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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set: 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}
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