If `a in R` and both the roots of `x^(2) - 6ax + 9a^(2) + 2a - 2 = 0` exceed 3, then a lies in the in-terval

Options

A. `(1, infty)`
B. `(2, infty)`
C. `(11//9, infty)`
D. `phi`

Answer

To solve the problem, we need to find the values of \( a \) such that both roots of the quadratic equation \( x^2 - 6ax + (9a^2 + 2a - 2) = 0 \) exceed 3.

### Step 1: Determine the Discriminant
For the roots to be real, the discriminant must be non-negative. The discriminant \( D \) of the quadratic equation \( ax^2 + bx + c = 0 \) is given by:

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

In our case:
- \( a = 1 \)
- \( b = -6a \)
- \( c = 9a^2 + 2a - 2 \)

Calculating the discriminant:

\[ D = (-6a)^2 - 4 \cdot 1 \cdot (9a^2 + 2a - 2) \]
\[ D = 36a^2 - 4(9a^2 + 2a - 2) \]
\[ D = 36a^2 - 36a^2 - 8a + 8 \]
\[ D = -8a + 8 \]

For the roots to be real, we need:

\[ -8a + 8 \geq 0 \]

### Step 2: Solve the Inequality
Rearranging the inequality:

\[ 8 \geq 8a \]
\[ 1 \geq a \quad \text{or} \quad a \leq 1 \]

### Step 3: Find the Roots
Now we need to find the roots of the quadratic equation:

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

Substituting \( b \) and \( D \):

\[ x = \frac{6a \pm \sqrt{-8a + 8}}{2} \]
\[ x = 3a \pm \frac{\sqrt{8(1 - a)}}{2} \]

### Step 4: Set Conditions for Roots to Exceed 3
We need both roots to exceed 3:

1. \( 3a + \frac{\sqrt{8(1 - a)}}{2} > 3 \)
2. \( 3a - \frac{\sqrt{8(1 - a)}}{2} > 3 \)

#### Condition 1:
\[ 3a + \frac{\sqrt{8(1 - a)}}{2} > 3 \]
Rearranging gives:
\[ \frac{\sqrt{8(1 - a)}}{2} > 3 - 3a \]
Multiplying both sides by 2:
\[ \sqrt{8(1 - a)} > 6 - 6a \]
Squaring both sides:
\[ 8(1 - a) > (6 - 6a)^2 \]
Expanding:
\[ 8 - 8a > 36 - 72a + 36a^2 \]
Rearranging gives:
\[ 36a^2 - 64a + 28 > 0 \]

#### Condition 2:
\[ 3a - \frac{\sqrt{8(1 - a)}}{2} > 3 \]
Rearranging gives:
\[ -\frac{\sqrt{8(1 - a)}}{2} > 3 - 3a \]
Multiplying both sides by -2 (flipping the inequality):
\[ \sqrt{8(1 - a)} < 6 - 6a \]
Squaring both sides:
\[ 8(1 - a) < (6 - 6a)^2 \]
This leads to the same quadratic inequality as before.

### Step 5: Solve the Quadratic Inequality
Now we solve the quadratic inequality:

\[ 36a^2 - 64a + 28 > 0 \]

Finding the roots using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ a = \frac{64 \pm \sqrt{(-64)^2 - 4 \cdot 36 \cdot 28}}{2 \cdot 36} \]
Calculating the discriminant:

\[ = 4096 - 4032 = 64 \]
Thus,
\[ a = \frac{64 \pm 8}{72} \]
Calculating the roots:
\[ a_1 = \frac{72}{72} = 1, \quad a_2 = \frac{56}{72} = \frac{7}{9} \]

### Step 6: Determine the Interval
The quadratic opens upwards (since the coefficient of \( a^2 \) is positive), so the solution to \( 36a^2 - 64a + 28 > 0 \) is:

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

### Step 7: Combine Conditions
From our earlier condition \( a \leq 1 \), we combine:

\[ \frac{7}{9} < a \leq 1 \]

### Final Answer
Thus, the interval for \( a \) is:

\[ \boxed{\left( \frac{7}{9}, 1 \right]} \]