Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 9

$2$ different imaginary-number solutions that are complex conjugates.

$ x^{2}+11=0\qquad$.... $a=1,\ b=0,\ c=11$ $ b^{2}-4ac\qquad$....substitute $b$ for $0,\ a$ for $1$ and $c$ for $11$ $=(0)^{2}-4\cdot 1\cdot 11$ $=0-44$ $=-44$ Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.