Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 68

Answer

$x_1 = 30, x_2 = -20$

Work Step by Step

Given: $x^2+10x-600$ Use the Quadratic Equation Formula: $x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$ $1x^2+10x-600$ $a = 1, b = 10, c = -600$ $x = \frac{-(10) \pm \sqrt {(10)^2- 4(1 \times -600)}}{2(1)}$ $x = \frac{-10 \pm \sqrt {100+ 2400}}{2}$ $x = \frac{-10 \pm \sqrt {2500}}{2}$ $x_1 = \frac{-10+50}{2}, x_2 = \frac{-10-50}{2}$ $x_1 = \frac{40}{2}, x_2 = \frac{-60}{2}$ $x_1 = 20, x_2 = -30$
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