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 257: 66


$45$ and $28$

Work Step by Step

Our equation is: $2\cdot35\cdot18=(35+x)(18+x)\\1260=630+x^2+53x\\x^2+53x-630=0\\(x+63)(x-10)=0$ Thus $x=10$ or $x=−63$ but $x$ has to be positive, thus $x=10$. Thus, the new dimensions are $45$ and $28$.
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.