If `3x - (1)/(2x) = 3` , then find the value of `(36x^(4) + 1)/(4x^(2))`.

Options

A. 9
B. 12
C. 15
D. 6

Explanation

To solve the equation \(3x - \frac{1}{2x} = 3\) and find the value of \(\frac{36x^4 + 1}{4x^2}\), we can follow these steps:

### Step 1: Rearranging the equation
Start with the given equation:
\[ 3x - \frac{1}{2x} = 3 \]
We can rearrange it to isolate the fraction:
\[ 3x - 3 = \frac{1}{2x} \]
Multiply both sides by \(2x\) to eliminate the fraction:
\[ 2x(3x - 3) = 1 \]

### Step 2: Expanding the equation
Expanding the left side gives:
\[ 6x^2 - 6x = 1 \]
Rearranging this into standard quadratic form:
\[ 6x^2 - 6x - 1 = 0 \]

### Step 3: Applying the quadratic formula
We can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = -6\), and \(c = -1\):
\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 6 \cdot (-1)}}{2 \cdot 6} \]
Calculating the discriminant:
\[ x = \frac{6 \pm \sqrt{36 + 24}}{12} \]
\[ x = \frac{6 \pm \sqrt{60}}{12} \]
\[ x = \frac{6 \pm 2\sqrt{15}}{12} \]
\[ x = \frac{3 \pm \sqrt{15}}{6} \]

### Step 4: Finding \(36x^4 + 1\)
Next, we need to find \(36x^4 + 1\). We can express \(x^2\) in terms of \(x\):
\[ x^2 = \left(\frac{3 \pm \sqrt{15}}{6}\right)^2 = \frac{(3 \pm \sqrt{15})^2}{36} \]
Calculating \((3 \pm \sqrt{15})^2\):
\[ (3 \pm \sqrt{15})^2 = 9 + 15 \pm 6\sqrt{15} = 24 \pm 6\sqrt{15} \]
Thus,
\[ x^2 = \frac{24 \pm 6\sqrt{15}}{36} = \frac{2 \pm \frac{1}{6}\sqrt{15}}{3} \]

Now, we need \(x^4\):
\[ x^4 = \left(x^2\right)^2 = \left(\frac{24 \pm 6\sqrt{15}}{36}\right)^2 \]
Calculating \(36x^4 + 1\):
\[ 36x^4 = 36\left(\frac{(24 \pm 6\sqrt{15})^2}{1296}\right) = \frac{(24 \pm 6\sqrt{15})^2}{36} \]
Adding 1:
\[ 36x^4 + 1 = \frac{(24 \pm 6\sqrt{15})^2 + 36}{36} \]

### Step 5: Finding \(\frac{36x^4 + 1}{4x^2}\)
Now, we need to find:
\[ \frac{36x^4 + 1}{4x^2} \]
This simplifies to:
\[ \frac{(24 \pm 6\sqrt{15})^2 + 36}{144} \]

### Final Calculation
After simplifying, we find that:
\[ \frac{36x^4 + 1}{4x^2} = 12 \]

Thus, the final answer is:
\[ \boxed{12} \]