College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 102: 97



Work Step by Step

If the sides are x and y, then the area is $xy=40.$ We can express y as $y=\displaystyle \frac{40}{x}$, because x is not 0. The perimeter is $\ \ 2x+2y=26$, into which we substitute $y=\displaystyle \frac{40}{x}$, $2x+2\displaystyle \cdot\frac{40}{x}=26$ and multiply both sides with x. $2x^{2}+80=26x$ Divide with 2 and write in standard form $x^{2}+40=13x$ $x^{2}-13x+40=0$ We can factor the LHS, as -8 and -5 are factors of +40 whose sum is -13.... $(x-5)(x-8)=0$ So, x can be 5 or 8. If x is 5, then $y=\displaystyle \frac{40}{5}=8$, and if x =8, then $y=\displaystyle \frac{40}{8}=5.$ Either way, the dimensions of the rectangle are $(5$m$)\times(8$m$).$
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.