<Question>
Let $p(x)$ be a quadratic polynomial with integer coefficients which has $4-\sqrt{11}$ as a root. Compute $\frac{p(3)}{p(4)}.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the value of p(3)/p(4) where p(x) is a quadratic polynomial with integer coefficients, and it has 4 - √11 as a root. Hmm, let me think. Since the polynomial has integer coefficients and 4 - √11 is a root, there must be another root that's its conjugate. Right? Because for polynomials with integer coefficients, irrational roots come in conjugate pairs.
So the other root should be 4 + √11. That makes sense. So the quadratic polynomial can be written as p(x) = a(x - (4 - √11))(x - (4 + √11)), where a is an integer.

Let me expand that. First, multiply the two factors. Let's see, (x - 4 + √11)(x - 4 - √11) is a difference of squares.

<prompt_1.1> Expand the polynomial using the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
So that would be (x - 4)^2 - (√11)^2. Calculating that, (x - 4)^2 is x² - 8x + 16, and (√11)^2 is 11. So subtracting 11 from that gives x² - 8x + 16 - 11 = x² - 8x + 5. So the quadratic polynomial is p(x) = a(x² - 8x + 5). Since the problem says it's a quadratic polynomial with integer coefficients, a must be an integer. But since they don't specify any leading coefficient, I think we can assume a = 1 for simplicity unless there's a reason to choose another value. So p(x) = x² - 8x + 5.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, I need to compute p(3) and p(4). Let's calculate p(3) first.
Plugging x = 3 into the polynomial: 3² - 8*3 + 5. That's 9 - 24 + 5. 9 - 24 is -15, plus 5 is -10. So p(3) = -10.

Next, p(4).
Plugging x = 4 into the polynomial: 4² - 8*4 + 5. That's 16 - 32 + 5. 16 - 32 is -16, plus 5 is -11. So p(4) = -11.

Therefore, the ratio p(3)/p(4) is (-10)/(-11) which simplifies to 10/11.

Wait, but let me double-check my calculations to make sure I didn't make any mistakes.

<prompt_2.1> Verify the calculations for p(3) and p(4) using the above plan. </prompt_2.1>

Is there another possible polynomial with a different leading coefficient a that could also have 4 - √11 as a root? For example, if a is 2, then the polynomial would be 2x² - 16x + 10.

<prompt_2.2> Consider the effect of different leading coefficients on p(3) and p(4) using the above plan. </prompt_2.2>

Therefore, the answer is 10/11. But just to make sure, let me verify the polynomial again.

<prompt_2.3> Re-verify the polynomial and the calculations using the above plan. </prompt_2.3>

Wait, but just to make sure, let me consider if there's another way to approach this problem. For example, suppose someone didn't construct the polynomial explicitly. Maybe using the properties of quadratic polynomials.

<prompt_2.4> Explore alternative methods to solve the problem using the above plan. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
First, p(3): 3 squared is 9, 8 times 3 is 24, so 9 - 24 is -15, plus 5 is -10. Correct. Then p(4): 4 squared is 16, 8 times 4 is 32, so 16 - 32 is -16, plus 5 is -11. Correct. So the ratio is -10 divided by -11, which is positive 10/11.
</execution_2.1>

<execution_2.2>
Then p(3) would be 2*(9) -16*3 +10 = 18 -48 +10 = -20, and p(4) would be 2*(16) -16*4 +10 = 32 -64 +10 = -22. Then the ratio would be (-20)/(-22) = 10/11. Oh, interesting! So even if the leading coefficient is different, as long as it's a quadratic with integer coefficients, the ratio p(3)/p(4) remains the same. Because when you factor out the leading coefficient a, both p(3) and p(4) would be multiplied by a, so a/a cancels out. Therefore, the ratio is independent of the leading coefficient. So regardless of the value of a (as long as it's non-zero integer), the ratio will always be 10/11.
</execution_2.2>

<execution_2.3>
If a is 2, then p(3) would be 2*(9) -16*3 +10 = 18 -48 +10 = -20, and p(4) would be 2*(16) -16*4 +10 = 32 -64 +10 = -22. Then the ratio would be (-20)/(-22) = 10/11. So same result. So even if a is different integers, the ratio remains the same. That's because when you scale the polynomial by a constant factor, both p(3) and p(4) are scaled by the same factor, so their ratio is unaffected. Therefore, regardless of the leading coefficient, the ratio p(3)/p(4) is 10/11.
</execution_2.3>

<execution_2.4>
Let’s denote the quadratic polynomial as p(x) = ax² + bx + c, with integer coefficients. Given that 4 - √11 is a root, then by the conjugate root theorem, 4 + √11 must also be a root. Therefore, the sum of the roots is (4 - √11) + (4 + √11) = 8. The product of the roots is (4 - √11)(4 + √11) = 16 - 11 = 5. Therefore, the quadratic can be written as p(x) = a(x² - (sum of roots)x + (product of roots)) = a(x² - 8x + 5). So this confirms the earlier result. Therefore, regardless of the value of a (as long as it's an integer), the ratio p(3)/p(4) would be the same because the a cancels out. For example, p(3) = a(9 - 24 + 5) = a(-10), and p(4) = a(16 - 32 + 5) = a(-11). Therefore, the ratio is (-10a)/(-11a) = 10/11. So even if a is different, the ratio remains the same. Therefore, the answer is indeed 10/11.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.4:
Alternatively, another way to think about this is that the ratio p(3)/p(4) depends only on the values of the polynomial evaluated at 3 and 4, but since the polynomial is defined up to a constant multiple, the ratio is invariant under scaling. Therefore, we can choose the simplest polynomial (with a = 1) to compute the ratio, and it will be the same for any other quadratic polynomial with the same roots.
Therefore, I can confidently say that the answer is 10/11.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the final answer is $\boxed{\frac{10}{11}}$.
</Final_answer>
</think>