Intermediate Algebra for College Students (7th Edition)

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

Chapter 8 - Review Exercises - Page 657: 10

Answer

{$\dfrac{-3 \pm \sqrt{41}}{4}$}

Work Step by Step

Given: $2x^2+3x-4=0$ or, $x^2+\dfrac{3}{2}x=2$ Add $(\dfrac{3}{4})^2$ on both sides to make a perfect square. Thus, $x^2+\dfrac{3}{2}x+(\dfrac{3}{4})^2=2+(\dfrac{3}{4})^2$ $x^2+2(x)(\dfrac{3}{4})+\dfrac{9}{16}=\dfrac{41}{16}$ or, $(x+\dfrac{3}{4})^2=\dfrac{41}{16}$ Apply Square root property. $x=-\dfrac{3}{4}\pm \dfrac{\sqrt{41}}{4}$ Hence,we have $x=${$\dfrac{-3 \pm \sqrt{41}}{4}$}
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