Algebra 2 (1st Edition)

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

Chapter 6 Rational Exponents and Radical Functions - 6.6 Solve Radical Equations - 6.6 Exercises - Skill Practice - Page 457: 50

Answer

3

Work Step by Step

We solve the equation using the rules of exponents and radicals. Note, if we wanted to check our solution, we could plug the value we get back into the original equation to see if it yields a true statement. Solving the equation, we find: $$\left(\sqrt{2x+3}+2\right)^2=\left(\sqrt{6x+7}\right)^2 \\ 4\sqrt{2x+3}+7=4x+7\\ \left(4\sqrt{2x+3}\right)^2=\left(4x\right)^2 \\ 32x+48=16x^2 \\ x=3,\:x=-1$$ Checking for extraneous solutions, we see that the solution $x=-1$ is extraneous.
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