College Algebra, Chapter 1, 1.3, Section 1.3, Problem 62

Solve $A = 2 \pi r^2 + 2 \pi rh$ for $r$.


$\begin{equation} \begin{aligned} A =& 2 \pi r^2 + 2 \pi rh && \text{Given} \\ \\ A =& 2 \pi (r^2 + rh) && \text{Factor out } 2 \pi \\ \\ \frac{A}{2 \pi} =& r^2 + rh && \text{Divide both sides by } 2 \pi \\ \\ \frac{A}{2 \pi} + \frac{h^2}{4} =& r^2 + rh + \frac{h^2}{4} && \text{Complete the square: add } \left( \frac{h}{2} \right)^2 = \frac{h^2}{4} \\ \\ \frac{A}{2 \pi} + \frac{h^2}{4} =& \left( r + \frac{h}{2} \right)^2 && \text{Perfect square} \\ \\ \pm \sqrt{\frac{2A + h^2}{4 \pi}} =& r + \frac{h}{2} && \text{Take the square root and simplify by using LCD} \\ \\ r =& \frac{-h}{2} \pm \sqrt{\frac{2A + h^2}{4 \pi}} && \text{Subtract } \frac{h}{2} \\ \\ r =& \frac{-h}{2} \pm \frac{1}{2} \left( \sqrt{\frac{2A + h^2}{\pi}} \right) && \text{Solve for } r \end{aligned} \end{equation}$