Given:
\((x - 1)^2 + 2kx + 11 = 0\) has no real roots, where \(k\) is the largest integer.
Step 1: Simplify the equation:
\((x - 1)^2 + 2kx + 11 = 0\)
\(x^2 - 2x + 1 + 2kx + 11 = 0\)
\(x^2 + (2k - 2)x + 12 = 0\)
Step 2: Apply the discriminant condition for no real roots.
For a quadratic equation \(ax^2 + bx + c = 0\), there are no real roots if \(D = b^2 - 4ac<0\).
Here, \(a=1\), \(b=2k-2\), and \(c=12\).
\((2k - 2)^2 - 4 \cdot 1 \cdot 12<0\)
\(4(k - 1)^2 - 48<0\)
\(4(k - 1)^2<48\)
\((k - 1)^2<12\)
Since \((k - 1)\) must be an integer, the largest integer whose square is less than 12 is 3.
\(k - 1 = 3\)
\(k = 4\)
Step 3: Determine the expression to minimize: \(\frac{k}{4y} + 9y\).
Substitute \(k=4\): \(\frac{4}{4y} + 9y = \frac{1}{y} + 9y\)
Step 4: Apply the AM-GM inequality to find the minimum value.
For positive numbers \(a\) and \(b\), \(\frac{a+b}{2} \ge \sqrt{ab}\), which implies \(a+b \ge 2\sqrt{ab}\).
Let \(a = \frac{1}{y}\) and \(b = 9y\).
\(\frac{1}{y} + 9y \ge 2\sqrt{\frac{1}{y} \cdot 9y}\)
\(\frac{1}{y} + 9y \ge 2\sqrt{9}\)
\(\frac{1}{y} + 9y \ge 2 \cdot 3\)
\(\frac{1}{y} + 9y \ge 6\)
Final Answer: The minimum value of the expression is 6.