Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 396: 52

Answer

$x = 23$

Work Step by Step

Rewrite exponents as radicals: $\sqrt {2x + 3} - 7 = 0$ Isolate the radical: $\sqrt {2x + 3} = 7$ Square both sides of the equation to eliminate the radical: $2x + 3 = 49$ $2x = 46$ Divide both sides of the equation by $2$: $x = 23$ To check if we have an extraneous solution, we substitute the solution into the original equation to see if the two sides equal one another. $\sqrt {2(23) + 3} - 7 = 0$ Simplify the radicand: $\sqrt {49} - 7 = 0$ Take the square root: $7 - 7 = 0$ Combine like terms: $0 = 0$ Both sides are equal; therefore, this solution is valid.
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