1. Differentiate the function

Given $f(x)=2x^3-9ax^2+12a^2x+1,\, a>0$

For local maxima/minima, we use $f'(x)=0.$

Differentiate: $f'(x)=6x^2-18ax+12a^2$ $=6(x^2-3ax+2a^2)$ $=6(x-a)(x-2a).$

So the critical points are: $x=a \quad \text{and} \quad x=2a.$

2. Identify which is maximum and which is minimum

Now, $f''(x)=12x-18a.$

Check at the critical points:

• At $x=a$: $f''(a)=12a-18a=-6a<0 \quad (a>0)$ Hence, $x=a$ is a local maximum.

• At $x=2a$: $f''(2a)=24a-18a=6a>0$ Hence, $x=2a$ is a local minimum.

Thus, $\alpha=a, \qquad \alpha^2=2a.$

3. Relate $\alpha$ and $\alpha^2$

Since $\alpha=a$, substitute into $\alpha^2=2a$: $\alpha^2=2\alpha.$

So, $\alpha(\alpha-2)=0.$

But $a>0$ and $\alpha=a$, so $\alpha>0$. Hence, $\alpha=2.$

Then $\alpha^2=4.$

4. Form the quadratic equation whose roots are $\alpha$ and $\alpha^2$

The roots are $2$ and $4$.

So the quadratic is: $x^2-(2+4)x+(2)(4)=0$ $x^2-6x+8=0.$

5. Check options

• A: $x^2-6x+8=0$ ✅
• B: $8x^2-6x+1=0$
• C: $8x^2+6x-1=0$
• D: $x^2+6x+8=0$

Therefore, the correct option is A.