College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 164: 154

Answer

Both pieces pieces are $4 \ in$ long.

Work Step by Step

$(\dfrac{x}{4})(\dfrac{x}{4}) + (\dfrac{8-x}{4})(\dfrac{8-x}{4}) = 2 \ in \longrightarrow \dfrac{x^2}{16} + \dfrac{(8-x)^2}{16} = 2 \longrightarrow \dfrac{x^2 + 8^2 - 2(8)(x) +x^2}{16} = 2 \longrightarrow \dfrac{2x^2 - 16x +64}{16} = 2 \longrightarrow \dfrac{x^2-8x+32}{8} = 2 \longrightarrow x^2-8x+32=16 \longrightarrow x^2-8x+32-16=0 \longrightarrow x^2-8x+16=0 \longrightarrow (x-4)^2=0 \longrightarrow \sqrt{(x-4)^2} = \sqrt{0} \longrightarrow x-4=0 \longrightarrow x = 4 \ in$ Since $x = 4 $ both pieces pieces are $4 \ in$ long.
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