Elementary Geometry for College Students (7th Edition)

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Appendix A - A.5 - The Quadratic Formula and Square Root Properties - Exercises - Page 562: 34


The length of CP is $~~1.56$

Work Step by Step

We can rearrange the given equation as a quadratic equation: $x~(x+5) = (x+1)~4$ $x^2+5x = 4x+4$ $x^2+x-4 = 0$ We can use the quadratic formula to find the solutions of the equation: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(1) \pm \sqrt{(1)^2-(4)(1)(-4)}}{(2)(1)}$ $x = \frac{-1 \pm \sqrt{1+16}}{2}$ $x = \frac{-1 \pm \sqrt{17}}{2}$ $x = \frac{-1 - \sqrt{17}}{2}~~$ or $~~x = \frac{-1 + \sqrt{17}}{2}$ $x = -2.56~~$ or $~~x = 1.56$ Since the length $x$ must be positive, the length of CP is $~~1.56$
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