Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 241: 22


x=5 (2x+3) = 13 (2x+2)= 12

Work Step by Step

By Pythagoras Theorem $(2x+3)^{2} = (2x+2)^{2} +x^{2}$ $4x^{2}+9+12x = 4x^{2} +4+8x+x^{2}$ $x^{2} -4x-5 = 0$ $x^{2} -(5-1)x-5 = 0$ $x^{2} -5x+x-5 = 0$ $x(x-5)+1(x-5) $ $(x-5)(x+1)=0$ x=5 that is correct x=-1 that is not possible because length can not be negative (2x+3) = 13 (2x+2)= 12
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