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: 75

Answer

9, 12

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-3 leg c=15 (x)^2+(x-3)^2=225 x^2+x^2-6x+9=225 2x^2-6x+9-225=0 Simplify 2x^2-6x-216=0 Using ac method a=-432 b=-6 Find two numbers that their products are -432 and their sums are -6. These two numbers are 18 and -24. Rewrite: 2x^2+18x-24x-216=0 2x(x+9)-24(x+9)=0 (2x-24)(x+9)=0 2x-24=0 2x=24 x=12 x+9=0 x=-9 a negative length does not make sense. So leg a=x a=12 leg b=x-3 b=12-3 b=9 leg c=15 proof (12)^2+(9)^2=(15)^2 144+81=225
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