Intermediate Algebra for College Students (7th Edition)

Published by Pearson

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 53

Answer

{$\dfrac{4}{3}-\dfrac{\sqrt{13}}{3},\dfrac{4}{3} + \dfrac{\sqrt{13}}{3}$} or, {$\dfrac{4-\sqrt{13}}{3},\dfrac{4+\sqrt{13}}{3}$}

Work Step by Step

Given: $3x^2-8x+1=0$ This can be re-written as: $x^2-\dfrac{8}{3}x=\dfrac{-1}{3}$ We will have to add both sides $(-\dfrac{4}{3})^2$ to complete the square. Thus, $x^2-\dfrac{8}{3}x+(-\dfrac{4}{3})^2=\dfrac{-1}{3}+(-\dfrac{4}{3})^2$ or, $(x-\dfrac{4}{3})^2=\dfrac{13}{9}$ or, $(x-\dfrac{4}{3})=\pm \dfrac{\sqrt{13}}{3}$ or, $x=\dfrac{4}{3} \pm \dfrac{\sqrt{13}}{3}$ Hence, our desired solution set is {$\dfrac{4}{3}-\dfrac{\sqrt{13}}{3},\dfrac{4}{3} + \dfrac{\sqrt{13}}{3}$} or, {$\dfrac{4-\sqrt{13}}{3},\dfrac{4+\sqrt{13}}{3}$}

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