Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 7

$2$ different irrational solutions that are conjugates.

$ x^{2}-7x+5=0\qquad$.... $a=1,\ b=-7,\ c=5$ $ b^{2}-4ac\qquad$....substitute $b$ for $-7,\ a$ for $1$ and $c$ for $5$ $=(-7)^{2}-4\cdot 1\cdot 5$ $=49-20$ $=29$ Since the discriminant is a positive number that is not a perfect square, there are $2$ different irrational solutions that are conjugates.