## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set: 16

#### Answer

$x=\pm\dfrac{\sqrt{15}}{3}$

#### Work Step by Step

Using the properties of equality, the given equation, $3x^2-5=0 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x^2=5 \\\\ x^2=\dfrac{5}{3} .\end{array} Taking the square root of both sides (the Square Root Property), the solutions to the given equation are \begin{array}{l}\require{cancel} x=\pm\sqrt{\dfrac{5}{3}} \\\\ x=\pm\sqrt{\dfrac{5}{3}\cdot\dfrac{3}{3}} \\\\ x=\pm\sqrt{\dfrac{15}{9}} \\\\ x=\pm\dfrac{\sqrt{15}}{\sqrt{9}} \\\\ x=\pm\dfrac{\sqrt{15}}{\sqrt{(3)^2}} \\\\ x=\pm\dfrac{\sqrt{15}}{3} .\end{array}

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