Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - Chapter Review - 10-4 Solving Radical Equations - Page 643: 43

Answer

$n=1.5$

Work Step by Step

$n\sqrt2 = \sqrt {9-3n}$ $(n\sqrt2)^2 = (\sqrt {9-3n})^2$ $n^2*2 = 9-3n$ $2n^2=-3n+9$ $2n^2+3n-9=-3n+9+3n-9$ $2n^2+3n-9 = 0$ $(2n-3)(n+3)=0$ $2n-3=0$ $2n=3$ $2n/2 = 3/2$ $n=1.5$ $n+3=0$ $n=-3$ $n=-3$ $n\sqrt2 = \sqrt {9-3n}$ $-3*\sqrt2 = \sqrt {9-3(-3)}$ $-3*\sqrt2 = \sqrt {9+9}$ $-3*\sqrt2 = \sqrt {18}$ $-3*\sqrt2 = \sqrt {3*3*2}$ $-3*\sqrt2 \ne 3\sqrt 2$ $n=1.5$ $n\sqrt2 = \sqrt {9-3n}$ $1.5*\sqrt2 = \sqrt {9-3*1.5}$ $1.5\sqrt2 = \sqrt {9-4.5}$ $1.5\sqrt2 = \sqrt {4.5}$ $1.5\sqrt2 = \sqrt 9/\sqrt2$ $1.5\sqrt2 = \sqrt 9/\sqrt2*(\sqrt2/\sqrt2)$ $1.5\sqrt2 = 3\sqrt2/2$ $1.5\sqrt2 =1.5\sqrt2$
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