Intermediate Algebra for College Students (7th Edition)

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

Chapter 8 - Section 8.2 - The Quadratic Formula - Exercise Set - Page 610: 86

Answer

The length of each piece is $4$ inches.

Work Step by Step

Set up the equation: $A = A_s + A_l$ $2 = (\frac{x}{4})^2+(\frac{8-x}{4})^2$ Simplify the equation $2 = \frac{x^2}{4^2}+\frac{(8-x)^2}{4^2}$ $2 = \frac{x^2}{16}+\frac{64-16x+x^2}{16}$ Multiply both sides by $16$. Thus, the equation becomes: $32=x^2+64-16x+x^2$ $0=2x^2-16x+32$ Divide both sides by $2$ $0=x^2-8x+16$ Factor. $0=(x-4)(x-4)$ $0=(x-4)$ $x=4$ The length of each piece is equal to their perimeters. $P_{square1}=4(\frac{x}{4})$ $P_{square1}=4(\frac{4}{4})$ $P_{square1}=4$ $inches$ $P_{square2}=4(\frac{8-x}{4})$ $P_{square2}=4(\frac{8-4}{4})$ $P_{square2}=4(\frac{4}{4})$ $P_{square2}=4$ $inches$
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