Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.4 Applications Involving Quadratic Equations - 11.4 Exercise Set - Page 726: 59

Answer

$\frac{-10.2+6\sqrt{-{{A}^{2}}+13A-39.36}}{A-6.5}$.

Work Step by Step

$A=6.5-\frac{20.4t}{{{t}^{2}}+36}$ Multiply $\left( {{t}^{2}}+36 \right)$ on both sides of the equation $A=6.5-\frac{20.4t}{{{t}^{2}}+36}$, $\begin{align} & A=6.5-\frac{20.4t}{{{t}^{2}}+36} \\ & A\times \left( {{t}^{2}}+36 \right)=\left( 6.5-\frac{20.4t}{{{t}^{2}}+36} \right)\times \left( {{t}^{2}}+36 \right) \\ & A{{t}^{2}}+36A=\left( {{t}^{2}}+36 \right)\times \left( 6.5 \right)-\left( {{t}^{2}}+36 \right)\left( \frac{20.4t}{{{t}^{2}}+36} \right) \\ & A{{t}^{2}}+36A=6.5{{t}^{2}}+234-20.4t \end{align}$ This gives: $\begin{align} & A{{t}^{2}}-6.5{{t}^{2}}+20.4t+36A-234=0 \\ & \left( A-6.5 \right){{t}^{2}}+20.4t+\left( 36A-234 \right)=0 \end{align}$ Using the quadratic formula $x=-b\pm \frac{\sqrt{{{b}^{2}}-4ac}}{2a}$ to solve for t, $\begin{align} & t=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & =\frac{-20.4\pm \sqrt{{{\left( 20.4 \right)}^{2}}-4\left( A-6.5 \right)\left( 36A-234 \right)}}{2\left( A-6.5 \right)} \\ & =\frac{-20.4\pm \sqrt{416.16-144{{A}^{2}}+1872A-6084}}{2\left( A-6.5 \right)} \\ & =\frac{-20.4\pm \sqrt{-144{{A}^{2}}+1872A-5667.84}}{2\left( A-6.5 \right)} \end{align}$ Simplify more, $\begin{align} & t=\frac{-20.4\pm \sqrt{144\left( -{{A}^{2}}+13A-39.36 \right)}}{2\left( A-6.5 \right)} \\ & =\frac{-20.4\pm 12\sqrt{-{{A}^{2}}+13A-39.36}}{2\left( A-6.5 \right)} \\ & =\frac{2\left( -10.2\pm 6\sqrt{-{{A}^{2}}+13A-39.36} \right)}{2\left( A-6.5 \right)} \\ & =\frac{-10.2\pm 6\sqrt{-{{A}^{2}}+13A-39.36}}{A-6.5} \end{align}$ Since we only consider the positive answer: $t=\frac{-10.2+6\sqrt{-{{A}^{2}}+13A-39.36}}{A-6.5}$.
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