If ` 2 x + (1)/( 4 x ) = 1 ` then the value of ` x^(2) + (1)/( 64 x^(2)) ` is

Options

A. (a) 0
B. (b) 1
C. (c) `(1)/(4)`
D. (d) 2

Answer

To solve the equation \( 2x + \frac{1}{4x} = 1 \) and find the value of \( x^2 + \frac{1}{64x^2} \), we can follow these steps:

### Step 1: Rearrange the Equation
Start with the equation:
\[ 2x + \frac{1}{4x} = 1 \]
We can rearrange this to isolate the terms involving \( x \):
\[ 2x = 1 - \frac{1}{4x} \]

### Step 2: Eliminate the Fraction
To eliminate the fraction, multiply both sides by \( 4x \):
\[ 4x(2x) = 4x\left(1 - \frac{1}{4x}\right) \]
This simplifies to:
\[ 8x^2 = 4x - 1 \]

### Step 3: Rearrange into Standard Form
Rearranging gives us a quadratic equation:
\[ 8x^2 - 4x + 1 = 0 \]

### Step 4: Use the Quadratic Formula
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 8 \), \( b = -4 \), and \( c = 1 \):
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 8 \cdot 1 = 16 - 32 = -16 \]
Since the discriminant is negative, there are no real solutions for \( x \).

### Step 5: Find \( x^2 + \frac{1}{64x^2} \)
However, we can find \( x^2 + \frac{1}{64x^2} \) using the original equation. We know:
\[ (2x + \frac{1}{4x})^2 = 1^2 \]
Expanding the left side:
\[ (2x)^2 + 2(2x)(\frac{1}{4x}) + \left(\frac{1}{4x}\right)^2 = 1 \]
This simplifies to:
\[ 4x^2 + 1 + \frac{1}{16x^2} = 1 \]
Rearranging gives:
\[ 4x^2 + \frac{1}{16x^2} = 0 \]

### Step 6: Divide by 4
Dividing the entire equation by 4 gives:
\[ x^2 + \frac{1}{64x^2} = 0 \]

### Conclusion
Thus, the value of \( x^2 + \frac{1}{64x^2} \) is:
\[ \boxed{0} \]