Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set: 76


12, 16

Work Step by Step

Based on Pythagorean theorem a^2 +b^2 = c^2 where a and b are the equal legs and c is the hypotenuse. leg a=x leg b=x+4 leg c=20 (x)^2+(x+4)^2=400 x^2+x^2+8x+16=400 2x^2+8x+16=400 Factor 2(x^2+8x+16=400) x^2+4x+8=200 Simplify x^2+4x-192=0 Using ac method a=-192 b=4 Find two numbers that their products are -192 and their sums are 4. These two numbers are 16 and -12. Rewrite: x^2+16x-12x-192=0 x(x+16)-12(x+16)=0 (x-12)(x+16)=0 x-12=0 x=12 x+16=0 x=-16 a negative length does not make sense. So leg a=x a=12 leg b=x+4 b=12+4 b=16 leg c=20 proof (12)^2+(16)^2=(20)^2 144+176=400
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