College Algebra (11th Edition)

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

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 28

Answer

$x=\left\{ -4\sqrt{3},4\sqrt{3} \right\}$

Work Step by Step

Using the properties of equality, the given equation, $ 48-x^2=0 ,$ is equivalent to \begin{array}{l}\require{cancel} -x^2=-48 \\\\ x^2=48 .\end{array} Taking the square root of both sides, the solutions to the equation above are \begin{array}{l}\require{cancel} x=\pm\sqrt{48} \\\\ x=\pm\sqrt{16\cdot3} \\\\ x=\pm\sqrt{(4)^2\cdot3} \\\\ x=\pm4\sqrt{3} .\end{array} Hence, the solution is $ x=\left\{ -4\sqrt{3},4\sqrt{3} \right\} .$
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