Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 502: 58



Work Step by Step

$x-\sqrt (17-4x) -3 = 0$ $x-3 = \sqrt (17-4x) $ Squaring on both sides. $(x-3)^{2} = 17-4x$ Using $(a-b)^{2} = a^{2} -2ab +b^{2} $ $(x-3)^{2} = 17-4x$ $x^{2}-6x+9= 17-4x$ $x^{2}-6x+9- 17+4x=0$ $x^{2}-2x-8=0$ By factoring, $(x-4)(x+2)=0$ $x=4$ or $x=-2$ Substituting $x=4$ and $x=-2$ in the given equation, $x=4$ only satisfies the equation. $x=-2$ does not satisfy. So, $x=4$ is the solution.
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