реши

Confirming behavior of discriminants for cubic roots

The discriminant of the quartic is always positive because it suggests no real roots, and the polynomial doesn’t cross zero. Interestingly, the cubic equation for s = 1 always seems to have three distinct real roots. After analyzing, I see the polynomial’s behavior is influenced by the ratio u = t/s, allowing us to simplify the analysis by setting s = 1.

Analyzing when cubic has three real roots for all u

I need to check if the polynomial has three real roots for all values of u. We’ve seen that the discriminant is always positive, suggesting three distinct real roots, but we still need a cleaner proof. Perhaps intermediate value theory will work—evaluating the polynomial at key points like -1, 1, and 2 might show that the sign changes and guarantees at least three real roots.

Determining the roots' behavior based on u

We can test the sign of the polynomial at key points to establish where the roots lie. For $u > 0$, we observe sign changes at $x = -1, 1, 2$, proving three roots in the intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, 2)$. For $u < 0$, signs reverse, showing one root for $u < 0$ after $x = 2$.

Confirming Three Distinct Real Roots

For $u > 0$, sign changes tell us there are three roots in intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, 2)$. For $u < 0$, we find roots in $(-1, 1)$, $(1, 2)$, and $(2, \infty)$. Since the polynomial is cubic, it has three distinct real roots, confirming they can't coincide based on these sign changes.

Ensuring Distinct Roots and Final Solution

The roots in disjoint intervals are distinct due to the Intermediate Value Theorem. For $u > 0$, we know there's a root in each interval: $(-\infty, -1)$, $(-1, 1)$, and $(1, 2)$. The polynomial cannot have a fourth root for degree 3. Thus, exactly three distinct roots exist.

Final conclusion on the three real roots

The cubic polynomial has three distinct real roots for any real value of $u$. For $u > 0$, there are roots in the intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, 2)$. For $u < 0$, the roots are in the intervals $(-1, 1)$, $(1, 2)$, and $(2, \infty)$, all confirmed with sign evaluations.

Thought for 3m 15s