Section 2.5 : Quadratic Equations - Part I
10. Use factoring to solve the following equation.
\[\frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\]Show All Steps Hide All Steps
Start SolutionThis 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 2We 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!