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


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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.