The set of values of a for which each on of the roots of `x^(2) - 4ax + 2a^(2) - 3a + 5 = 0` is greater than 2, is

Options

A. `a in (1, oo)`
B. `a = 1`
C. `a in (-oo, 1)`
D. `a in (9//2, oo)`

Explanation

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

\[ x^2 - 4ax + (2a^2 - 3a + 5) = 0 \]

are greater than 2. We will use three conditions to find the valid range of \( a \).

### Step 1: Calculate the Discriminant
The first condition is that the discriminant \( D \) of the quadratic must be non-negative for real roots to exist. The discriminant \( D \) is given by:

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

For our quadratic, \( a = 1 \), \( b = -4a \), and \( c = 2a^2 - 3a + 5 \). Thus,

\[ D = (-4a)^2 - 4(1)(2a^2 - 3a + 5) \]

Calculating this gives:

\[ D = 16a^2 - 4(2a^2 - 3a + 5) = 16a^2 - 8a^2 + 12a - 20 \]
\[ D = 8a^2 + 12a - 20 \]

We need \( D \geq 0 \):

\[ 8a^2 + 12a - 20 \geq 0 \]

Dividing the entire inequality by 4:

\[ 2a^2 + 3a - 5 \geq 0 \]

### Step 2: Factor the Quadratic
Next, we factor \( 2a^2 + 3a - 5 \):

\[ (2a + 5)(a - 1) \geq 0 \]

### Step 3: Solve the Inequality
To find the intervals where this inequality holds, we find the roots:

- \( 2a + 5 = 0 \) gives \( a = -\frac{5}{2} \)
- \( a - 1 = 0 \) gives \( a = 1 \)

Now we test the intervals:

1. \( a < -\frac{5}{2} \)
2. \( -\frac{5}{2} < a < 1 \)
3. \( a > 1 \)

Testing these intervals:

- For \( a < -\frac{5}{2} \), choose \( a = -3 \): \( (2(-3) + 5)(-3 - 1) = (-6 + 5)(-4) = -1 \cdot -4 > 0 \) (True)
- For \( -\frac{5}{2} < a < 1 \), choose \( a = 0 \): \( (2(0) + 5)(0 - 1) = 5 \cdot -1 < 0 \) (False)
- For \( a > 1 \), choose \( a = 2 \): \( (2(2) + 5)(2 - 1) = (4 + 5)(1) = 9 > 0 \) (True)

Thus, the solution from this condition is:

\[ a \leq -\frac{5}{2} \quad \text{or} \quad a \geq 1 \]

### Step 4: Vertex Condition
The second condition is that the x-coordinate of the vertex must be greater than 2. The x-coordinate of the vertex for the quadratic \( ax^2 + bx + c \) is given by:

\[ x = -\frac{b}{2a} = \frac{4a}{2} = 2a \]

We require:

\[ 2a > 2 \implies a > 1 \]

### Step 5: Function Value Condition
The third condition is that \( f(2) > 0 \):

\[ f(2) = 2^2 - 4a(2) + (2a^2 - 3a + 5) = 4 - 8a + 2a^2 - 3a + 5 \]
\[ = 2a^2 - 11a + 9 > 0 \]

Factoring this quadratic:

\[ (2a - 1)(a - 9) > 0 \]

Finding the roots:

- \( 2a - 1 = 0 \) gives \( a = \frac{1}{2} \)
- \( a - 9 = 0 \) gives \( a = 9 \)

Testing the intervals:

1. \( a < \frac{1}{2} \)
2. \( \frac{1}{2} < a < 9 \)
3. \( a > 9 \)

Testing these intervals:

- For \( a < \frac{1}{2} \), choose \( a = 0 \): \( (2(0) - 1)(0 - 9) = -1 \cdot -9 > 0 \) (True)
- For \( \frac{1}{2} < a < 9 \), choose \( a = 1 \): \( (2(1) - 1)(1 - 9) = 1 \cdot -8 < 0 \) (False)
- For \( a > 9 \), choose \( a = 10 \): \( (2(10) - 1)(10 - 9) = 19 \cdot 1 > 0 \) (True)

Thus, the solution from this condition is:

\[ a < \frac{1}{2} \quad \text{or} \quad a > 9 \]

### Step 6: Combine Conditions
Now we combine all conditions:

1. From the discriminant: \( a \leq -\frac{5}{2} \) or \( a \geq 1 \)
2. From the vertex condition: \( a > 1 \)
3. From the function value condition: \( a < \frac{1}{2} \) or \( a > 9 \)

The only overlapping condition is:

\[ a > 9 \]

### Final Answer
Thus, the set of values of \( a \) for which each root of the quadratic is greater than 2 is:

\[ \boxed{(9, \infty)} \]