AM-GM Form of a Quadratic

For the quadratic
$$x^{2} - (t^{2}-5t+6)x + 1 = 0$$
let the two distinct roots be $$\alpha$$ and $$\beta$$.

By Vieta’s relations
$$\alpha + \beta = t^{2}-5t+6 \quad -(1)$$
$$\alpha\beta = 1 \quad -(2)$$

Define the sequence $$a_{n} = \alpha^{n} + \beta^{n}\;(n \ge 0).$$

Case 1: Recurrence for $$a_n$$
Using $$\alpha\beta = 1$$ we obtain a standard identity: for all $$n \ge 0$$
$$\alpha^{\,n+2} + \beta^{\,n+2} = (\alpha+\beta)\,(\alpha^{\,n+1} + \beta^{\,n+1}) - (\alpha\beta)\,(\alpha^{\,n} + \beta^{\,n}).$$
Hence
$$a_{\,n+2} = (\alpha+\beta)\,a_{\,n+1} - a_{\,n}\quad -(3)$$

Case 2: Evaluating the required ratio
Set $$n = 2023$$ in the recurrence $$-(3)$$:
$$a_{2025} = (\alpha+\beta)\,a_{2024} - a_{2023}\quad -(4)$$

Add $$a_{2023}$$ to both sides of $$-(4)$$:
$$a_{2023} + a_{2025} = (\alpha+\beta)\,a_{2024}$$

Therefore
$$\frac{a_{2023} + a_{2025}}{a_{2024}} = \alpha + \beta$$

From $$-(1)$$ the ratio equals
$$\alpha + \beta = t^{2} - 5t + 6\quad -(5)$$

Case 3: Minimising the expression
We must find the minimum of the quadratic function
$$f(t) = t^{2} - 5t + 6$$
for real $$t$$, under the condition that the roots $$\alpha, \beta$$ are distinct.

Write $$f(t)$$ in completed-square form:
$$f(t) = \bigl(t - \tfrac{5}{2}\bigr)^{2} - \tfrac{1}{4}$$

The square term is always non-negative, so the minimum value occurs at $$t = \tfrac{5}{2}$$ and equals
$$f_{\min} = -\tfrac{1}{4}$$

Case 4: Checking the conditions
(i) Distinct roots: the discriminant of the given quadratic in $$x$$ is
$$\Delta = (\alpha+\beta)^{2} - 4\alpha\beta = f(t)^{2} - 4$$

At $$t = \tfrac{5}{2}$$, $$f(t) = -\tfrac14$$, so
$$\Delta = \bigl(-\tfrac14\bigr)^{2} - 4 = \tfrac1{16} - 4 = -\tfrac{63}{16} \neq 0.$$
Thus the roots are indeed distinct (they form a non-real conjugate pair).

(ii) Denominator non-zero: with distinct roots, $$a_{2024} = \alpha^{2024} + \beta^{2024}$$ cannot vanish for all real $$t$$; in particular it is non-zero at $$t = \tfrac{5}{2}$$, so the ratio is well defined at the minimum.

Hence the minimum possible value of $$\dfrac{a_{2023}+a_{2025}}{a_{2024}}$$ is
$$-\tfrac14.$$

Option B is correct.

The equation is $$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$$.

Divide both sides by $$e^{2x}$$:

$$e^{2x} + 4e^x - 58 + 4e^{-x} + e^{-2x} = 0$$

$$(e^{2x} + e^{-2x}) + 4(e^x + e^{-x}) - 58 = 0$$

Let $$t = e^x + e^{-x}$$, where $$t \geq 2$$ (by AM-GM). Then $$t^2 = e^{2x} + 2 + e^{-2x}$$, so $$e^{2x} + e^{-2x} = t^2 - 2$$.

Substituting:

$$(t^2 - 2) + 4t - 58 = 0$$

$$t^2 + 4t - 60 = 0$$

$$(t + 10)(t - 6) = 0$$

Since $$t \geq 2$$, we have $$t = 6$$.

Now solving $$e^x + e^{-x} = 6$$:

$$e^{2x} - 6e^x + 1 = 0$$

$$e^x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$$

Both values $$3 + 2\sqrt{2} > 0$$ and $$3 - 2\sqrt{2} > 0$$ are valid.

$$x = \ln(3 + 2\sqrt{2})$$ or $$x = \ln(3 - 2\sqrt{2})$$

Note that $$(3+2\sqrt{2})(3-2\sqrt{2}) = 1$$, so the two solutions are negatives of each other, confirming they are distinct.

Hence the number of real solutions is $$\boxed{2}$$.