Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 8 Rational Functions - 8.1 Model Inverse and Joint Variation - 8.1 Exercises - Mixed Review - Page 557: 55



Work Step by Step

Given: $x-6=\sqrt 3x$ Square both sides: $(x-6)^2=(\sqrt 3x)^2\\x^2-12x+36=3x\\x^2-15x+36=0$ Solve for x: $x=12 \lor x=3$ Check: $12-6=6=\sqrt 3(12)=6$ $3-6=-3\ne\sqrt 3(3)=3$ Hence, the solution is $x=12$; there is an extraneous solution.
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