If the roots of Quadratic `i^(2)x^(2)+5ix-6` are `a+ib` and `c +id` (`a,b,c,d in R` and `i = sqrt(-1)`) then, `(ac-bd) + (ad+bc)i = ?`

Options

A. 6
B. -6
C. 6i
D. None of these

Answer

To solve the given quadratic equation \( i^2 x^2 + 5ix - 6 = 0 \) and find the value of \( (ac - bd) + (ad + bc)i \), we will follow these steps:

### Step 1: Rewrite the quadratic equation
The equation can be rewritten using the fact that \( i^2 = -1 \):
\[ -i x^2 + 5ix - 6 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ x^2 - 5ix + 6 = 0 \]

### Step 2: Use the quadratic formula
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -5i \), and \( c = 6 \). Plugging in these values:
\[ x = \frac{-(-5i) \pm \sqrt{(-5i)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
\[ x = \frac{5i \pm \sqrt{-25 - 24}}{2} \]
\[ x = \frac{5i \pm \sqrt{-49}}{2} \]
\[ x = \frac{5i \pm 7i}{2} \]

### Step 3: Calculate the roots
Calculating the two roots:
1. \( x_1 = \frac{5i + 7i}{2} = \frac{12i}{2} = 6i \)
2. \( x_2 = \frac{5i - 7i}{2} = \frac{-2i}{2} = -i \)

### Step 4: Express the roots in the form \( a + ib \)
The roots can be expressed as:
- \( x_1 = 0 + 6i \) (where \( a = 0, b = 6 \))
- \( x_2 = 0 - 1i \) (where \( c = 0, d = -1 \))

### Step 5: Substitute into the expression \( (ac - bd) + (ad + bc)i \)
Now substituting \( a, b, c, d \) into the expression:
\[ (ac - bd) + (ad + bc)i \]
Calculating \( ac - bd \):
\[ ac = 0 \cdot 0 = 0 \]
\[ bd = 6 \cdot (-1) = -6 \]
Thus,
\[ ac - bd = 0 - (-6) = 6 \]

Calculating \( ad + bc \):
\[ ad = 0 \cdot (-1) = 0 \]
\[ bc = 6 \cdot 0 = 0 \]
Thus,
\[ ad + bc = 0 + 0 = 0 \]

### Step 6: Final result
Combining these results:
\[ (ac - bd) + (ad + bc)i = 6 + 0i = 6 \]

### Final Answer
The value of \( (ac - bd) + (ad + bc)i \) is \( \boxed{6} \).