Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.5 - Solving Problems Using Quadratic Equations - Problem Set 10.5 - Page 462: 2


26, 27

Work Step by Step

We take the first whole number as $x$. Since the next number will be one more than this number, the other whole number is taken as $(x+1)$. Since the product of the consecutive whole numbers is $702$, we write the following equation and solve it: $x(x+1)=702$ $x^{2}+x-702=0$ $x^{2}-26x+27x-702=0$ $x(x-26)+27(x-26)=0$ $(x-26)(x+27)=0$ $(x-26)=0$ or $(x+27)=0$ $x=26$ or $x=-27$ Disregarding the negative answer, we find that the smaller whole number is 26. This means that the larger whole number is 27.
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