Question

Let \(k\) be the largest integer such that the equation \((x-1)^2+2kx+11=0\) has no real roots. If \(y\) is a positive real number, then the least possible value of \(\frac{k}{4y}+9y\) is

Answer 1

Given:
\((x - 1)^2 + 2kx + 11 = 0\) has no real roots, where \(k\) is the largest integer.

Step 1: Simplify the expression: 
\((x - 1)^2 + 2kx + 11 = 0\)
⇒ \(x^2 - 2x + 1 + 2kx + 11 = 0\)
⇒ \(x^2 - 2(k - 1)x + 12 = 0\)

Step 2: Use discriminant condition for no real roots:
For a quadratic to have no real roots, \(D < 0\)
⇒ \(b^2 - 4ac < 0\)
⇒ \([2(k - 1)]^2 - 4 \cdot 1 \cdot 12 < 0\)
⇒ \(4(k - 1)^2 < 48\)
⇒ \((k - 1)^2 < 12\)

As \((k - 1)\) is an integer, the maximum integer satisfying this is 3.
⇒ \(k - 1 = 3\) ⇒ \(k = 4\)

Step 3: Minimize the expression \(\frac{k}{4y} + 9y\)
⇒ \(\frac{4}{4y} + 9y = \frac{1}{y} + 9y\)

Step 4: Use AM-GM inequality:
\(\frac{1}{y} + 9y \ge 2\sqrt{9y \cdot \frac{1}{y}} = 2\sqrt{9} = 6\)

Final Answer: The least possible value is 6.