Quadratic Equation from Sum and Product

Given $$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$$ with $$a > 0$$. Differentiating: $$f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a)$$. The critical points are $$x = a$$ and $$x = 2a$$.

To classify these, we use the second derivative $$f''(x) = 12x - 18a$$. At $$x = a$$: $$f''(a) = 12a - 18a = -6a$$. Since $$a > 0$$, we have $$f''(a) < 0$$, confirming a local maximum at $$x = a$$. At $$x = 2a$$: $$f''(2a) = 24a - 18a = 6a > 0$$, confirming a local minimum at $$x = 2a$$.

We are told the local maximum occurs at $$x = \alpha$$ and the local minimum at $$x = \alpha^2$$. So $$\alpha = a$$ and $$\alpha^2 = 2a$$. Substituting $$a = \alpha$$ into the second equation: $$\alpha^2 = 2\alpha$$, giving $$\alpha(\alpha - 2) = 0$$. Since $$\alpha = a > 0$$, we get $$\alpha = 2$$, and therefore $$\alpha^2 = 4$$.

The required equation has roots $$2$$ and $$4$$. Their sum is $$6$$ and product is $$8$$, so the equation is $$x^2 - 6x + 8 = 0$$, which is $$\boxed{\text{Option (A)}}$$.

Given $$x^2 + 2\sqrt{2}x - 1 = 0$$ with roots $$\alpha, \beta$$.

By Vieta's: $$\alpha + \beta = -2\sqrt{2}$$, $$\alpha\beta = -1$$.

$$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 8 + 2 = 10$$

$$\alpha^4 + \beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = 100 - 2 = 98$$

$$\alpha^6 + \beta^6 = (\alpha^2+\beta^2)^3 - 3(\alpha\beta)^2(\alpha^2+\beta^2) = 1000 - 3(1)(10) = 970$$

So the two roots of the required equation are: $$\alpha^4 + \beta^4 = 98$$ and $$\frac{1}{10}(\alpha^6+\beta^6) = \frac{970}{10} = 97$$.

Sum of roots = $$98 + 97 = 195$$.

Product of roots = $$98 \times 97 = 9506$$.

The equation is $$x^2 - 195x + 9506 = 0$$.

The correct answer is Option 3.

We are given $$p + q = 2$$ and $$p^4 + q^4 = 272$$, and we need to find the quadratic equation whose roots are $$p$$ and $$q$$.

We know that $$p^2 + q^2 = (p + q)^2 - 2pq = 4 - 2pq$$.

Now, $$p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2$$. Substituting, we get $$(4 - 2pq)^2 - 2(pq)^2 = 272$$.

Let $$s = pq$$. Then $$16 - 16s + 4s^2 - 2s^2 = 272$$, which gives $$2s^2 - 16s + 16 = 272$$.

So $$2s^2 - 16s - 256 = 0$$, or $$s^2 - 8s - 128 = 0$$.

Using the quadratic formula, $$s = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm \sqrt{576}}{2} = \frac{8 \pm 24}{2}$$.

This gives $$s = 16$$ or $$s = -8$$.

The quadratic equation with sum of roots $$p + q = 2$$ and product $$pq = 16$$ is $$x^2 - 2x + 16 = 0$$, and with $$pq = -8$$ it is $$x^2 - 2x - 8 = 0$$. Among the given options, only $$x^2 - 2x + 16 = 0$$ is present.

Hence, the correct answer is Option D.