7.6 Quadratic Equations

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \\ \\ \\ \text{Multiply the binomials.}\hfill \\ \\ \\ \text{Write the quadratic equation in standard form.}\hfill \\ \\ \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & \begin{array}{ccc}\hfill \left(3x-8\right)\left(x-1\right)& =\hfill & 3x\hfill \\ \\ \\ \hfill 3x^2-11x+8& =\hfill & 3x\hfill \\ \\ \\ \hfill 3x^2-14x+8& =\hfill & 0\hfill \\ \\ \\ \hfill \left(3x-2\right)\left(x-4\right)& =\hfill & 0\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \\ \\ \text{Solve each equation.}\hfill \\ \\ \\ \end{array}\hfill & & & \begin{array}{}\\ \\ \\ \hfill 3x-2& =\hfill & 0\hfill & & & \hfill x-4& =\hfill & 0\hfill \\ \\ \\ \hfill 3x& =\hfill & 2\hfill & & & \hfill x& =\hfill & 4\hfill \\ \\ \\ \hfill x& =\hfill & \frac{2}{3}\hfill \end{array}\hfill \\ \\ \\ \text{Check your answers.}\hfill & & & \text{The check is left to you!}\hfill \end{array}$

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \\ \\ \\ \text{Write the quadratic equation in standard form.}\hfill \\ \\ \\ \text{Factor the greatest common factor first.}\hfill \\ \\ \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & \begin{array}{ccc}\hfill 4x^2& =\hfill & 16x+84\hfill \\ \\ \\ \hfill 4x^2-16x-84& =\hfill & 0\hfill \\ \\ \\ \hfill 4\left(x^2-4x-21\right)& =\hfill & 0\hfill \\ \\ \\ \hfill 4\left(x-7\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Use the Zero Product Property to set each factor to 0.}\hfill \\ \\ \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & \begin{array}{ccccccccccccc}\hfill 4& \ne \hfill & 0\hfill & & & \hfill x-7& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \\ \\ \hfill 4& \ne \hfill & 0\hfill & & & \hfill x& =\hfill & 7\hfill & & & \hfill x& =\hfill & \mathrm{-3}\hfill \end{array}\hfill \\ \\ \\ \text{Check your answers.}\hfill & & & \text{The check is left to you.}\hfill \end{array}$

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

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

$\begin{array}{cccccc}\hfill \text{If the first integer is}n=11& & & & & \hfill \text{If the first integer is}n=\mathrm{-12}\\ \\ \\ \hfill \text{then the next integer is}n+1& & & & & \hfill \text{then the next integer is}n+1\\ \\ \\ \hfill 11+1& & & & & \hfill \mathrm{-12}+1\\ \\ \\ \hfill 12& & & & & \hfill \mathrm{-11}\end{array}$

Step 6. Check the answer.

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

Step 7. Answer the question. The consecutive integers are $11,12$ and $\mathrm{-11},\mathrm{-12}$.