Algebra 2 Common Core

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

Chapter 9 - Sequences and Series - 9-1 Mathematical Patterns - Practice and Problem-Solving Exercises - Page 571: 72

Answer

$x = 12$

Work Step by Step

First, isolate the radical by adding $3$ to both sides of the equation: $\sqrt {4x - 23} = 5$ Square both sides of the equation to eliminate the radical: $(\sqrt {4x - 23})^2 = 5^2$ $4x - 23 = 25$ Add $23$ to both sides of the equation to isolate the $x$ term: $4x = 48$ Divide both sides of the equation by $4$ to solve for $x$: $x = 12$ Check if the solution is valid by substituting $12$ to $x$ in the given equation: $\begin{align*} \sqrt{4x-23}-3&=2\\ \sqrt{4(12)-23}-3&=2\\ \sqrt{25}-3&=2\\ 5-3&=2\\ 2&=2 \end{align*}$ Thus, the solution is $x=12$.
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