Because $p(x)$ has integer coefficients (in particular, because it has rational coefficients), the other root of $p(x)$ must be the radical conjugate of $4-\sqrt{11},$ which is $4+\sqrt{11}.$ Then, $p(x)$ must take the form \[p(x) = A(x-(4-\sqrt{11}))(x-(4+\sqrt{11}))\]for some nonzero constant $A$. This means that \[p(3) = A(-1+\sqrt{11})(-1-\sqrt{11}) = -10A\]and \[p(4) = A(\sqrt{11})(-\sqrt{11}) = -11A,\]so \[\frac{p(3)}{p(4)} = \frac{-10A}{-11A} = \boxed{\frac{10}{11}}.\]
Question:
Let $p(x)$ be a quadratic polynomial with integer coefficients which has $4-\sqrt{11}$ as a root. Compute $\frac{p(3)}{p(4)}.$
Answer:
Let's think step by step.

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\frac{10}{11}
Let $p(x)$ be a quadratic polynomial with integer coefficients which has $4-\sqrt{11}$ as a root. Compute $\frac{p(3)}{p(4)}.$
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