Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 113: 120


Eastbound plane: $550$ miles/hour Northbound plane: $600$ miles/hour

Work Step by Step

Let's note: $x$=the speed of the eastbound plane Then the speed of the northbound plane is: $x+50$ In 3 hours the two planes traveled $3x$ miles and $3(x+50)$ miles. Use the Pythagorean Theorem: $(3x)^2+[3(x+50)]^2=2440^2$ $9x^2+9(x^2+100x+2500)=5,953,600$ $9x^2+9x^2+900x+22,500-5,953,600=0$ $18x^2+900x-5,931,100=0$ Use the quadratic formula to determine $x$: $x=\dfrac{-900\pm\sqrt{900^2-4(18)(-5,931,100)}}{2(18}$ $=\dfrac{-900\pm\sqrt{427,849,200}}{36}$ $=\dfrac{-900\pm 20,684.5}{36}$ $x_1=\dfrac{-900-20,684.5}{36}\approx -600$ $x_2=\dfrac{-900+20,684.5}{36}\approx 550$ As $x\gt 0$, the only solution is: $x=550$ miles/hour The speed of the northbound plane is: $550+50=600$ miles/hour
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