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 - Got It? - Page 391: 1

Answer

$x=6$

Work Step by Step

Add $5$ to both sides: $\sqrt{4x+1}-5+5=0+5$ $\sqrt{4x+1}=5$ Square both sides: $(\sqrt{4x+1})^2=(5)^2$ $4x+1=25$ Subtract $1$ from both sides: $4x+1-1=25-1$ $4x=24$ Divide both sides by $4$: $\dfrac{4x}{4}=\dfrac{24}{4}$ $x=6$ Substitute into original problem to check for extraneous solutions: $\sqrt{4(6)+1}-5=0$ $\sqrt{24+1}-5=0$ $\sqrt{25}-5=0$ $5-5=0$ $0=0$
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