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### Section 2.5 : Quadratic Equations - Part I

9. Use factoring to solve the following equation.

${t^5} = 9{t^3}$

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Do 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 2

Now 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 : $$\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0,\,\,t = 3\,\,{\mbox{and }}t = - 3}}$$