Find the roots of equation `2x^4+x^3-11x^2+x+2=0`

Options

A. `(-8-sqrt5)/2,(-8+sqrt5)/2,1/2,2`
B. `(-3-sqrt5)/2,(-3+sqrt5)/2,-2,2`
C. `(-3-sqrt7)/2,(-3+sqrt7)/2,1/2,2`
D. `(-3-sqrt5)/2,(-3+sqrt5)/2,1/2,2`

Answer

To find the roots of the equation \( 2x^4 + x^3 - 11x^2 + x + 2 = 0 \), we can follow these steps:

### Step 1: Rewrite the Equation
The given equation is:
\[ 2x^4 + x^3 - 11x^2 + x + 2 = 0 \]

### Step 2: Divide by \( x^2 \)
To simplify the equation, we can divide the entire equation by \( x^2 \) (assuming \( x \neq 0 \)):
\[ \frac{2x^4}{x^2} + \frac{x^3}{x^2} - \frac{11x^2}{x^2} + \frac{x}{x^2} + \frac{2}{x^2} = 0 \]
This simplifies to:
\[ 2x^2 + x - 11 + \frac{1}{x} + \frac{2}{x^2} = 0 \]

### Step 3: Combine Terms
We can rewrite the equation as:
\[ 2x^2 + x - 11 + \left( x + \frac{1}{x} \right) + \left( \frac{2}{x^2} \right) = 0 \]
Let \( y = x + \frac{1}{x} \). Then, we have:
\[ x^2 + \frac{1}{x^2} = y^2 - 2 \]
Thus, the equation becomes:
\[ 2(y^2 - 2) + y - 11 = 0 \]
This simplifies to:
\[ 2y^2 + y - 15 = 0 \]

### Step 4: Factor the Quadratic Equation
Now we need to factor or use the quadratic formula to solve for \( y \):
\[ 2y^2 + y - 15 = 0 \]
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 1 + 120 = 121 \]
Thus:
\[ y = \frac{-1 \pm 11}{4} \]
This gives us two values for \( y \):
1. \( y = \frac{10}{4} = \frac{5}{2} \)
2. \( y = \frac{-12}{4} = -3 \)

### Step 5: Solve for \( x \)
Now we will solve for \( x \) in both cases.

**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:
\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 25 - 16 = 9 \]
Thus:
\[ x = \frac{5 \pm 3}{4} \]
This gives us:
1. \( x = 2 \)
2. \( x = \frac{1}{2} \)

**Case 2: \( y = -3 \)**
\[ x + \frac{1}{x} = -3 \]
Multiplying through by \( x \):
\[ x^2 + 3x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 9 - 4 = 5 \]
Thus:
\[ x = \frac{-3 \pm \sqrt{5}}{2} \]

### Final Roots
The roots of the original equation \( 2x^4 + x^3 - 11x^2 + x + 2 = 0 \) are:
1. \( x = 2 \)
2. \( x = \frac{1}{2} \)
3. \( x = \frac{-3 + \sqrt{5}}{2} \)
4. \( x = \frac{-3 - \sqrt{5}}{2} \)