If `2x-(1)/(2x)=6` then what is the value of `x^(2)+(1)/(16x^(2))` ?

Options

A. `(19)/(2)`
B. `(17)/(2)`
C. `(18)/(3)`
D. `(15)/(2)`

Answer

To solve the equation \(2x - \frac{1}{2x} = 6\) and find the value of \(x^2 + \frac{1}{16x^2}\), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 2x - \frac{1}{2x} = 6 \]
To eliminate the fraction, we can multiply both sides by \(2x\):
\[ 2x(2x) - 1 = 6(2x) \]
This simplifies to:
\[ 4x^2 - 1 = 12x \]

### Step 2: Rearrange into standard quadratic form
Now, we can rearrange the equation to bring all terms to one side:
\[ 4x^2 - 12x - 1 = 0 \]

### Step 3: Use the quadratic formula
To solve for \(x\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 4\), \(b = -12\), and \(c = -1\). Plugging in these values:
\[ x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ x = \frac{12 \pm \sqrt{144 + 16}}{8} \]
\[ x = \frac{12 \pm \sqrt{160}}{8} \]
\[ x = \frac{12 \pm 4\sqrt{10}}{8} \]
\[ x = \frac{3 \pm \sqrt{10}}{2} \]

### Step 4: Calculate \(x^2 + \frac{1}{16x^2}\)
Now we need to find \(x^2 + \frac{1}{16x^2}\). First, we can find \(x^2\):
\[ x^2 = \left(\frac{3 \pm \sqrt{10}}{2}\right)^2 = \frac{(3 \pm \sqrt{10})^2}{4} = \frac{9 \pm 6\sqrt{10} + 10}{4} = \frac{19 \pm 6\sqrt{10}}{4} \]

Now, we calculate \(\frac{1}{16x^2}\):
\[ \frac{1}{16x^2} = \frac{1}{16 \cdot \frac{19 \pm 6\sqrt{10}}{4}} = \frac{1}{4(19 \pm 6\sqrt{10})} = \frac{1}{76 \pm 24\sqrt{10}} \]

To find \(x^2 + \frac{1}{16x^2}\), we can simplify:
\[ x^2 + \frac{1}{16x^2} = \frac{19 \pm 6\sqrt{10}}{4} + \frac{1}{76 \pm 24\sqrt{10}} \]

However, we can also use the identity:
\[ \left(x - \frac{1}{4x}\right)^2 = x^2 + \frac{1}{16x^2} - 2 \cdot \frac{1}{4} \]
From our earlier steps, we know:
\[ x - \frac{1}{4x} = 3 \]
Thus,
\[ 3^2 = x^2 + \frac{1}{16x^2} - \frac{1}{2} \]
\[ 9 = x^2 + \frac{1}{16x^2} - \frac{1}{2} \]
\[ x^2 + \frac{1}{16x^2} = 9 + \frac{1}{2} = \frac{19}{2} \]

### Final Answer
Thus, the value of \(x^2 + \frac{1}{16x^2}\) is:
\[ \frac{19}{2} \]