Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Appendix A - A.5 - The Quadratic Formula and Square Root Properties - Exercises - Page 555: 39


$a = 3.83$

Work Step by Step

We can find rearrange the equation as a quadratic equation: $c^2 = a^2+b^2$ $(a+4)^2 = a^2+(a+3)^2$ $a^2+8a+16 = a^2+(a^2+6a+9)$ $a^2+8a+16 = 2a^2+6a+9$ $a^2-2a-7=0$ We can use the quadratic formula to find the solutions of the equation: $a = \frac{-B \pm \sqrt{B^2-4AC}}{2A}$ $a = \frac{-(-2) \pm \sqrt{(-2)^2-(4)(1)(-7)}}{(2)(1)}$ $a = \frac{2 \pm \sqrt{4+28}}{2}$ $a = \frac{2 \pm \sqrt{32}}{2}$ $a = \frac{2 - \sqrt{32}}{2}~~$ or $~~a = \frac{2 + \sqrt{32}}{2}$ $a = -1.83~~$ or $~~a = 3.83$ Since the length $a$ must be positive, the value of $a$ is $~~3.83$
Update this answer!


Update this answer
