College Algebra (6th Edition)

Published by Pearson

Chapter 5 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-5) - Page 588: 34

Answer

Width: $4m$ Length: $9m$

Work Step by Step

The exercise describes the rectangle as having a length that is "one meter more than twice the width". In other words, if we let $l$ be length and $w$ be width, we can write the previous statement in the following mathematical expression: $$l = 1 + 2w$$ Since the exercise asks us for the $area$ of the rectangle, which is given as $36 m^{2}$, we can use the formula for a rectangle's area ($A = lw$) and arrive at the following: $$Area = lw = (1 + 2w)w$$ We can now use the value given by the exercise to simplify: $$36 m^{2} = (1 + 2w)w = w + 2w^{2}$$ And now, solve for $w$: $$0 = 2w^{2} + w - 36$$ Using the quadratic formula ($x = \frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$), we can continue solving for $w$: $w = \frac{-1 \frac{+}{} \sqrt {1^{2} - 4(2)(-36)}}{2(2)} = \frac{-1 \frac{+}{} \sqrt {1 + 288}}{4} = \frac{-1 \frac{+}{} \sqrt {289}}{4} = \frac{-1 \frac{+}{} 17}{4}$ where $w$ has two solutions: $w = \frac{-1 + 17}{4} = \frac{16}{4} = 4$, or $w = \frac{-1 - 17}{4} = \frac{-18}{4}$. The latter solution, however, doesn't make sense within the context of the exercise since the width of a rectangle cannot have a negative value. Therefore, we conclude that the width of the rectangle in question is $w = 4m$. Using the first equation that we derived, we can now end by finding the value of $l$: $l = 1 + 2w$ $l = 1 + 2(4)$ $l = 1 + 8 = 9$

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