Consider the equation $x^4 - (k-1)x^2 + (2-k) = 0$. The complete set of possible values of real $k$ for which the equation has 3 distinct real roots is:
Step-by-Step Solution
Key Concept: A quartic with $x^2$ symmetry has 3 distinct real roots only when the substitution $y = x^2$ yields a quadratic with roots 0 and a positive value, producing the roots $0, \pm\sqrt{y_2}$.
Let $y = x^2$ to get $y^2 - (k-1)y + (2-k) = 0$. For the original equation to have exactly 3 distinct real roots, one root of this quadratic must be 0 and the other must be positive. If $y = 0$: $2 - k = 0$, so $k = 2$. When $k = 2$, the quadratic becomes $y^2 - y = 0$, giving $y(y-1) = 0$, so $y = 0$ or $y = 1$. Since $y = x^2$, we get $x = 0$ (one root) and $x = \pm 1$ (two roots), totaling 3 distinct real roots. Verification shows $k = 2$ is the only value satisfying this condition.
Correct Answer: 1