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 - Chapter Test - Page 427: 15

Answer

$2$

Work Step by Step

$\sqrt {x+4}=\sqrt{3x}$ Square both sides: $(\sqrt {x+4})^2=(\sqrt{3x})^2$ $x+4=3x$ Subtract $x$ from both sides: $x+4-x=3x-x$ $4=2x$ Divide both sides by $2$: \begin{align*} \frac42&=\frac{2x}2\\\\ 2&=x\end{align*} Substitute into original problem to check for extraneous solutions: $\sqrt {(2)+4}=\sqrt{3(2)}$ $\sqrt {6}=\sqrt{6}$ $6=6$
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