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Section 2.5 : Quadratic Equations - Part I
9. Use factoring to solve the following equation.
\[{t^5} = 9{t^3}\]Show All Steps Hide All Steps
Start SolutionDo not let the fact that this equation is not a quadratic equation convince you that you can’t do it! Note that we move both terms to one side we can factor a \({t^3}\)out of the equation. Doing that gives,
\[\begin{align*}{t^5} - 9{t^3} & = 0\\ {t^3}\left( {{t^2} - 9} \right) & = 0\end{align*}\]The quantity in the parenthesis is a quadratic and we can factor it. The full factoring of the equation is then,
\[{t^3}\left( {t - 3} \right)\left( {t + 3} \right) = 0\] Show Step 2Now all we need to do is use the zero factor property to get,
\[\begin{array}{*{20}{c}}{{t^3} = 0}\\{t = 0}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{t - 3 = 0}\\{t = 3}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{t + 3 = 0}\\{t = - 3}\end{array}\]Therefore the three solutions are : \(\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0,\,\,t = 3\,\,{\mbox{and }}t = - 3}}\)