Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.3 Solve x(squared) + bx + c = 0 - 4.3 Exercises - Problem Solving - Page 258: 69


See below.

Work Step by Step

Our equation is according to the text of the exercise: $3\cdot18\cdot15=(18+x)(15+x)\\810=270+x^2+3x\\x^2+33x-540=0\\(x+45)(x-12)=0$ Thus $x=-45$ or $x=12$ but $x$ has to be positive, thus $x=12$. Thus the new dimensions are $30$ and $27$, and the length of rope needed is $2(30+27)=114ft$
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.