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 58: 137

Answer

$$ 132.6 \space ft$$

Work Step by Step

Total time = $t_1 + t_2 = 3$ $$3 = \frac{\sqrt d}4 + \frac d {1090}$$ $$ \frac{\sqrt d}4 + \frac d {1090} = 3$$ $$ \frac{\sqrt d}4 =3 - \frac{d}{1090}$$ $$\sqrt d =4(3 - \frac d{1090}) = 12 - \frac{4}{1090}d$$ $$d = (12 - \frac{4}{1090}d)^2$$ $$d = 144 - \frac{96}{1090}d + \frac{16}{1188100}d^2$$ $$0 = 144 - \frac{96}{1090}d+ d + \frac{16}{1188100}d^2$$ $$0 = 144 - \frac{1186}{1090}d + \frac{16}{1188100}d^2$$ $$0 = 144 - 1.08807339d + 0.00001346688d^2$$ $$\sqrt{\Delta} = \sqrt {( -1.08807339)^2 - 4(144)(0.00001346688)}$$ $$\sqrt{\Delta} = 1.08450301$$ $$d_1 = \frac{- (-1.08807339) + 1.08450301}{2(0.00001346688)} = 80663.7 $$ $$d_2 = \frac{- (-1.08807339) - 1.08450301}{2(0.00001346688)} = 132.6 $$ Substituting $d_1$ in the original equation, we can see that it does not give the correct result. But, if we do the same for $d_2$, we will get 3 seconds, proving that $d_2$ is the correct answer.
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