If both the roots of the equation `x^(2)-6ax+2-2a+9a^(2)=0` exceed 3, then

Options

A. `a gt (11)/(9)`
B. `a lt (11)/(9)`
C. `a gt (9)/(11)`
D. `a lt (9)/(11)`

Answer

To solve the problem, we need to analyze the quadratic equation given by:

\[ x^2 - 6ax + (2 - 2a + 9a^2) = 0 \]

We want to determine the conditions under which both roots of this equation exceed 3.

### Step 1: Identify the coefficients
The coefficients of the quadratic equation are:
- \( A = 1 \)
- \( B = -6a \)
- \( C = 2 - 2a + 9a^2 \)

### Step 2: Use the sum of the roots
The sum of the roots \( \alpha + \beta \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:

\[ \alpha + \beta = -\frac{B}{A} = -\frac{-6a}{1} = 6a \]

Since both roots exceed 3, we have:

\[ \alpha + \beta > 6 \quad \text{(because } 3 + 3 = 6\text{)} \]

Thus, we can write:

\[ 6a > 6 \]

Dividing both sides by 6 gives:

\[ a > 1 \]

### Step 3: Use the product of the roots
The product of the roots \( \alpha \beta \) is given by:

\[ \alpha \beta = \frac{C}{A} = 2 - 2a + 9a^2 \]

Since both roots are greater than 3, we have:

\[ \alpha \beta > 9 \quad \text{(because } 3 \times 3 = 9\text{)} \]

Thus, we can write:

\[ 2 - 2a + 9a^2 > 9 \]

### Step 4: Rearranging the inequality
Rearranging the inequality gives:

\[ 9a^2 - 2a + 2 - 9 > 0 \]

This simplifies to:

\[ 9a^2 - 2a - 7 > 0 \]

### Step 5: Finding the roots of the quadratic inequality
To solve the quadratic inequality \( 9a^2 - 2a - 7 = 0 \), we can use the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 9, b = -2, c = -7 \):

\[ a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 9 \cdot (-7)}}{2 \cdot 9} \]

Calculating the discriminant:

\[ = \frac{2 \pm \sqrt{4 + 252}}{18} = \frac{2 \pm \sqrt{256}}{18} = \frac{2 \pm 16}{18} \]

Calculating the roots:

1. \( a_1 = \frac{18}{18} = 1 \)
2. \( a_2 = \frac{-14}{18} = -\frac{7}{9} \)

### Step 6: Analyzing the intervals
The roots of the quadratic \( 9a^2 - 2a - 7 = 0 \) are \( a = 1 \) and \( a = -\frac{7}{9} \). We will test the intervals:

1. \( (-\infty, -\frac{7}{9}) \)
2. \( (-\frac{7}{9}, 1) \)
3. \( (1, \infty) \)

Testing a value from each interval:

- For \( a = -1 \) (in \( (-\infty, -\frac{7}{9}) \)):
\[ 9(-1)^2 - 2(-1) - 7 = 9 + 2 - 7 = 4 > 0 \quad \text{(satisfied)} \]

- For \( a = 0 \) (in \( (-\frac{7}{9}, 1) \)):
\[ 9(0)^2 - 2(0) - 7 = -7 < 0 \quad \text{(not satisfied)} \]

- For \( a = 2 \) (in \( (1, \infty) \)):
\[ 9(2)^2 - 2(2) - 7 = 36 - 4 - 7 = 25 > 0 \quad \text{(satisfied)} \]

### Step 7: Conclusion
The solution to the inequality \( 9a^2 - 2a - 7 > 0 \) is:

\[ a < -\frac{7}{9} \quad \text{or} \quad a > 1 \]

Since we also have the condition \( a > 1 \), the final answer is:

\[ a > 1 \]