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:

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:

Accepted Answer

### Question Statement:
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:

### Options:
Not applicable (integer-type question)

### Correct Answer:
6

### Title:
Quadratic Equation Optimization

### Difficulty:
Medium

### Section:
Quantitative Aptitude

### Category:
Algebra

### Topic:
Quadratic Equations

### Tags:
Discriminant, Real Roots, Optimization, AM-GM Inequality

### Solution Language:
To solve the problem:

1. **Condition for No Real Roots:**  
   The quadratic equation is:  
   $$
   (x - 1)^2 + 2kx + 11 = 0.
   $$  
   Expanding it:  
   $$
   x^2 - 2x + 1 + 2kx + 11 = 0,
   $$  
   which simplifies to:  
   $$
   x^2 + 2(k - 1)x + 12 = 0.
   $$

For the equation to have no real roots, the discriminant must be negative:  
   $$
   b^2 - 4ac < 0,
   $$  
   where $$ a = 1 $$, $$ b = 2(k - 1) $$, and $$ c = 12 $$.

Substituting into the discriminant:  
   $$
   (2(k - 1))^2 - 4(1)(12) < 0.
   $$  
   Simplify:  
   $$
   4(k - 1)^2 < 48,
   $$  
   or:  
   $$
   (k - 1)^2 < 12.
   $$

2. **Solve for $$ k $$:**  
   Taking square roots:  
   $$
   -\sqrt{12} < k - 1 < \sqrt{12}.
   $$  
   Approximate $$ \sqrt{12} \approx 3.46 $$:  
   $$
   -3.46 < k - 1 < 3.46.
   $$  
   Adding $$ 1 $$:  
   $$
   -2.46 < k < 4.46.
   $$

The largest integer value of $$ k $$ is **4**.

3. **Minimizing Expression:**  
   Substitute $$ k = 4 $$ into the expression to minimize:  
   $$
   f(y) = \frac{k}{4y} + 9y = \frac{4}{4y} + 9y = \frac{1}{y} + 9y.
   $$

4. **Use AM-GM Inequality:**  
   By AM-GM inequality, for two positive numbers $$ \frac{1}{y} $$ and $$ 9y $$:  
   $$
   AM \geq GM,
   $$  
   where:  
   $$
   AM = \frac{\frac{1}{y} + 9y}{2},\quad GM = \sqrt{\frac{1}{y} \cdot (9y)} = \sqrt{9}.
   $$

5. **Simplify:**  
   Calculate the geometric mean:  
   $$
   GM = \sqrt{9} = 3.
   $$

6. **Equality Condition:**  
   Equality in AM-GM occurs when both terms are equal, i.e., when:  
   $$
   \frac{1}{y} = 9y.
   $$

7. **Solve for y:**  
    Rearrange:  
    $$
    y^2 = \frac{1}{9}.
    $$

Thus, $$ y = \frac{1}{3}. $$

8. **Calculate Minimum Value:**  
    Substitute into the expression:  
    $$
    f(y) = \frac{1}{y} + 9y = \frac{1}{\frac{1}{3}} + 9\cdot\frac{1}{3} = 3 + 3 = 6.
    $$

Thus, the least possible value of the expression is **6**.

Citations:
[1] https://pplx-res.cloudinary.com/image/upload/v1743433476/user_uploads/BfGmZoqaOnnwWOJ/Screenshot-2025-03-31-203351.jpg
[2] https://pplx-res.cloudinary.com/image/upload/v1743433476/user_uploads/NqNvKYFCznxrxGy/Screenshot-2025-03-31-203411.jpg

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Answer from Perplexity: pplx.ai/share