If `-1 +i` is a root of ` x^4 + 4x^3 + 5x^2 + k=0` then its real roots are

Options

A. `-1,-1`
B. `-1/2 ,-3/2`
C. `-1 +- sqrt(2)`
D. `1+- sqrt(2)`

Answer

To solve the problem, we need to find the real roots of the polynomial equation \( x^4 + 4x^3 + 5x^2 + k = 0 \) given that one of its roots is \( -1 + i \).

### Step 1: Identify the conjugate root
Since \( -1 + i \) is a root, its conjugate \( -1 - i \) must also be a root. This is because complex roots always come in conjugate pairs.

### Step 2: Form the quadratic equation from the roots
We can form a quadratic equation from these two roots. The sum of the roots \( (-1 + i) + (-1 - i) = -2 \) and the product of the roots \( (-1 + i)(-1 - i) = (-1)^2 - (i)^2 = 1 + 1 = 2 \).

Thus, the quadratic equation with roots \( -1 + i \) and \( -1 - i \) is:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
Substituting the values:
\[ x^2 + 2x + 2 = 0 \]

### Step 3: Divide the original polynomial by the quadratic factor
Next, we will divide the original polynomial \( x^4 + 4x^3 + 5x^2 + k \) by the quadratic \( x^2 + 2x + 2 \) to find the other quadratic factor.

Perform polynomial long division:
1. Divide \( x^4 \) by \( x^2 \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x^2 + 2x + 2 \) to get \( x^4 + 2x^3 + 2x^2 \).
3. Subtract this from the original polynomial:
\[ (x^4 + 4x^3 + 5x^2 + k) - (x^4 + 2x^3 + 2x^2) = 2x^3 + 3x^2 + k \]
4. Divide \( 2x^3 \) by \( x^2 \) to get \( 2x \).
5. Multiply \( 2x \) by \( x^2 + 2x + 2 \) to get \( 2x^3 + 4x^2 + 4x \).
6. Subtract:
\[ (2x^3 + 3x^2 + k) - (2x^3 + 4x^2 + 4x) = -x^2 - 4x + k \]
7. Divide \( -x^2 \) by \( x^2 \) to get \( -1 \).
8. Multiply \( -1 \) by \( x^2 + 2x + 2 \) to get \( -x^2 - 2x - 2 \).
9. Subtract:
\[ (-x^2 - 4x + k) - (-x^2 - 2x - 2) = -2x + k + 2 \]

### Step 4: Set the remainder to zero
Since \( x^2 + 2x + 2 \) is a factor, the remainder must be zero:
\[ -2x + k + 2 = 0 \]
From this, we can solve for \( k \):
\[ k = 2x - 2 \]

### Step 5: Find the other quadratic factor
The quotient from the division is \( x^2 + 2x - 1 \). We need to find the roots of this quadratic equation:
\[ x^2 + 2x - 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]

### Final Answer
The real roots of the equation are:
\[ -1 + \sqrt{2} \quad \text{and} \quad -1 - \sqrt{2} \]