Solve the following equations :
`2(x^(2)+(1)/(x^(2)))-3(x+(1)/(x))=1`

Answer

To solve the equation \( 2\left(x^2 + \frac{1}{x^2}\right) - 3\left(x + \frac{1}{x}\right) = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2\left(x^2 + \frac{1}{x^2}\right) - 3\left(x + \frac{1}{x}\right) = 1 \]
We can express \( x^2 + \frac{1}{x^2} \) in terms of \( x + \frac{1}{x} \). Recall that:
\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \]
Let \( y = x + \frac{1}{x} \). Then, we can rewrite \( x^2 + \frac{1}{x^2} \) as:
\[ x^2 + \frac{1}{x^2} = y^2 - 2 \]
Substituting this back into the equation gives:
\[ 2(y^2 - 2) - 3y = 1 \]

### Step 2: Simplify the equation
Now, simplify the equation:
\[ 2y^2 - 4 - 3y = 1 \]
Adding 4 to both sides:
\[ 2y^2 - 3y - 5 = 0 \]

### Step 3: Solve the quadratic equation
Now we need to solve the quadratic equation \( 2y^2 - 3y - 5 = 0 \). We can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = -3 \), and \( c = -5 \). Plugging in these values:
\[ y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \]
\[ y = \frac{3 \pm \sqrt{9 + 40}}{4} \]
\[ y = \frac{3 \pm \sqrt{49}}{4} \]
\[ y = \frac{3 \pm 7}{4} \]

### Step 4: Find the values of \( y \)
Calculating the two possible values for \( y \):
1. \( y = \frac{10}{4} = \frac{5}{2} \)
2. \( y = \frac{-4}{4} = -1 \)

### Step 5: Solve for \( x \)
Now we will solve for \( x \) using \( y = x + \frac{1}{x} \).

**Case 1:** \( y = \frac{5}{2} \)
\[ x + \frac{1}{x} = \frac{5}{2} \]
Multiplying through by \( 2x \):
\[ 2x^2 - 5x + 2 = 0 \]
Using the quadratic formula:
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ x = \frac{5 \pm \sqrt{25 - 16}}{4} \]
\[ x = \frac{5 \pm 3}{4} \]
This gives:
1. \( x = \frac{8}{4} = 2 \)
2. \( x = \frac{2}{4} = \frac{1}{2} \)

**Case 2:** \( y = -1 \)
\[ x + \frac{1}{x} = -1 \]
Multiplying through by \( x \):
\[ x^2 + x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{-3}}{2} \]
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]

### Final Solutions
Thus, the solutions for \( x \) are:
1. \( x = 2 \)
2. \( x = \frac{1}{2} \)
3. \( x = \frac{-1 + i\sqrt{3}}{2} \)
4. \( x = \frac{-1 - i\sqrt{3}}{2} \)