Questions: Select the correct answer. The complex solutions of which equation have a real component of 4 ? A. x^2 + 4x + 4 = -25 B. x^2 - 4x + 4 = -25 C. x^2 + 8x + 16 = -21 D. x^2 - 8x + 16 = -21

Solution Steps

To find the complex solutions with a real component of 4, we need to solve each equation for \( x \) and check if the real part of the solution is 4. We will use the quadratic formula to find the roots of each equation and then verify the real part of the solutions.

We need to determine which of the given equations has complex solutions with a real component of \( 4 \). The equations are:

A. \( x^{2} + 4x + 4 = -25 \)
B. \( x^{2} - 4x + 4 = -25 \)
C. \( x^{2} + 8x + 16 = -21 \)
D. \( x^{2} - 8x + 16 = -21 \)

We can rearrange each equation into standard quadratic form \( ax^{2} + bx + c = 0 \):

A. \( x^{2} + 4x + 29 = 0 \)
B. \( x^{2} - 4x + 29 = 0 \)
C. \( x^{2} + 8x + 37 = 0 \)
D. \( x^{2} - 8x + 37 = 0 \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), we calculate the roots for each equation:

• For equation A:
  \( a = 1, b = 4, c = 29 \)
  \( x = \frac{-4 \pm \sqrt{4^{2} - 4 \cdot 1 \cdot 29}}{2 \cdot 1} = \frac{-4 \pm \sqrt{-100}}{2} = -2 \pm 5i \)

• For equation B:
  \( a = 1, b = -4, c = 29 \)
  \( x = \frac{4 \pm \sqrt{(-4)^{2} - 4 \cdot 1 \cdot 29}}{2 \cdot 1} = \frac{4 \pm \sqrt{-100}}{2} = 2 \pm 5i \)

• For equation C:
  \( a = 1, b = 8, c = 37 \)
  \( x = \frac{-8 \pm \sqrt{8^{2} - 4 \cdot 1 \cdot 37}}{2 \cdot 1} = \frac{-8 \pm \sqrt{-84}}{2} = -4 \pm 3i\sqrt{7} \)

• For equation D:
  \( a = 1, b = -8, c = 37 \)
  \( x = \frac{8 \pm \sqrt{(-8)^{2} - 4 \cdot 1 \cdot 37}}{2 \cdot 1} = \frac{8 \pm \sqrt{-84}}{2} = 4 \pm 3i\sqrt{7} \)

From the calculations, we find the real components of the solutions:

• Equation A: \( -2 \)
• Equation B: \( 2 \)
• Equation C: \( -4 \)
• Equation D: \( 4 \)

The only equation with a real component of \( 4 \) is equation D.

Final Answer