Algebra - Quadratic Equations

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Section 2.5 : Quadratic Equations - Part I

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10. Use factoring to solve the following equation.

\[\frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\]

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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 \(w + 2\) in this case. Also, note that we now know that we must avoid \(w = - 2\) so we do not get division by zero.

Multiplying by the LCD and doing some basic simplification gives,

\[\begin{align*}\left( {w + 2} \right)\left( {\frac{{{w^2} - 10}}{{w + 2}} + w - 4} \right) & = \left( {w - 3} \right)\left( {w + 2} \right)\\ {w^2} - 10 + \left( {w - 4} \right)\left( {w + 2} \right) & = \left( {w - 3} \right)\left( {w + 2} \right)\\ {w^2} - 10 + {w^2} - 2w - 8 & = {w^2} - w - 6\\ {w^2} - w - 12 & = 0\end{align*}\] Show Step 2

We can now factor the quadratic to get,

\[\left( {w - 4} \right)\left( {w + 3} \right) = 0\]

The zero factor property now tells us,

\[\begin{array}{*{20}{c}}{w - 4 = 0}\\{w = 4}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{w + 3 = 0}\\{w = - 3}\end{array}\]

Therefore the two solutions are : \(\require{bbox} \bbox[2pt,border:1px solid black]{{w = 4\,\,{\mbox{and }}w = - 3}}\).

Note as well that because neither of these are \(w = - 2\) we know that we won’t get division by zero. Do not forget this important part of the solution process for equations involving rational expressions!