College Algebra, Chapter 1, 1.3, Section 1.3, Problem 74

Solve $b^2 x^2 - 5bx + 4 = 0 (b \neq 0)$ for $x$.


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\begin{equation}
\begin{aligned}

b^2 x^2 - 5bx + 4 =& 0
&& \text{Given}
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b^2 x^2 - 5bx =& -4
&& \text{Subtract 4}
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x^2 - \frac{5x}{b} =& \frac{-4}{b^2}
&& \text{Divide both sides by } b^2
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x^2 - \frac{5x}{b} + \frac{25}{4b^2} =& \frac{-4}{b^2} + \frac{25}{4b^2}
&& \text{Complete the square: add } \left( \frac{\displaystyle \frac{-5}{b}}{2} \right)^2 = \frac{25}{4b^2}
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\left( x - \frac{5}{2b} \right)^2 =& \frac{-4}{b^2} + \frac{25}{4b^2}
&& \text{Perfect square}
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x - \frac{5}{2b} =& \pm \sqrt{\frac{-16 + 25}{4b^2}}
&& \text{Take the square root, then simplify the right side of the equation by using LCD}
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x =& \frac{5}{2b} \pm \sqrt{\frac{9}{4b^2}}
&& \text{Add } \frac{5}{2b}, \text{ simplify the right side of the equation}
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x =& \frac{5 + 3}{2b} \text{ and } x = \frac{5 - 3}{2b}
&& \text{Solve for } x
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x =& \frac{4}{b} \text{ and } x = \frac{1}{b}
&& \text{Simplify}



\end{aligned}
\end{equation}
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