Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - Chapter Review - Page 655: 43


The solution is $n=\frac{3}{2}$.

Work Step by Step

$n\sqrt2=\sqrt 9-3n$ Square both sides: $2n^2=9-3n$ $2n^2+3n-9=0$ $(2n-3)(n+3)=0$ $2n-3=0$ or $n+3=0$ $n=\frac{3}{2}$ or $n=-3$ Check: $n\sqrt2=\sqrt 9-3(\frac{3}{2})$ $\frac{3}{2}\sqrt2=\frac{3\sqrt 2}{2}$ $(-3)\sqrt2=\sqrt 9-3(-3)$ $-3\sqrt2=3\sqrt2$ The solution is $n=\frac{3}{2}$.
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