Solve in R 2x4 + 9x3 + 8x2 + 9x + 2 = 0

Substitute $x=0$ into the equation: $2(0)^4 + 9(0)^3 + 8(0)^2 + 9(0) + 2 = 2$. Since $2 \neq 0$, $x=0$ is not a solution.

Since $x \neq 0$, we can divide by $x^2$:

Factor out the common coefficients:

Let $y = x + \frac{1}{x}$.
Then, square both sides: $y^2 = \left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2}$.
So, $x^2 + \frac{1}{x^2} = y^2 - 2$.

We can factor this quadratic equation:

This gives two possible values for $y$:
$y + 4 = 0 \implies y = -4$
$2y + 1 = 0 \implies y = -\frac{1}{2}$

Case 1: $y = -4$

Multiply by $x$ to clear the denominator:

Rearrange into a standard quadratic form:

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

These are real solutions.

Case 2: $y = -\frac{1}{2}$

Multiply by $2x$ to clear the denominators:

Rearrange into a standard quadratic form:

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Since the discriminant is negative ($\sqrt{-15}$), these solutions are complex numbers and are not in $\mathbb{R}$.

The real solutions are $x = -2 + \sqrt{3}$ and $x = -2 - \sqrt{3}$.