To find the largest integer whose square root is an irrational number between 3 and 4, we start by identifying the possible values for \(x\) where \(3 < \sqrt{x} < 4\). Squaring all parts of this inequality, we get:
\[ 3^2 < x < 4^2 \]
Calculating the squares:
\[ 9 < x < 16 \]
This tells us that \(x\) must be an integer between 9 and 16. The integers in this range are:
\[ 10, 11, 12, 13, 14, 15 \]
Next, we need to identify which of these integers has an irrational square root. The square roots of the integers can be categorized as follows:
- \(\sqrt{10}\) is irrational.
- \(\sqrt{11}\) is irrational.
- \(\sqrt{12} = 2\sqrt{3}\) is irrational.
- \(\sqrt{13}\) is irrational.
- \(\sqrt{14}\) is irrational.
- \(\sqrt{15}\) is irrational.
Since all the square roots of the integers from 10 to 15 are irrational, we now need to find the largest integer in that range, which is 15.
Therefore, the largest integer whose square root is an irrational number between 3 and 4 is:
\[ \boxed{15} \]
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