Solve `6(x^(2)+(1)/(x^(2)))-25(x-(1)/(x))+12=0`

Answer

To solve the equation \( 6\left(x^2 + \frac{1}{x^2}\right) - 25\left(x - \frac{1}{x}\right) + 12 = 0 \), we can follow these steps:

### Step 1: Substitute Variables
Let \( k = x - \frac{1}{x} \). Then, we can express \( x^2 + \frac{1}{x^2} \) in terms of \( k \):
\[ x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 = k^2 + 2 \]

### Step 2: Rewrite the Equation
Substituting \( x^2 + \frac{1}{x^2} \) into the original equation gives:
\[ 6(k^2 + 2) - 25k + 12 = 0 \]
Expanding this, we have:
\[ 6k^2 + 12 - 25k + 12 = 0 \]
This simplifies to:
\[ 6k^2 - 25k + 24 = 0 \]

### Step 3: Factor the Quadratic Equation
Now we need to factor the quadratic equation \( 6k^2 - 25k + 24 = 0 \). We look for two numbers that multiply to \( 6 \times 24 = 144 \) and add to \( -25 \). The numbers are \( -16 \) and \( -9 \):
\[ 6k^2 - 16k - 9k + 24 = 0 \]
Grouping the terms:
\[ 2k(3k - 8) - 3(3k - 8) = 0 \]
Factoring out \( (3k - 8) \):
\[ (3k - 8)(2k - 3) = 0 \]

### Step 4: Solve for \( k \)
Setting each factor to zero gives:
1. \( 3k - 8 = 0 \) → \( k = \frac{8}{3} \)
2. \( 2k - 3 = 0 \) → \( k = \frac{3}{2} \)

### Step 5: Solve for \( x \)
Recall that \( k = x - \frac{1}{x} \). We will solve for \( x \) for both values of \( k \).

#### Case 1: \( k = \frac{8}{3} \)
\[ x - \frac{1}{x} = \frac{8}{3} \]
Multiplying through by \( x \):
\[ x^2 - \frac{8}{3}x - 1 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 3x^2 - 8x - 3 = 0 \]
Using the quadratic formula:
\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]
\[ x = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm 10}{6} \]
This gives:
1. \( x = \frac{18}{6} = 3 \)
2. \( x = \frac{-2}{6} = -\frac{1}{3} \)

#### Case 2: \( k = \frac{3}{2} \)
\[ x - \frac{1}{x} = \frac{3}{2} \]
Multiplying through by \( x \):
\[ x^2 - \frac{3}{2}x - 1 = 0 \]
Multiplying through by 2:
\[ 2x^2 - 3x - 2 = 0 \]
Using the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \]
This gives:
1. \( x = \frac{8}{4} = 2 \)
2. \( x = \frac{-2}{4} = -\frac{1}{2} \)

### Final Solutions
Thus, the solutions for \( x \) are:
\[ x = 3, \quad x = -\frac{1}{3}, \quad x = 2, \quad x = -\frac{1}{2} \]