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Section 2.5 : Quadratic Equations - Part I

11. Use factoring to solve the following equation.

\[\frac{{4z}}{{z + 1}} + \frac{5}{z} = \frac{{6z + 5}}{{{z^2} + z}}\]

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This is an equation containing rational expressions so we know that the first step is to clear out the denominator by multiplying by the LCD, which is \(z\left( {z + 1} \right)\) in this case. Also, note that we now know that we must avoid \(z = 0\) and \(z = - 1\) so we do not get division by zero.

Multiplying by the LCD and doing some basic simplification gives,

\[\begin{align*}z\left( {z + 1} \right)\left( {\frac{{4z}}{{z + 1}} + \frac{5}{z}} \right) & = \left( {\frac{{6z + 5}}{{{z^2} + z}}} \right)z\left( {z + 1} \right)\\ \left( z \right)\left( {4z} \right) + 5\left( {z + 1} \right) & = 6z + 5\\ 4{z^2} + 5z + 5 & = 6z + 5\\ 4{z^2} - z & = 0\end{align*}\] Show Step 2

We can now factor the quadratic to get,

\[z\left( {4z - 1} \right) = 0\]

The zero factor property now tells us,

\[\begin{array}{*{20}{c}}{z = 0}\\{}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{4z - 1 = 0}\\{\displaystyle z = \frac{1}{4}}\end{array}\]

Note that we cannot use the first potential solution since that would give us division by zero! Therefore the only solution is : \[\require{bbox} \bbox[2pt,border:1px solid black]{{z = \frac{1}{4}}}\] .

When dealing with equations that have rational expressions do not forget to verify that you do not get division by zero with any of the potential solutions! As we saw in this case if we had not checked we would have gotten a value of \(z\) that seemed to be a solution but in fact was not because of the division by zero issue.