College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 160: 34


The solutions are $x=-4\pm\dfrac{3}{2}\sqrt{3}$

Work Step by Step

$(2x+8)^{2}=27$ Take the square root of both sides: $\sqrt{(2x+8)^{2}}=\pm\sqrt{27}$ $2x+8=\pm\sqrt{27}$ Substitute $27$ by $9\cdot3$ and simplify: $2x+8=\pm\sqrt{9\cdot3}$ $2x+8=\pm3\sqrt{3}$ Solve for $x$: $2x=-8\pm3\sqrt{3}$ $x=-\dfrac{8}{2}\pm\dfrac{3}{2}\sqrt{3}$ $x=-4\pm\dfrac{3}{2}\sqrt{3}$
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