The objective is to solve the equation \( (x - 1)^2 + 2kx + 11 = 0 \) under the condition that it has no real roots. Additionally, the minimum value of the expression \( \frac{k}{4y} + 9y \) for the determined value of \(k\) needs to be found.
The given equation is:
\((x - 1)^2 + 2kx + 11 = 0\)
Expanding the equation yields:
\((x^2 - 2x + 1) + 2kx + 11 = 0\)
Simplification results in:
\(x^2 + (2k - 2)x + 12 = 0\)
For a quadratic equation to have no real roots, its discriminant \(Δ\) must be less than 0. The discriminant for a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( Δ = b^2 - 4ac \).
In this equation, \(a = 1\), \(b = 2k - 2\), and \(c = 12\). Therefore, the discriminant is:
\(Δ = (2k - 2)^2 - 4(1)(12)\)
Simplifying, we get:
\(Δ = (2k - 2)^2 - 48\)
The condition for no real roots is \( Δ<0 \), which means:
\((2k - 2)^2 < 48\)
Solving the inequality:
\(-\sqrt{48}<2k - 2<\sqrt{48}\)
Since \( \sqrt{48} = 4\sqrt{3} \), the inequality becomes:
\(-4\sqrt{3}<2k - 2<4\sqrt{3}\)
Adding 2 to all parts of the inequality:
\(-4\sqrt{3} + 2<2k<4\sqrt{3} + 2\)
Dividing by 2:
\(1 - 2\sqrt{3}<k<1 + 2\sqrt{3}\)
Given that \(k\) is an integer, and \(1 + 2\sqrt{3} \approx 1 + 2(1.732) = 1 + 3.464 = 4.464\), the largest integer value of \(k\) that satisfies this condition is 4.
\(k = 4\)
Substituting \(k = 4\) into the expression gives:
\(f(y) = \frac{4}{4y} + 9y = \frac{1}{y} + 9y\)
To find the minimum value of \(f(y)\), we calculate the derivative with respect to \(y\) and set it to zero:
\(f'(y) = -\frac{1}{y^2} + 9\)
Setting \(f'(y) = 0\):
\(-\frac{1}{y^2} + 9 = 0\)
Solving for \(y\):
\(\frac{1}{y^2} = 9\)
\(y^2 = \frac{1}{9}\)
Thus, \( y = \frac{1}{3} \) (assuming \(y>0\) for the expression to be defined in a typical context, or considering the context where AM-GM might be implicitly applied).
Substituting \(y = \frac{1}{3}\) back into \(f(y)\):
\(f\left(\frac{1}{3}\right) = \frac{1}{\frac{1}{3}} + 9\left(\frac{1}{3}\right) = 3 + 3 = 6\)
The minimum value of \( \frac{k}{4y} + 9y \) is 6.