Question

Which equation has exactly two real and two nonreal solutions?

Answer 1

Given:

Equations are given.

Option B and D only has the 4 solutions.

First taking the option B

$\begin{gathered} x^4-12x^2-64=0 \\ x^4+4x^2-16x^2-64=0 \\ x^2(x^2+4)-16(x^2+4)=0 \\ (x^2+4)(x^2-16)=0 \end{gathered}$$\begin{gathered} x^2+4=0 \\ x^2=-4 \\ x=\pm2i \end{gathered}$$\begin{gathered} x^2-16=0 \\ x^2=16 \\ x=\pm4 \end{gathered}$

Option B is the final answer.