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

Published by Pearson

# Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 706: 12

#### Answer

$y=\pm\sqrt{3}$

#### Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $4y^2=12 ,$ are \begin{array}{l}\require{cancel} y^2=\dfrac{12}{4} \\\\ y^2=3 \\\\ y=\pm\sqrt{3} .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.