College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Review Exercises: 64

Answer

$x=\left\{ -\dfrac{\sqrt{2}}{2},0,\dfrac{\sqrt{2}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^2-2x^4=0 ,$ use factoring. Then equate each factor to zero and solve for the value of the variable using the Square Root Principle. $\bf{\text{Solution Details:}}$ Factoring the $GCF=x^2,$ the factored form of the given expression is \begin{array}{l}\require{cancel} x^2(1-2x^2)=0 .\end{array} Equating each factor to zero (Zero Product Property) results to \begin{array}{l}\require{cancel} x^2=0 \\\\\text{OR}\\\\ 1-2x^2=0 .\end{array} Isolating the squared variable results to \begin{array}{l}\require{cancel} x^2=0 \\\\\text{OR}\\\\ -2x^2=-1 \\\\ x^2=\dfrac{-1}{-2} \\\\ x^2=\dfrac{1}{2} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} x^2=0 \\\\ x=\pm\sqrt{0} \\\\ x=0 \\\\\text{OR}\\\\ x^2=\dfrac{1}{2} \\\\ x=\pm\sqrt{\dfrac{1}{2}} \\\\ x=\pm\sqrt{\dfrac{1}{2}\cdot\dfrac{2}{2}} \\\\ x=\pm\sqrt{\dfrac{1}{4}\cdot2} \\\\ x=\pm\sqrt{\left( \dfrac{1}{2}\right)^2\cdot2} \\\\ x=\pm\dfrac{1}{2}\sqrt{2} \\\\ x=\pm\dfrac{\sqrt{2}}{2} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{\sqrt{2}}{2},0,\dfrac{\sqrt{2}}{2} \right\} .$
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