Solving Applications Modeled By Quadratic Equations

The product of two consecutive integers is \(132\). Find the integers.

Solution

\(\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{We are looking for two consecutive integers.}\hfill \\ \mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & & & \text{Let}\phantom{\rule{0.2em}{0ex}}n=\text{the first integer}\hfill \\ & & & & & n+1=\phantom{\rule{0.2em}{0ex}}\text{the next consecutive integer}\hfill \\ \mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into an equation. Restate the}\hfill & & & & & \text{The product of the two consecutive integers is 132.}\hfill \\ \text{problem in a sentence.}\hfill & & & & & & \\ & & & & & \text{The first integer times the next integer is 132.}\hfill \\ \begin{array}{c}\text{Translate to an equation.}\hfill \\ \mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the equation.}\hfill \\ \text{Bring all the terms to one side.}\hfill \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & & & \begin{array}{ccc}\hfill n\left(n+1\right)& =\hfill & 132\hfill \\ \hfill {n}^{2}+n& =\hfill & 132\hfill \\ \hfill {n}^{2}+n-132& =\hfill & 0\hfill \\ \hfill \left(n-11\right)\left(n+12\right)& =\hfill & 0\hfill \end{array}\hfill \\ \begin{array}{c}\text{Use the zero product property.}\hfill \\ \text{Solve the equations.}\hfill \end{array}\hfill & & & & & \phantom{\rule{4em}{0ex}}\begin{array}{cccccccccc}\hfill n-11& =\hfill & 0\hfill & & & & & \hfill n+12& =\hfill & 0\hfill \\ \hfill n& =\hfill & 11\hfill & & & & & \hfill n& =\hfill & -12\hfill \end{array}\hfill \end{array}\)

There are two values for \(n\) that are solutions to this problem. So there are two sets of consecutive integers that will work.

\(\begin{array}{cccccc}\hfill \text{If the first integer is}\phantom{\rule{0.2em}{0ex}}n=11& & & & & \hfill \text{If the first integer is}\phantom{\rule{0.2em}{0ex}}n=-12\\ \hfill \text{then the next integer is}\phantom{\rule{0.2em}{0ex}}n+1& & & & & \hfill \text{then the next integer is}\phantom{\rule{0.2em}{0ex}}n+1\\ \hfill 11+1& & & & & \hfill -12+1\\ \hfill 12& & & & & \hfill -11\end{array}\)

Step 6. Check the answer.

The consecutive integers are \(11,12\) and \(-11,-12\). The product \(11·12=132\) and the product \(-11\left(-12\right)=132\). Both pairs of consecutive integers are solutions.

Step 7. Answer the question. The consecutive integers are \(11,12\) and \(-11,-12\).