Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Section 11.1 - Solving Quadratic Equations by Completing the Square - Practice - Page 759: 8

Answer

$y=(\frac{5+\sqrt 17}{2}, \frac{5-\sqrt 17}{2})$

Work Step by Step

Step-1 : Since there is no constant side on the right side, we subtract 2 from both sides. Therefore the equation becomes $y^2-5y=-2$ Step -2 : Add the square of half of the co-efficient of x to both sides. Co-efficient of y = -5 Half of -5 = $\frac{1}{2}\times-5 = \frac{-5}{2}$ Square of $\frac{-5}{2}$ is $\frac{-5}{2} \times\frac{-5}{2} = \frac{25}{4}$ The equation becomes $y^2-5y+\frac{25}{4}=-2 +\frac{25}{4}$ Step-3 Factor the trinomial and simplify the right hand side. $(y-\frac{5}{2})^2= \frac{-8+25}{4}$ $(y-\frac{5}{2})^2= \frac{-17}{4}$ Step-4 Use the square root property and solve for y $y-\frac{5}{2}=±\sqrt \frac{17}{4}$ $y-\frac{5}{2}=±\frac{\sqrt 17}{2}$ Step-5 Add $\frac{5}{2}$ on both the sides $y=\frac{5}{2}±\frac{\sqrt 17}{2}$ $y=\frac{5±\sqrt 17}{2}$ Therefore the solution set is $(\frac{5+\sqrt 17}{2}, \frac{5-\sqrt 17}{2})$
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