Precalculus (6th Edition) Blitzer

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

Chapter 8 - Cumulative Review Exercises - Page 952: 3


The solution of the equation $\sqrt{2x-1}+2=x$ is $\left\{ 5 \right\}$ .

Work Step by Step

Consider the equation, $\sqrt{2x-1}+2=x$ To isolate the radical to one side of the equation, subtract $2$ from both sides of the equation, $\begin{align} & \sqrt{2x-1}+2-2=x-2 \\ & \sqrt{2x-1}=x-2 \end{align}$ Square both sides of the equation to eliminate the radical. So, ${{\left( \sqrt{2x-1} \right)}^{2}}={{\left( x-2 \right)}^{2}}$ Use the formula ${{\left( A-B \right)}^{2}}={{A}^{2}}-2AB+{{B}^{2}}$ to expand the right side of the equation, $2x-1={{x}^{2}}-4x+4$ Subtract $2x$ and add $1$ to form the equation in quadratic form. So, $\begin{align} & 2x-1-2x+1={{x}^{2}}-4x+4-2x+1 \\ & 0={{x}^{2}}-6x+5 \end{align}$ Factorize the quadratic equation, $\begin{align} & {{x}^{2}}-6x+5=0 \\ & \left( x-1 \right)\left( x-5 \right)=0 \\ & x=1\text{ or }x=5 \end{align}$ Now, verify the answer by substituting the obtained solutions into the original equation. So, first substitute $x=1$ in the equation $\sqrt{2x-1}+2=x$. Therefore, $\begin{align} \sqrt{2\left( 1 \right)-1}+2\overset{?}{\mathop{=}}\,1 & \\ \sqrt{1}+2\overset{?}{\mathop{=}}\,1 & \\ 1+2\overset{?}{\mathop{=}}\,1 & \\ 3\overset{?}{\mathop{=}}\,1 & \\ \end{align}$ This is a false statement. Hence, the solution of the equation cannot be $x=1$. Now, substitute $x=5$ into the original equation $\sqrt{2x-1}+2=x$. $\begin{align} \sqrt{2\left( 5 \right)-1}+\overset{?}{\mathop{=}}\,5 & \\ \sqrt{9}+2\overset{?}{\mathop{=}}\,5 & \\ 3+2\overset{?}{\mathop{=}}\,5 & \\ 5=5 & \\ \end{align}$ This statement is true if the solution of the equation is $x=5$. Hence, the solution set of the equation $\sqrt{2x-1}+2=x$ is $\left\{ 5 \right\}$.
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