Algebra: A Combined Approach (4th Edition)

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

Chapter 11 - Section 11.2 - Solving Quadratic Equations by Completing the Square - Exercise Set: 15

Answer

$y=1$ and $y=-\dfrac{3}{2}$

Work Step by Step

$\dfrac{2}{5}y^{2}+\dfrac{1}{5}y=\dfrac{3}{5}$ Multiply the whole equation by $5$ to avoid working with fractions: $5\Big(\dfrac{2}{5}y^{2}+\dfrac{1}{5}y=\dfrac{3}{5}\Big)$ $2y^{2}+y=3$ Take the $3$ to the left side of the equation: $2y^{2}+y-3=0$ Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=2$, $b=1$ and $c=-3$ Substitute: $y=\dfrac{-1\pm\sqrt{1^{2}-4(2)(-3)}}{2(2)}=\dfrac{-1\pm\sqrt{1+24}}{4}=\dfrac{-1\pm\sqrt{25}}{4}=...$ $...=\dfrac{-1\pm5}{4}$ Our two solutions are: $y=\dfrac{-1+5}{4}=1$ $y=\dfrac{-1-5}{4}=-\dfrac{3}{2}$
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.