Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 11 - Section 11.3 - Applications Involving Quadratic Equations - Exercise - Page 370: 4

Answer

Length = $9.66$ ft Width = $1.66$ ft

Work Step by Step

First, we can say that the width of the rectangular sheet metal is represented by $x$, and the length can then be expressed as $(x+8)$ ft as it is $8$ ft longer than the width. So, now we know that: Length = $(x+8)$ ft Width = $x$ ft Area = $16 ft^2$ We also know that the formula for the area of a rectangle is Length $\times$ Width. Using this formula, we can substitute the values mentioned earlier to form a quadratic equation. Area = Length $\times$ Width $16 = (x+8) \times x$ $16 = x^2+8x$ (Subtract 16 from each side to make the equation=0) $x^2+8x-16=0$ Now that we have our equation, we can solve it using the quadratic formula! $x=\frac{-b±\sqrt {b^2-4ac}}{2a}$ $x_{1}=\frac{-8+\sqrt {8^2-4(1)(-16)}}{2(1)}$, $x_{2}=\frac{-8-\sqrt {8^2-4(1)(-16)}}{2(1)}$ $x_{1}=\frac{-8+\sqrt {128}}{2}$, $x_{2}=\frac{-8-\sqrt {128}}{2}$ $x_{1}=1.66$, $x_{2}=-9.66$ We want 1 solution only in this case, and it cannot be negative as we are finding the width of a rectangle, which cannot be below 0. So, $x=1.66$. Length = $(1.66) +8$ =$9.66$ ft Width = $1.66$ 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.