Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 788: 55

Answer

5 mph for 3 miles, 4 mph for 4 miles

Work Step by Step

Speed at 3 miles: x mph Speed at 4 miles: (x-1) mph $3/x + 4/(x-1) = 8/5$ $3*x/x + 4*x/(x-1) = 8*x/5$ $3*5 + 4*x*5/(x-1) = 8*x*5/5$ $15 + 20x/(x-1) = 8x$ $15*(x-1) + 20x*(x-1)/(x-1) = 8x*(x-1)$ $15*(x-1) + 20x = 8x*(x-1)$ $15x-15+20x=8x^2-8x$ $35x-15=8x^2-8x$ $8x^2-43x+15=0$ $x=(-b±\sqrt {b^2-4ac})/2a$ $x=(-(-43)±\sqrt {(-43)^2-4*8*15})/2*8$ $x=(43±\sqrt {(-43)^2-60*8})/16$ $x=(43±\sqrt {1849-480})/16$ $x=(43±\sqrt {1369})/16$ $x=(43±37)/16$ $x=(43±37)/16$ $x=(43+37)/16$ $x=80/16$ $x=5$ $x=(43±37)/16$ $x=(43-37)/16$ $x=6/16$ $x=3/8$ A speed of 3/8 mph would provide a negative denominator for the second fraction. Thus, we would have a negative speed. $x-1$ $5-1$ $4$
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