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Section 2.8 : Applications of Quadratic Equations

3. Two people can paint a house in 14 hours. Working individually one of the people takes 2 hours more than it takes the other person to paint the house. How long would it take each person working individually to paint the house?

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First, let Person A be the faster of the two painters and let t be the amount of time it takes to paint the house by himself. Next, let Person B be the slower of the two painters and so it will take this person \(t + 2\) hours to paint the house by himself.

Show Step 2

Working together they can paint the house in 14 hours so we have the following word equation for them working together to paint the house.

\[\left( \begin{array}{c}{\mbox{Portion of job }}\\ {\mbox{ done by Person A}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Portion of job}}\\ {\mbox{done by Person B}}\end{array} \right) = 1{\mbox{ Job}}\]

We know that Portion of Job = Work Rate X Work Time so this gives the following word equation.

\[\begin{align*}\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person A}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{of Person A}}\end{array} \right) + \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person B}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{of Person B}}\end{array} \right) & = 1\\\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person A}}\end{array} \right)\left( {14} \right) + \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person B}}\end{array} \right)\left( {14} \right) & = 1\end{align*}\] Show Step 3

Now we need the work rate of each person which we can get from their individual painting times as follows,

\[\begin{align*}\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person A}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{of Person A}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person A}}\end{array} \right)\left( t \right) & = 1\\ & \Rightarrow \hspace{0.25in}{\mbox{Work Rate of Person A = }}\frac{1}{t}\end{align*}\] \[\begin{align*}\left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person B}}\end{array} \right)\left( \begin{array}{c}{\mbox{Work Time}}\\ {\mbox{of Person B}}\end{array} \right) = \left( \begin{array}{c}{\mbox{Work Rate}}\\ {\mbox{of Person B}}\end{array} \right)\left( {t + 2} \right) & = 1 \\ & \Rightarrow \hspace{0.25in}{\mbox{Work Rate of Person B = }}\frac{1}{{t + 2}}\end{align*}\] Show Step 4

Plugging these into the word equation from Step 2 we arrive at the following equation.

\[\begin{align*}\left( {\frac{1}{t}} \right)\left( {14} \right) + \left( {\frac{1}{{t + 2}}} \right)\left( {14} \right) & = 1\\ \frac{{14}}{t} + \frac{{14}}{{t + 2}} & = 1\end{align*}\] Show Step 5

To solve this we know that we’ll need to multiply by the LCD, \(t\left( {t + 2} \right)\) in this case, to clear the denominators. Doing this gives,

\[\begin{align*}14\left( {t + 2} \right) + 14t & = t\left( {t + 2} \right)\\ 28t + 28 & = {t^2} + 2t\\ {t^2} - 26t - 28 & = 0\end{align*}\]

After some simplification we arrive a fairly simple quadratic equation to solve. Using the quadratic formula gives,

\[L = \frac{{ - \left( { - 26} \right) \pm \sqrt {{{\left( { - 26} \right)}^2} - 4\left( 1 \right)\left( { - 28} \right)} }}{{2\left( 1 \right)}} = \frac{{26 \pm \sqrt {788} }}{2}\] Show Step 6

Reducing the two values we got in the previous steps to decimals we arrive at the following two solutions to the quadratic equation from Step 2.

\[t = \frac{{26 - \sqrt {788} }}{2} = - 1.0357\hspace{0.25in}\hspace{0.25in}t = \frac{{26 + \sqrt {788} }}{2} = 27.0357\]

The first solution to the equation doesn’t make any sense since it is negative (we are working with time and so it’s safe to assume we are starting at \(t = 0\) after all!) so that means the second is the answer we need.

This means that Person A can paint the house in 27.0357 hours while Person B can paint the house in 29.0357 hours (two hours more than Person A).