Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Test - Page 832: 6


$x_{1}=\dfrac{3 + \sqrt{29}}{2}$ and $x_{2}=\dfrac{3- \sqrt{29}}{2}$

Work Step by Step

Given $y^2-3y=5 \longrightarrow y^2-3y-5=0$ $a=1, \ b=-3, \ c=-5$ Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have: $\dfrac{-(-3)\pm \sqrt{(-3)^2-4\times 1\times (-5)}}{2\times 1} = \dfrac{3\pm \sqrt{9+20}}{2} = \dfrac{3\pm \sqrt{29}}{2}$ Therefore the solutions are $x_{1}=\dfrac{3 + \sqrt{29}}{2}$ and $x_{2}=\dfrac{3- \sqrt{29}}{2}$
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