If $\alpha \neq \beta$ but, $\alpha^2 = 4\alpha - 2$ and $\beta^2 = 4\beta - 2$, then the quadratic equation with roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is
Step-by-Step Solution
Key Concept: When an equation involves an absolute value or a conditional expression based on a discriminant, split the solution into cases based on the condition
The equation $(x-1)[x^2 - 4x + 3] + 2x^2 + 5x - 5 = 0$ is analyzed by cases. When $x^2 - 4x + 3 \geq 0$ (i.e., $x \in (-\infty, 1] \cup [3, \infty)$), we have $(x-1)(x^2 - 3x + 3) + 2x^2 + 5x - 5 = 0$, which simplifies to $(x-1)(x^2 - 2x + 5) = 0$. Since the quadratic has $D < 0$, only $x = 1$ is valid. When $x^2 - 4x + 3 < 0$ (i.e., $x \in (1, 3)$), the equation becomes $(x-1)(-x^2 + 4x - 3 + 2x^2 + 5x - 5) = 0$, which gives $x = 1$ (boundary, excluded) or no additional solutions.
Correct Answer: 1