For ` k gt 0` if ` k sqrt(-1)` is a root of the equation `x^4 + 6x^3 - 16x^2 + 24 x- 80 =0` then ` k^2` =

Options

A. 2
B. 3
C. 4
D. 6

Answer

To solve the problem, we need to find the value of \( k^2 \) given that \( k\sqrt{-1} \) is a root of the polynomial equation:

\[ x^4 + 6x^3 - 16x^2 + 24x - 80 = 0 \]

### Step 1: Substitute \( x = k\sqrt{-1} \)

Since \( \sqrt{-1} = i \), we can substitute \( x = ki \) into the equation:

\[ (ki)^4 + 6(ki)^3 - 16(ki)^2 + 24(ki) - 80 = 0 \]

### Step 2: Calculate each term

Now, we will calculate each term in the equation:

1. \( (ki)^4 = k^4 i^4 = k^4 \cdot 1 = k^4 \)
2. \( 6(ki)^3 = 6k^3 i^3 = 6k^3 \cdot (-i) = -6k^3 i \)
3. \( -16(ki)^2 = -16k^2 i^2 = -16k^2 \cdot (-1) = 16k^2 \)
4. \( 24(ki) = 24k i \)
5. The constant term remains as \(-80\).

Putting these together, we have:

\[ k^4 - 6k^3 i + 16k^2 + 24k i - 80 = 0 \]

### Step 3: Group real and imaginary parts

Now, we can group the real and imaginary parts:

- Real part: \( k^4 + 16k^2 - 80 \)
- Imaginary part: \( (-6k^3 + 24k)i \)

Setting both parts equal to zero gives us two equations:

1. \( k^4 + 16k^2 - 80 = 0 \) (Real part)
2. \( -6k^3 + 24k = 0 \) (Imaginary part)

### Step 4: Solve the imaginary part equation

From the imaginary part:

\[ -6k^3 + 24k = 0 \]

Factoring out \( 6k \):

\[ 6k(-k^2 + 4) = 0 \]

This gives us:

\[ k = 0 \quad \text{or} \quad k^2 = 4 \]

Since \( k > 0 \), we have:

\[ k^2 = 4 \]

### Step 5: Solve the real part equation

Now, we can check the real part equation:

\[ k^4 + 16k^2 - 80 = 0 \]

Substituting \( k^2 = 4 \):

\[ (4)^2 + 16(4) - 80 = 16 + 64 - 80 = 0 \]

This confirms that \( k^2 = 4 \) is indeed a solution.

### Final Answer

Thus, the value of \( k^2 \) is:

\[ \boxed{4} \]