Given equation, $4(z^2 + \frac{1}{z^2}) + 16(z + \frac{1}{z}) - 57 = 0$. Find $x$ such that $z$ is rational.
Step-by-Step Solution
Key Concept: Substitute $z + \frac{1}{z} = y$ to reduce to a quadratic, then solve for $z$ from the resulting linear equations.
Let $z + \frac{1}{z} = y$, then $z^2 + \frac{1}{z^2} = y^2 - 2$. The equation becomes $4y^2 + 16y - 65 = 0$, giving $y = \frac{-16 \pm \sqrt{256 + 1040}}{8} = \frac{-16 \pm 36}{8}$, so $y = \frac{5}{2}$ or $y = -\frac{13}{2}$. When $y = \frac{5}{2}$, we have $z + \frac{1}{z} = \frac{5}{2}$, giving $2z^2 - 5z + 2 = 0$, so $z = 2$ or $z = \frac{1}{2}$. Since $z$ is rational, $x = 2$ or $x = \frac{1}{2}$.
Correct Answer: 2