Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.6 Solving Quadratic Equations by Using the Quadratic Formula - 4.6 Exercises - Page 373: 63

Answer

$d=-\frac{45}{7}$

Work Step by Step

This is a linear function. Solve for $d$. $$ \begin{aligned} -1.5(d+5)-7 & =2 d+8 \\ -1.5(d+5) & =2 d+8+7 \\ -1.5(d+5) & =2 d+15 \\ -1.5 d-7.5 & =2 d+15 \\ -1.5 d-2 d & =15+7.5 \\ -3.5 d & =22.5 \\ d & =-\frac{22.5}{3.5}\\ & = -\frac{225}{35}\\ &= -\frac{45}{7}\\ &\approx -6.429. \end{aligned} $$ The solution is $$ d=-\frac{45}{7}.$$ Define the following version of the function and plot it. We see that the zero is where it should be. $$ f(d) = 2d+8+1.5(d+5)+7=3.5d+22.5. $$
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