`3+4i` is a root of `x^(2) + Ax + B = 0` and `sqrt(3) -2` is a root of `x^(2) + Cx + D = 0` then

Options

A. a) `A lt Clt D lt B`
B. b) `A lt D lt C lt B`
C. c) `A gt C gt D gt B`
D. d) `A gt D gt C gt B`

Answer

To solve the problem, we need to find the values of \( A \), \( B \), \( C \), and \( D \) based on the given roots of the quadratic equations. Let's go through the steps systematically.

### Step 1: Determine the value of \( A \) and \( B \)

Given that \( 3 + 4i \) is a root of the equation \( x^2 + Ax + B = 0 \), we know that the other root must be its conjugate, \( 3 - 4i \), because the coefficients of the polynomial are rational.

#### Sum of roots:
The sum of the roots \( (3 + 4i) + (3 - 4i) = 6 \).

Using Vieta's formulas:
\[ A = -\text{(sum of roots)} = -6 \]

#### Product of roots:
The product of the roots is given by:
\[ (3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 - (-16) = 9 + 16 = 25 \]
Thus,
\[ B = \text{(product of roots)} = 25 \]

### Step 2: Determine the value of \( C \) and \( D \)

Now, we have the root \( \sqrt{3} - 2 \) for the equation \( x^2 + Cx + D = 0 \). The other root must be \( -\sqrt{3} - 2 \).

#### Sum of roots:
The sum of the roots \( (\sqrt{3} - 2) + (-\sqrt{3} - 2) = -4 \).

Using Vieta's formulas:
\[ C = -\text{(sum of roots)} = 4 \]

#### Product of roots:
The product of the roots is given by:
\[ (\sqrt{3} - 2)(-\sqrt{3} - 2) = -(\sqrt{3} - 2)(\sqrt{3} + 2) = -((\sqrt{3})^2 - 2^2) = -(3 - 4) = -(-1) = 1 \]
Thus,
\[ D = \text{(product of roots)} = 1 \]

### Step 3: Summary of values
Now we have:
- \( A = -6 \)
- \( B = 25 \)
- \( C = 4 \)
- \( D = 1 \)

### Step 4: Arrange the values
Now we can arrange \( A \), \( B \), \( C \), and \( D \) in ascending order:
\[ A < D < C < B \]
This gives us:
\[ -6 < 1 < 4 < 25 \]

### Conclusion
The correct order is \( A < D < C < B \).