If ` 2 (x^(2) + (1)/( x^(2))) - ( x - (1)/( x)) - 7 = 0 ` then two values of x are

Options

A. 1,2
B. `2 , - (1)/(2)`
C. 0,1
D. `(1)/(2),1`

Answer

To solve the equation \( 2 \left( x^2 + \frac{1}{x^2} \right) - \left( x - \frac{1}{x} \right) - 7 = 0 \), we can follow these steps:

### Step 1: Simplify the equation
Start by rewriting the equation:
\[ 2 \left( x^2 + \frac{1}{x^2} \right) - \left( x - \frac{1}{x} \right) - 7 = 0 \]

### Step 2: Substitute \( y = x - \frac{1}{x} \)
We know that:
\[ x^2 + \frac{1}{x^2} = \left( x - \frac{1}{x} \right)^2 + 2 = y^2 + 2 \]
Thus, the equation becomes:
\[ 2 \left( y^2 + 2 \right) - y - 7 = 0 \]

### Step 3: Expand and rearrange
Expanding the equation gives:
\[ 2y^2 + 4 - y - 7 = 0 \]
This simplifies to:
\[ 2y^2 - y - 3 = 0 \]

### Step 4: Factor the quadratic equation
Next, we can factor the quadratic equation:
\[ 2y^2 - 3y + 2y - 3 = 0 \]
Grouping gives:
\[ (2y^2 - 3y) + (2y - 3) = 0 \]
Factoring out common terms:
\[ y(2y - 3) + 1(2y - 3) = 0 \]
This can be factored as:
\[ (2y - 3)(y + 1) = 0 \]

### Step 5: Solve for \( y \)
Setting each factor to zero gives:
1. \( 2y - 3 = 0 \) → \( y = \frac{3}{2} \)
2. \( y + 1 = 0 \) → \( y = -1 \)

### Step 6: Substitute back to find \( x \)
Recall that \( y = x - \frac{1}{x} \):
1. For \( y = \frac{3}{2} \):
\[ x - \frac{1}{x} = \frac{3}{2} \]
Multiplying through by \( x \) gives:
\[ x^2 - \frac{3}{2}x - 1 = 0 \]
Multiplying by 2 to eliminate the fraction:
\[ 2x^2 - 3x - 2 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4} \]
This gives:
\[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-2}{4} = -\frac{1}{2} \]

2. For \( y = -1 \):
\[ x - \frac{1}{x} = -1 \]
Multiplying through by \( x \):
\[ x^2 + x - 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{5}}{2} \]

### Final Values of \( x \)
The two values of \( x \) are:
1. \( x = 2 \)
2. \( x = -\frac{1}{2} \)
3. \( x = \frac{-1 \pm \sqrt{5}}{2} \)