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 Review - Page 425: 48

Answer

\begin{align*} x&=-1 &\text{or}& &x=2 \end{align*}

Work Step by Step

Add $1$ to both sides: \begin{align*} \sqrt{3x+3}&=x+1 \end{align*} Square both sides. Note that $(x+a)^2=x^2+2ax+a^2$. \begin{align*} \left(\sqrt{3x+3}\right)^2&=(x+1)^2\\ 3x+3&=x^2+2x+1 \end{align*} Put all terms on the right side: \begin{align*} 0&=x^2+2x+1-(3x+3)\\ 0&=x^2+2x+1-3x-3\\ 0&=x^2-x-2\\ \end{align*} Factor the trinomial to obtain: \begin{align*} 0&=(x-2)(x+1) \end{align*} Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain: \begin{align*} x-2&=0 &\text{or}& &x+1=0\\ x&=2 & \text{or}& &x=-1 \end{align*}
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