Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 426: 59

Answer

$x=2$

Work Step by Step

The provided radical equation is $\sqrt{x+7}-1=x$. Isolate the term $\sqrt{x+7}$ by adding $1$ to both sides of the equation. This gives $\begin{align} & \sqrt{x+7}-1+1=x+1 \\ & \sqrt{x+7}=x+1 \end{align}$ Square both sides of the equation. ${{\left( \sqrt{x+7} \right)}^{2}}={{\left( x+1 \right)}^{2}}$ Use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to solve the above equation further as, $\begin{align} & x+7={{x}^{2}}+2\times x\times 1+{{\left( 1 \right)}^{2}} \\ & x+7={{x}^{2}}+2x+1 \end{align}$ Combine all the like terms and solve as, $\begin{align} & x+7={{x}^{2}}+2x+1 \\ & {{x}^{2}}+2x+1-x-7=0 \\ & {{x}^{2}}+x-6=0 \end{align}$ Then writing the factors as, $\begin{align} & {{x}^{2}}+3x-2x-6=0 \\ & \left( x+3 \right)\left( x-2 \right)=0 \end{align}$ Equating each factor to 0 and solving further, $x=2\text{ or -3}$ To check the solutions: Substituting $x=2$ in the equation $\sqrt{x+7}-1=x$ $\begin{matrix} \sqrt{x+7}-1=x \\ \sqrt{2+7}-1\overset{?}{\mathop{=}}\,2 \\ 3-1\overset{?}{\mathop{=}}\,2 \\ 2=2,\text{ true} \\ \end{matrix}$ Now, substituting $x=-3$ in the equation $\sqrt{x+7}-1=x$ $\begin{matrix} \sqrt{x+7}-1=x \\ \sqrt{-3+7}-1\overset{?}{\mathop{=}}\,-3 \\ 2-1\overset{?}{\mathop{=}}\,-3 \\ 1\ne -3,\text{ False} \\ \end{matrix}$ Thus, $x=2$ is the right solution for the equation $\sqrt{x+7}-1=x$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.