Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Skills Handbook - Squaring Numbers and Finding Square Roots - Exercises - Page 889: 16


$x\approx2.4\text{ or }x\approx-2.4$

Work Step by Step

We start with the given equation: $5+x^2=11$ In order to get the term with $x$ by itself, we subtract $5$ from both sides of the equation: $$5-5+x^2=11-5$$$$x^2=6$$To solve the equation $x^2=6$, we must perform the opposite operation of squaring, which is taking the square root, in order to clear the exponent: $$x^2=6$$$$\sqrt {x^2}=\sqrt {6}$$$$x\approx\pm2.4$$$$x\approx2.4\text{ or }x\approx-2.4$$
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