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 1.6 - Page 178: 12



Work Step by Step

Square both sides and put all terms on one side to create a quadratic equation. Then use trial and error to factor it, and set the values found equal to zero. One of the solutions does not work. $\sqrt {20-8x}=x$ $(\sqrt {20-8x})^2=x^2$ $20-8x=x^2$ $x^2+8x-20=0$ $(x+10)(x-2)=0$ $x=-10,2$ $\sqrt {20-8\times(-10)}\ne-10$ $x=2$
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