Irrational roots of the equaiton `2x^(4) + 9x^(3) + 8x^(2) + 9x + 2 = 0` are

Options

A. `- 2 - sqrt(3) , 2 + sqrt(3)`
B. `2 - sqrt(3), 2 + sqrt(3)`
C. `- 2 + sqrt(3) , - 2 -sqrt(3)`
D. none of these

Answer

To find the irrational roots of the equation \(2x^4 + 9x^3 + 8x^2 + 9x + 2 = 0\), we can follow these steps:

### Step 1: Substitute and Simplify
We start with the given equation:
\[ 2x^4 + 9x^3 + 8x^2 + 9x + 2 = 0 \]
We can divide the entire equation by \(x^2\) (assuming \(x \neq 0\)):
\[ 2x^2 + 9x + 8 + \frac{9}{x} + \frac{2}{x^2} = 0 \]
This can be rearranged to:
\[ 2x^2 + 9x + 8 + 9\frac{1}{x} + 2\frac{1}{x^2} = 0 \]

### Step 2: Introduce a New Variable
Let \(y = x + \frac{1}{x}\). Then we can express \(x^2 + \frac{1}{x^2}\) in terms of \(y\):
\[ x^2 + \frac{1}{x^2} = y^2 - 2 \]
Now, substituting into the equation, we get:
\[ 2(y^2 - 2) + 9y + 8 = 0 \]
This simplifies to:
\[ 2y^2 + 9y + 4 = 0 \]

### Step 3: Solve the Quadratic Equation
Now we will solve the quadratic equation \(2y^2 + 9y + 4 = 0\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a = 2\), \(b = 9\), and \(c = 4\):
\[ y = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 9^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49 \]
Thus,
\[ y = \frac{-9 \pm \sqrt{49}}{4} = \frac{-9 \pm 7}{4} \]
Calculating the two possible values for \(y\):
1. \(y_1 = \frac{-9 + 7}{4} = \frac{-2}{4} = -\frac{1}{2}\)
2. \(y_2 = \frac{-9 - 7}{4} = \frac{-16}{4} = -4\)

### Step 4: Find \(x\) from \(y\)
Now we will find \(x\) for both values of \(y\).

**For \(y_1 = -\frac{1}{2}\):**
\[ x + \frac{1}{x} = -\frac{1}{2} \]
Multiplying by \(x\):
\[ x^2 + \frac{1}{2}x + 1 = 0 \]
The discriminant is:
\[ \left(\frac{1}{2}\right)^2 - 4 \cdot 1 \cdot 1 = \frac{1}{4} - 4 = \frac{1}{4} - \frac{16}{4} = -\frac{15}{4} \]
Since the discriminant is negative, the roots are non-real.

**For \(y_2 = -4\):**
\[ x + \frac{1}{x} = -4 \]
Multiplying by \(x\):
\[ x^2 + 4x + 1 = 0 \]
The discriminant is:
\[ 4^2 - 4 \cdot 1 \cdot 1 = 16 - 4 = 12 \]
Thus,
\[ x = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \]
These roots, \(-2 + \sqrt{3}\) and \(-2 - \sqrt{3}\), are irrational.

### Final Answer
The irrational roots of the equation \(2x^4 + 9x^3 + 8x^2 + 9x + 2 = 0\) are:
\[ x = -2 + \sqrt{3} \quad \text{and} \quad x = -2 - \sqrt{3} \]