Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 11


The solution of the equation is $\underline{x=\left\{ 3,4 \right\}}$

Work Step by Step

Let us consider the equation, $\sqrt{{{x}^{2}}-3x}=2x-6$ By squaring both sides, we obtain: $\begin{align} & {{\left( \sqrt{{{x}^{2}}-3x} \right)}^{2}}={{\left( 2x-6 \right)}^{2}} \\ & {{x}^{2}}-3x={{\left( 2x \right)}^{2}}-2\left( 2x \right)\left( -6 \right)+{{\left( 6 \right)}^{2}} \\ & {{x}^{2}}-3x=4{{x}^{2}}-24x+36 \\ & 3{{x}^{2}}-21x+36=0 \end{align}$ And rearrange the equation as given below: $\begin{align} & {{x}^{2}}-7x+12=0 \\ & {{x}^{2}}-3x-4x+12=0 \\ & x\left( x-3 \right)-4\left( x-3 \right)=0 \\ & \left( x-3 \right)\left( x-4 \right)=0 \\ & x=3,4 \end{align}$ Thus, the value of $ x=3,4$.
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