Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 38


The length is $3$

Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$ Then we can factor the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ We can rearrange the given equation: $x~(x+5)=24$ $x^2+5x=24$ $x^2+5x-24 = 0$ To factor the left side of the equation, we need to find two numbers $r$ and $s$ such that $r+s = 5$ and $r\times s = -24$. We can see that $(8)+(-3) = 5~$ and $(8)\times (-3) = -24$ We can solve the equation as follows: $x~(x+5)=24$ $x^2+5x=24$ $x^2+5x-24 = 0$ $(x+8)~(x-3) = 0$ $x+8 = 0~~$ or $~~x-3 = 0$ $x = -8~~$ or $~~x = 3$ Since the length must be a positive number, the length is $3$
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