Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 571: 28

Answer

Width $=10\;ft$. Length $=25\;ft$.

Work Step by Step

Let the width of the deck be $x$. The length of the deck is $2x+5$. Area of the deck $=(x)(2x+5)$. The given area of the deck is $=250\;ft^2$. Equate both areas. $\Rightarrow (x)(2x+5)=250$. Clear the parentheses. $\Rightarrow 2x^2+5x=250$. Subtract $250$ from both sides. $\Rightarrow 2x^2+5x-250=250-250$ Simplify. $\Rightarrow 2x^2+5x-250=0$ Rewrite the middle term $5x$ as $25x-20x$. $\Rightarrow 2x^2+25x-20x-250=0$ Factor out the common terms. $\Rightarrow x(2x+25)-10(2x+25)=0$ Factor out $(2x+25)$. $\Rightarrow (2x+25)(x-10)=0$ Use the zero product property. $ 2x+25=0$ or $x-10=0$ Solve for $x$. $ x=\frac{-25}{2}$ or $x=10$ Width of the deck: $10\;ft$ Length of the deck: $2x+5=2(10)+5=20+5=25\;ft$.
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.