Find the roots of the equation `x^(2) + ix - 1 - i = 0`

Options

A. 1 & i
B. i & -i
C. `1- i & 1`
D. `1 & -i - 1`

Answer

To find the roots of the equation \( x^2 + ix - 1 - i = 0 \), we will use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 1: Identify coefficients
In the given equation, we can identify:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = i \) (coefficient of \( x \))
- \( c = - (1 + i) \) (constant term)

### Step 2: Substitute coefficients into the quadratic formula
Now we substitute \( a \), \( b \), and \( c \) into the quadratic formula:

\[ x = \frac{-i \pm \sqrt{i^2 - 4 \cdot 1 \cdot (- (1 + i))}}{2 \cdot 1} \]

### Step 3: Calculate the discriminant
Next, we need to calculate the discriminant \( b^2 - 4ac \):

\[ b^2 = i^2 = -1 \]
\[ 4ac = 4 \cdot 1 \cdot - (1 + i) = -4(1 + i) = -4 - 4i \]

Now, substituting these values into the discriminant:

\[ b^2 - 4ac = -1 + 4 + 4i = 3 + 4i \]

### Step 4: Substitute back into the formula
Now we substitute back into the formula:

\[ x = \frac{-i \pm \sqrt{3 + 4i}}{2} \]

### Step 5: Simplify the square root
To simplify \( \sqrt{3 + 4i} \), we can express it in the form \( a + bi \) and solve for \( a \) and \( b \):

Let \( \sqrt{3 + 4i} = a + bi \). Then squaring both sides gives:

\[ 3 + 4i = a^2 - b^2 + 2abi \]

This gives us two equations:

1. \( a^2 - b^2 = 3 \)
2. \( 2ab = 4 \) → \( ab = 2 \)

From \( ab = 2 \), we can express \( b = \frac{2}{a} \) and substitute into the first equation:

\[ a^2 - \left(\frac{2}{a}\right)^2 = 3 \]

Multiplying through by \( a^2 \) to eliminate the fraction:

\[ a^4 - 4 = 3a^2 \]

Rearranging gives:

\[ a^4 - 3a^2 - 4 = 0 \]

Letting \( y = a^2 \), we have:

\[ y^2 - 3y - 4 = 0 \]

### Step 6: Solve the quadratic for \( y \)
Using the quadratic formula:

\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \]

This gives us:

\[ y = 4 \quad \text{or} \quad y = -1 \]

Since \( y = a^2 \), we discard \( y = -1 \) and take \( y = 4 \), thus \( a = 2 \).

Now substituting back to find \( b \):

\[ ab = 2 \implies 2b = 2 \implies b = 1 \]

So, \( \sqrt{3 + 4i} = 2 + i \).

### Step 7: Substitute back into the formula
Now substituting back:

\[ x = \frac{-i \pm (2 + i)}{2} \]

This gives us two cases:

1. \( x = \frac{-i + 2 + i}{2} = \frac{2}{2} = 1 \)
2. \( x = \frac{-i - 2 - i}{2} = \frac{-2i - 2}{2} = -1 - i \)

### Final Roots
Thus, the roots of the equation \( x^2 + ix - 1 - i = 0 \) are:

\[ x = 1 \quad \text{and} \quad x = -1 - i \]