Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set: 14

Answer

$x=-1$

Work Step by Step

Isolate the radical on the right side to obtain: $3x=\sqrt{3x+7} -5 \\3x+5 = \sqrt{3x+7}$ Square both sides of the equation to obtain: $(3x+5)^2=3x+7$ Use the rule $(a+b)^2 = a^2 + 2ab+b^2$ where $a=3x$ and $b=5$ to obtain: $\\(3x)^2+2(3x)(5)+5^2=3x+7 \\9x^2+30x+25=3x+7$ Move all terms to the left side of the equation. Note that when a term is moved to the other side of the equation, its sign changes to its opposite. $9x^2+30x+25-3x-7=0 \\9x^2+27x+18=0$ Factor out $9$ to obtain: $9(x^2+3x+2)=0$ Divide 9 on both sides to obtain: $x^2+3x+2=0$ Factor the trinomial to obtain: $(x+2)(x+1)=0$ Equate each factor to zero then solve each equation to obtain: $x+2 = 0 \text{ or } x+1 = 0 \\x = -2 \text{ or } x = -1$ Note that when $x=-2$, $3(-2) - \sqrt{3(-2) + 7}\stackrel{?}{=}-5 \\-6 - \sqrt{-6+7}\stackrel{?}{=}-5 \\-6 - \sqrt{1} \stackrel{?}{=}-5 \\-6-1 \stackrel{?}{=}-5 \\-7 \ne -5$ Thus, $-2$ is an extraneous solution. Therefore he only solution is $x=-1$.
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