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 - Exercise Set - Page 765: 66



Work Step by Step

$5y^{2}-15y=1$ Take out common factor $5$ from the left side of the equation: $5(y^{2}-3y)=1$ Take the $5$ to divide the right side of the equation: $y^{2}-3y=\dfrac{1}{5}$ Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-3$ $y^{2}-3y+\Big(\dfrac{-3}{2}\Big)^{2}=\dfrac{1}{5}+\Big(\dfrac{-3}{2}\Big)^{2}$ $y^{2}-3y+\dfrac{9}{4}=\dfrac{1}{5}+\dfrac{9}{4}$ $y^{2}-3y+\dfrac{9}{4}=\dfrac{49}{20}$ Factor the expression on the left side of the equation, which is a perfect square trinomial: $\Big(y-\dfrac{3}{2}\Big)^{2}=\dfrac{49}{20}$ Take the square root of both sides of the equation: $\sqrt{\Big(y-\dfrac{3}{2}\Big)^{2}}=\sqrt{\dfrac{49}{20}}$ $y-\dfrac{3}{2}=\pm\dfrac{7}{\sqrt{20}}$ Solve for $y$: $y=\dfrac{3}{2}\pm\dfrac{7}{\sqrt{20}}=\dfrac{3}{2}\pm\dfrac{7\sqrt{20}}{20}=\dfrac{3}{2}\pm\dfrac{7\sqrt{5}}{10}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.