Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 12

Answer

$2$ different irrational solutions that are conjugates.

Work Step by Step

$x^{2}-7=0\qquad$.... $a=1,\ b=0,\ c=-7$ $b^{2}-4ac\qquad$....substitute $b$ for $0,\ a$ for $1$ and $c$ for $-7$ $=(0)^{2}-4\cdot 1\cdot(-7)$ $=0+28$ $=28$ Since the discriminant is a positive number that is not a perfect square, there are $2$ different irrational solutions that are conjugates.

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