0 \]

### Step 4: Rearranging the inequality
Rearranging the inequality gives:

\[ 4\log_3 N \geq -16 \]

Dividing both sides by 4:

\[ \log_3 N \geq -4 \]

### Step 5: Exponentiating to eliminate the logarithm
Using the property of logarithms, we can rewrite this as:

\[ N \geq 3^{-4} \]

Calculating \( 3^{-4} \):

\[ N \geq \frac{1}{3^4} = \frac{1}{81} \]

### Conclusion
Thus, the minimum value of \( N \) such that the roots of the equation are real is:

\[ \boxed{\frac{1}{81}} \]