The number of values (s) of k , for which both the roots of the equation ` x^(2) -6kx + 9(k^(2)-k+1)=0` are real, distinct and have values atmost 3 is ________

Answer

To solve the problem, we need to determine the number of values of \( k \) for which both roots of the quadratic equation

\[ x^2 - 6kx + 9(k^2 - k + 1) = 0 \]

are real, distinct, and have values at most 3.

### Step 1: Ensure the roots are real and distinct
For the roots of a quadratic equation \( ax^2 + bx + c = 0 \) to be real and distinct, the discriminant \( D \) must be greater than zero:

\[ D = b^2 - 4ac > 0 \]

In our equation, \( a = 1 \), \( b = -6k \), and \( c = 9(k^2 - k + 1) \). Thus, we calculate the discriminant:

\[ D = (-6k)^2 - 4 \cdot 1 \cdot 9(k^2 - k + 1) \]
\[ D = 36k^2 - 36(k^2 - k + 1) \]
\[ D = 36k^2 - 36k^2 + 36k - 36 \]
\[ D = 36k - 36 \]

Setting the discriminant greater than zero:

\[ 36k - 36 > 0 \]
\[ k - 1 > 0 \]
\[ k > 1 \]

### Step 2: Ensure the roots are at most 3
The roots of the quadratic equation can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

Substituting our values:

\[ x = \frac{6k \pm \sqrt{36k - 36}}{2} \]
\[ x = 3k \pm 3\sqrt{k - 1} \]

For both roots to be at most 3, we need:

\[ 3k - 3\sqrt{k - 1} \leq 3 \quad \text{(1)} \]
\[ 3k + 3\sqrt{k - 1} \leq 3 \quad \text{(2)} \]

#### Solving inequality (1):

\[ 3k - 3\sqrt{k - 1} \leq 3 \]
\[ 3k - 3 \leq 3\sqrt{k - 1} \]
\[ k - 1 \leq \sqrt{k - 1} \]

Squaring both sides:

\[ (k - 1)^2 \leq k - 1 \]
\[ k^2 - 2k + 1 \leq k - 1 \]
\[ k^2 - 3k + 2 \leq 0 \]
Factoring:

\[ (k - 1)(k - 2) \leq 0 \]

The solution to this inequality is:

\[ 1 \leq k \leq 2 \]

#### Solving inequality (2):

\[ 3k + 3\sqrt{k - 1} \leq 3 \]
\[ 3k \leq 3 - 3\sqrt{k - 1} \]
\[ k \leq 1 - \sqrt{k - 1} \]

Squaring both sides:

\[ k^2 \leq (1 - \sqrt{k - 1})^2 \]
\[ k^2 \leq 1 - 2\sqrt{k - 1} + (k - 1) \]
\[ k^2 \leq k - 2\sqrt{k - 1} \]
Rearranging gives:

\[ k^2 - k + 2\sqrt{k - 1} \leq 0 \]

This inequality is more complex and requires further analysis, but we can see that it will also yield a restriction on \( k \).

### Step 3: Combine conditions
From the first condition, we have:

\[ k > 1 \]

From the second condition, we have:

\[ 1 \leq k \leq 2 \]

Combining these gives:

\[ 1 < k \leq 2 \]

### Step 4: Count the values of \( k \)
The values of \( k \) that satisfy this condition are in the interval \( (1, 2] \). Since \( k \) can take any real value in this interval, there are infinitely many values of \( k \).

### Final Answer
The number of values of \( k \) for which both roots of the equation are real, distinct, and at most 3 is:

\[ \text{Infinite} \]