Assertion (A) : If one root of `x^(2)- (3+2i)x+ (1 + 3i) = 0` is 2+ i then the other root is 2-i
Reason (R) : Imaginary roots (if occur) of a quadratic equation occurs in conjugate pairs only if the coefficients Q.E. are all real.

Options

A. Both A, R are true and R explain Assertion
B. Both A, R are true but R does't explain A
C. A is true R is false
D. A is false R is true.

Answer

To solve the problem, we need to analyze the assertion and the reason given in the question regarding the quadratic equation \( x^2 - (3 + 2i)x + (1 + 3i) = 0 \).

### Step 1: Identify the roots of the quadratic equation
We are given that one root of the equation is \( \alpha = 2 + i \). We need to find the other root \( \beta \).

### Step 2: Use Vieta's Formulas
According to Vieta's formulas, the sum of the roots \( \alpha + \beta \) can be expressed as:
\[ \alpha + \beta = -\frac{b}{a} \]
where \( b = -(3 + 2i) \) and \( a = 1 \).

### Step 3: Calculate the sum of the roots
Substituting the values of \( b \) and \( a \):
\[ \alpha + \beta = -(-3 - 2i) = 3 + 2i \]

### Step 4: Substitute the known root
Now substituting \( \alpha = 2 + i \):
\[ (2 + i) + \beta = 3 + 2i \]

### Step 5: Solve for the other root \( \beta \)
To find \( \beta \), we rearrange the equation:
\[ \beta = (3 + 2i) - (2 + i) \]
Calculating this gives:
\[ \beta = 3 - 2 + 2i - i = 1 + i \]

### Step 6: Conclusion about the roots
Thus, the other root \( \beta \) is \( 1 + i \), not \( 2 - i \) as stated in the assertion. Therefore, the assertion that the other root is \( 2 - i \) is false.

### Step 7: Evaluate the reason
The reason states that imaginary roots occur in conjugate pairs only if the coefficients of the quadratic equation are all real. Since the coefficients of our equation are not all real (they contain imaginary parts), this statement is true.

### Final Conclusion
- Assertion (A) is **false**.
- Reason (R) is **true**.