Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Test - Page 402: 39

Answer

Length : $5$ yards. Width: $3$ yards.

Work Step by Step

Let's note the length of the rectangular floor by $x$. The width of the rectangular floor is $2x-7$. The area of the rectangular floor is $(x)(2x-7)$. We are given that the area of the bedroom is $15$ square yards. Equate both. $\Rightarrow (x)(2x-7)=15$ Apply the distributive property. $\Rightarrow x\cdot 2x-x\cdot 7=15$ Simplify. $\Rightarrow 2x^2- 7x=15$ Subtract $15$ from both sides. $\Rightarrow 2x^2- 7x-15=15-15$ Add like terms. $\Rightarrow 2x^2- 7x-15=0$ Rewrite the middle term $-7x$ as $-10x+3x$ $\Rightarrow 2x^2- 10x+3x-15=0$ Group the terms. $\Rightarrow (2x^2- 10x)+(3x-15)=0$ Factor each group. $\Rightarrow 2x(x- 5)+3(x-5)=0$ Factor out $(x-5)$. $\Rightarrow (x- 5)(2x+3)=0$ Set each factor equal to zero. $x-5=0$ and $2x+3=0$ Isolate $x$. $x=5$ and $x=-\frac{3}{2}$ Take positive value because $x$ represents a dimension and has to be positive. Length $x=5$ yrads. Width $2x-7=2(5)-7=10-7=3$ yards.
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