If one root of the quadratic equation `ix^2-2(i+1)x +(2-i)=0,i =sqrt(-1)` is `2-i`, the other root is

Options

A. `3+i`
B. `3+sqrt(-1)`
C. `-1+i`
D. `-1-i`

Explanation

To find the other root of the quadratic equation \( ix^2 - 2(i+1)x + (2-i) = 0 \) given that one root is \( 2 - i \), we can follow these steps:

### Step 1: Identify the coefficients of the quadratic equation
The quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = i \)
- \( b = -2(i + 1) = -2i - 2 \)
- \( c = 2 - i \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Given that one root \( \alpha = 2 - i \), we need to find the other root \( \beta \).

### Step 3: Calculate the sum of the roots
Using the formula for the sum of the roots:
\[ \alpha + \beta = -\frac{b}{a} = -\frac{-2(i + 1)}{i} = \frac{2(i + 1)}{i} \]
Calculating this gives:
\[ \frac{2(i + 1)}{i} = \frac{2i + 2}{i} = 2 + \frac{2}{i} \]
Since \( \frac{1}{i} = -i \) (because multiplying by \( i \) gives \( -1 \)):
\[ \frac{2}{i} = -2i \]
Thus:
\[ \alpha + \beta = 2 - 2i \]

### Step 4: Solve for the other root \( \beta \)
Now we can substitute \( \alpha \) into the equation:
\[ (2 - i) + \beta = 2 - 2i \]
Rearranging gives:
\[ \beta = (2 - 2i) - (2 - i) = -2i + i = -i \]

### Conclusion
The other root \( \beta \) is:
\[ \beta = -i \]