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

To solve the equation $2x^4 + 9x^3 + 8x^2 + 9x + 2 = 0$ in $\mathbb{R}$, we recognize it as a reciprocal equation because the coefficients are symmetric.

Step 1: Check if $x=0$ is a solution. 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.

Step 2: Divide the entire equation by $x^2$. Since $x \neq 0$, we can divide by $x^2$: $\frac{2x^4}{x^2} + \frac{9x^3}{x^2} + \frac{8x^2}{x^2} + \frac{9x}{x^2} + \frac{2}{x^2} = 0$ $2x^2 + 9x + 8 + \frac{9}{x} + \frac{2}{x^2} = 0$

Step 3: Group terms with common coefficients. $\left(2x^2 + \frac{2}{x^2}\right) + \left(9x + \frac{9}{x}\right) + 8 = 0$ Factor out the common coefficients: $2\left(x^2 + \frac{1}{x^2}\right) + 9\left(x + \frac{1}{x}\right) + 8 = 0$

Step 4: Make a substitution. 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, $x^2 + \frac{1}{x^2} = y^2 - 2$.

Step 5: Substitute $y$ into the equation. $2(y^2 - 2) + 9y + 8 = 0$ $2y^2 - 4 + 9y + 8 = 0$ $2y^2 + 9y + 4 = 0$

Step 6: Solve the quadratic equation for $y$. We can factor this quadratic equation: $2y^2 + y + 8y + 4 = 0$ $y(2y + 1) + 4(2y + 1) = 0$ $(y + 4)(2y + 1) = 0$ This gives two possible values for $y$: $y + 4 = 0 \implies y = -4$ $2y + 1 = 0 \implies y = -\frac{1}{2}$

Step 7: Substitute back $y = x + \frac{1}{x}$ and solve for $x$.

Case 1: $y = -4$ $x + \frac{1}{x} = -4$ Multiply by $x$ to clear the denominator: $x^2 + 1 = -4x$ Rearrange into a standard quadratic form: $x^2 + 4x + 1 = 0$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)}$ $x = \frac{-4 \pm \sqrt{16 - 4}}{2}$ $x = \frac{-4 \pm \sqrt{12}}{2}$ $x = \frac{-4 \pm 2\sqrt{3}}{2}$ $x = -2 \pm \sqrt{3}$ These are real solutions.

Case 2: $y = -\frac{1}{2}$ $x + \frac{1}{x} = -\frac{1}{2}$ Multiply by $2x$ to clear the denominators: $2x^2 + 2 = -x$ Rearrange into a standard quadratic form: $2x^2 + x + 2 = 0$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-1 \pm \sqrt{1^2 - 4(2)(2)}}{2(2)}$ $x = \frac{-1 \pm \sqrt{1 - 16}}{4}$ $x = \frac{-1 \pm \sqrt{-15}}{4}$ 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 $x = -2 - \sqrt{3}$.

The final answer is $\boxed{x = -2 \pm \sqrt{3}}$. That's 2 down. 3 left today — send the next one.