Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Appendix A - Equations and Inequalities - Page 420: 46


{$\pm4\sqrt 3$}

Work Step by Step

$48-x^{2}=0$ can be rearranged as $x^{2}=48$ According to the square root property, if $x^{2}=48$ then $x=\sqrt (48)=\sqrt (16\times3)=\sqrt 16\sqrt 3=4\sqrt 3$ or $x=-\sqrt (48)=-\sqrt (16\times3)=-\sqrt 16\sqrt 3=-4\sqrt 3$ Therefore, the solution set is {$-4\sqrt 3,+4\sqrt 3$} or {$\pm4\sqrt 3$}
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