Algebra - Quadratic Equations

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Section 2.5 : Quadratic Equations - Part I

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3. Solve the following quadratic equation by factoring.

\[{y^2} = 11y - 28\]

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The first thing we need to do is get everything on one side of the equation and then factor the quadratic.

\[\begin{align*}{y^2} - 11y + 28 & = 0\\ \left( {y - 4} \right)\left( {y - 7} \right) & = 0\end{align*}\] Show Step 2

Now all we need to do is use the zero factor property to get,

\[\begin{array}{*{20}{c}}{y - 4 = 0}\\{y = 4}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{y - 7 = 0}\\{y = 7}\end{array}\]

Therefore the two solutions are : \(\require{bbox} \bbox[2pt,border:1px solid black]{{y = 4\,\,{\mbox{and }}y = 7}}\)

We’ll leave it to you to verify that they really are solutions if you’d like to by plugging them back into the equation.