Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 39

Answer

Since the length must be a positive number, 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 the next step is to rewrite the trinomial as follows: $~x^2+bx+c = (x+r)~(x+s)$ We can rewrite the given equation: $a^2 +(a+1)^2 = 25$ $a^2 +(a^2+2a+1) = 25$ $2a^2 +2a+1 = 25$ $2a^2 +2a-24=0$ $(2)~(a^2 +a-12)=0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = 1~$ and $~r\times s = -12$. We can see that $(4)+(-3) = 1~$ and $(4)\times (-3) = -12$ We can solve the equation as follows: $a^2 +(a+1)^2 = 25$ $a^2 +(a^2+2a+1) = 25$ $2a^2 +2a+1 = 25$ $2a^2 +2a-24=0$ $(2)~(a^2 +a-12)=0$ $(2)~(a+4)~(a-3)=0$ $a+4=0~~$or $~~a-3=0$ $a = -4~~$ or $~~a=3$ Since the length must be a positive number, the length is $3$
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