Intermediate Algebra for College Students (7th Edition)

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

Chapter 11 - Cumulative Review Exercises - Page 872: 1


The solution set is$\underline{\left\{ 22 \right\}}$.

Work Step by Step

$\sqrt{2x+5}-\sqrt{x+3}=2$ Add$\sqrt{x+3}$ to both sides: $\sqrt{2x+5}-\sqrt{x+3}+\sqrt{x+3}=2+\sqrt{x+3}$ Simplify: $\sqrt{2x+5}=2+\sqrt{x+3}$ Square both the sides: ${{\left( \sqrt{2x+5} \right)}^{2}}={{\left( 2+\sqrt{x+3} \right)}^{2}}$ Simplify, square eliminates the square root and using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. $\begin{align} & 2x+5={{2}^{2}}+2\times 2\times \sqrt{x+3}+{{\left( \sqrt{x+3} \right)}^{2}} \\ & 2x+5=4+4\sqrt{x+3}+x+3 \\ & 2x+5=7+4\sqrt{x+3}+x \end{align}$ Subtract$7$ from both sides: $2x+5-7=7-7+4\sqrt{x+3}+x$ Simplify: $2x-2=4\sqrt{x+3}+x$ Subtract x from both sides: $2x-x-2=4\sqrt{x+3}+x-x$ Simplify: $x-2=4\sqrt{x+3}$ Again, square both the sides: ${{\left( x-2 \right)}^{2}}={{\left( 4\sqrt{x+3} \right)}^{2}}$ Simplify, square eliminates the square root and using${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. ${{x}^{2}}-2\times x\times 2+4=16\left( x+3 \right)$ . Use distributive, ${{x}^{2}}-4x+4=16x+48$ Subtract$16x+48$ from both sides: \[\begin{align} & {{x}^{2}}-4x+4-\left( 16x+48 \right)=16x+48-\left( 16x+48 \right) \\ & {{x}^{2}}-4x+4-16x-48=16x+48-16x-48 \\ & {{x}^{2}}-20x-44=0 \end{align}\] Factorize, $\begin{align} & {{x}^{2}}-22x+2x-44=0 \\ & x\left( x-22 \right)+2\left( x-22 \right)=0 \\ & \left( x-22 \right)\left( x+2 \right)=0 \end{align}$ Compare each factor separately: \[\begin{align} & x-22=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+2=0 \\ & x=22\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-2 \\ \end{align}\] Check: $x=22$ $\sqrt{2x+5}-\sqrt{x+3}=2$ Substitute and proceed from L.H.S: $\begin{align} & \sqrt{2\left( 22 \right)+5}-\sqrt{22+3} \\ & \sqrt{44+5}-\sqrt{25} \\ & \sqrt{49}-\sqrt{25} \\ & 7-5 \\ & 2 \\ \end{align}$ which is true. Check: $x=-2$ $\sqrt{2x+5}-\sqrt{x+3}=2$ Substitute and proceed from L.H.S $\begin{align} & \sqrt{2\left( -2 \right)+5}-\sqrt{-2+3} \\ & \sqrt{-4+5}-\sqrt{1} \\ & \sqrt{1}-1 \\ & 1-1 \\ & 0 \\ \end{align}$ which is not true. Hence, the solution is $x=22$only. Hence, the solution set is$\left\{ 22 \right\}$.
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