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 723: 10

Answer

$\sqrt{\frac{k{{Q}_{1}}{{Q}_{2}}}{N}}$.

Work Step by Step

$N=\frac{k{{Q}_{1}}{{Q}_{2}}}{{{s}^{2}}}$ Multiply $\left( {{s}^{2}} \right)$ on both sides of the equation. $\begin{align} & N\times {{s}^{2}}=\frac{k{{Q}_{1}}{{Q}_{2}}}{{{s}^{2}}}\times {{s}^{2}} \\ & N{{s}^{2}}=k{{Q}_{1}}{{Q}_{2}} \end{align}$ Then, divide by $N$ on both sides of the equation, $\begin{align} & N{{s}^{2}}=k{{Q}_{1}}{{Q}_{2}} \\ & \frac{N{{s}^{2}}}{N}=\frac{k{{Q}_{1}}{{Q}_{2}}}{N} \\ & {{s}^{2}}=\frac{k{{Q}_{1}}{{Q}_{2}}}{N} \end{align}$ Taking square roots on both sides of the equation, $\begin{align} & \sqrt{{{s}^{2}}}=\sqrt{\frac{k{{Q}_{1}}{{Q}_{2}}}{N}} \\ & s=\sqrt{\frac{k{{Q}_{1}}{{Q}_{2}}}{N}} \end{align}$ Thus, the value is $s=\sqrt{\frac{k{{Q}_{1}}{{Q}_{2}}}{N}}$.
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