College Algebra, Chapter 1, 1.3, Section 1.3, Problem 64

Solve $\displaystyle \frac{1}{r} + \frac{2}{1 - r} = \frac{4}{r^2}$ for $r$.


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\begin{equation}
\begin{aligned}

\frac{1}{r} + \frac{2}{1 - r} =& \frac{4}{r^2}
&& \text{Given}
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\frac{(1 - r) + 2(r)}{r - r^2} =& \frac{4}{r^2}
&& \text{Get the LCD of the left side}
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\frac{1 + r}{r - r^2} =& \frac{4}{r^2}
&& \text{Simplify the numerator}
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r^2 (1 + r) =& 4(r - r^2)
&& \text{Apply cross multiplication}
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r^2 + r^3 =& 4r - 4r^2
&& \text{Apply Distributive Property}
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r^3 + 5r^2 - 4r =& 0
&& \text{Combine like terms}
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r(r^2 + 5r - 4) =& 0
&& \text{Factor out $r$, then eliminate}
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r^2 + 5r =& 4
&& \text{Add 4}
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r^2 + 5r + \frac{25}{4} =& 4 + \frac{25}{4}
&& \text{Complete the square: add } \left( \frac{5}{2} \right)^2 = \frac{25}{4}
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\left( r + \frac{5}{2} \right)^2 =& \frac{41}{4}
&& \text{Perfect square}
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r + \frac{5}{2} =& \pm \sqrt{\frac{41}{4}}
&& \text{Take the square root}
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r =& \frac{-5}{2} \pm \sqrt{\frac{41}{4}}
&& \text{Subtract } \frac{5}{2}
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r =& \frac{-5 + \sqrt{41}}{2} \text{ and } r = \frac{-5 - \sqrt{41}}{2}
&& \text{Solve for } r



\end{aligned}
\end{equation}
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