4.5 Solve Applications of Quadratic Equations

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for two consecutive odd integers.

Step 3: Name what we are looking for.

Let [latex]n=[/latex] the first odd integer.

Let [latex]n+2=[/latex] the next odd integer

Step 4: Translate into an equation. State the problem in one sentence.

“The product of two consecutive odd integers is [latex]195[/latex].”

The product of the first odd integer and the second odd integer is [latex]195[/latex].

Translate into the equation

[latex]\begin{eqnarray*}n(n+2)&=&195\\n^2+2n&=&195\end{eqnarray*}[/latex]

Step 5: Solve the equation. Distribute. Write the equation in standard form. Factor.

[latex]\begin{eqnarray*}n^2+2n-195&=&0\\{(n+15)}{(n-13)}&=&0\end{eqnarray*}[/latex]

Use the Zero Product Property. Solve each equation.

[latex]\begin{eqnarray*}\textstyle n+15&=&0\\n&=&-15\\\\n-13&=&0\\n&=&13\end{eqnarray*}[/latex]

There are two values of [latex]n[/latex] that are solutions. This will give us two pairs of consecutive odd integers for our solutions.

[latex]\begin{eqnarray*}\textstyle{\color{red}{\text{First}}}\;\text{odd integer is}\;{\color{red}{n}}{\color{red}{}}&=&{\color{red}{13}}.\\\text{Next odd integer is}\;{\color{blue}{n}}{\color{blue}{+}}{\color{blue}{2}}.\\{\color{red}{13}}{\color{blue}{+}}{\color{blue}{2}}&=&15\\\\{\color{red}{\text{First}}}\text{ odd integer is}\;{\color{red}{n}}{\color{red}{}}&=&{\color{red}{-}}{\color{red}{15}}.\\\text{Next odd integer is}\;{\color{blue}{n}}{\color{blue}{+}}{\color{blue}{2}}.\\{\color{red}{-}}{\color{red}{15}}{\color{blue}{+}}{\color{blue}{2}}&=&-13\end{eqnarray*}[/latex]

Step 6: Check the answer. Do these pairs work?

Are they consecutive odd integers?

[latex]\textstyle13,\;15\;=\;\text{yes}\;\;\;\;\;\;-13,-15\;=\;\text{yes}[/latex]

Is their product [latex]195[/latex]?

[latex]\begin{eqnarray*}\textstyle13\times15&=&195,\;\text{yes}.\\-13(-15)&=&195,\;\text{yes.}\end{eqnarray*}[/latex]

Step 7: Answer the question. 

Two consecutive odd integers whose product is [latex]195[/latex] are [latex]13[/latex], [latex]15[/latex] and [latex]-13[/latex], [latex]-15[/latex].

Step 1: Read the problem.

Draw a picture.

Step 2: Identify what we are looking for.

We are looking for the base and height.

Step 3: Name what we are looking for.

Let [latex]h[/latex] = the height of the triangle.

Let [latex]2h+4[/latex] = the base of the triangle

Step 4: Translate into an equation.

We know the area. Write the formula for the area of a triangle.

[latex]A=\frac{1}{2}bh[/latex]

Step 5: Solve the equation.

Substitute in the values.

[latex]120=\frac{1}{2}\left(2h+4\right)h[/latex]

Distribute.

[latex]120=h^2+2h[/latex]

This is a quadratic equation, rewrite it in standard form.

[latex]h^2+2h-120=0[/latex]

Factor.

[latex]\left(h-10\right)\left(h+12\right)=0[/latex]

Use the Zero Product Property.

[latex]h-10=0\phantom{\rule{1em}{0ex}}h+12=0[/latex]

Simplify.

[latex]h=10,\phantom{\rule{1.9em}{0ex}}\overline{)h=-12}[/latex]

Since [latex]h[/latex] is the height of a window, a value of [latex]h=−12[/latex] does not make sense.

The height of the triangle [latex]h=10[/latex].

The base of the triangle [latex]2h+4[/latex].

[latex]2\times10+4=24[/latex]

Step 6: Check the answer.

Does a triangle with height [latex]10[/latex] and base [latex]24[/latex] have area [latex]120[/latex]? Yes.

Step 7: Answer the question.

The height of the triangular window is [latex]10[/latex] feet and the base is [latex]24[/latex] feet.

Step 1: Read the problem.

Draw a picture.

Step 2: Identify what we are looking for.

We are looking for the length and width.

Step 3: Name what we are looking for.

Let [latex]w=[/latex] the width of the rectangle.

Let [latex]3w-1=[/latex] the length of the rectangle.

Step 4: Translate into an equation.

We know the area. Write the formula for the area of a rectangle.

[latex]A=L\times W[/latex]

Step 5: Solve the equation. Substitute in the values.

[latex]150 = (3w - 1)w[/latex]

Distribute.

[latex]150 = 3w^2 - w[/latex]

This is a quadratic equation; rewrite it in standard form.

Solve the equation using the Quadratic Formula.

[latex]\begin{eqnarray}{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{c}}{\color{red}{}}&=&{\color{red}{0}}\\3w^2-w-150&=&0\end{eqnarray}[/latex]

Identify the [latex]a,b,c[/latex] values.

[latex]{a\;=\;}{\color{blue}{3}}{\;,\;b\;=\;{\color{red}{-}}{\color{red}{1}}\;,\;c\;=\;{\color{green}{-}}}{\color{green}{150}}[/latex]

Write the Quadratic Formula.

[latex]w\;=\;\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex]

Then substitute in the values of [latex]a,b,c[/latex].

[latex]{w\;=\;}\frac{-{\color{red}{(-1}}{\color{red}{)}}\pm\sqrt{{{\color{red}{(}}{\color{red}{-}}{\color{red}{1}}{\color{red}{)}}}^2-4\times{\color{blue}{3}}\times({\color{green}{-150}}{)}}}{{2\times}{\color{blue}{3}}}[/latex]

Simplify.

[latex]\begin{eqnarray*}w&=&\frac{1\pm\sqrt{1+1800}}6\\w&=&\frac{1\pm\sqrt{1801}}6\end{eqnarray*}[/latex]

Rewrite to show two solutions.

[latex]w\;=\;\frac{1+\sqrt{1801}}6,\;w\;=\;\frac{1-\sqrt{1801}}6[/latex]

Approximate the answers using a calculator.

We eliminate the negative solution for the width.

[latex]\begin{eqnarray}w\;&\approx&\;7.2,\;\cancel{w\;\approx\;-6.9}\\\text{width}\;w\;&\approx&\;7.2\;\\\text{length}&\approx&\;3w\;-\;1\;\\&\approx&\;3(7.2)\;-\;1\\&\approx&\;20.6\end{eqnarray}[/latex]

Step 6: Check the answer.

Make sure that the answers make sense. Since the answers are approximate, the area will not come out exactly to [latex]150[/latex]

Step 7: Answer the question.

The width of the rectangle is approximately [latex]7.2[/latex] feet and the length is approximately [latex]20.6[/latex] feet.

Step 1: Read the problem.

Draw a picture.

Step 2: Identify what we are looking for.

We are looking for the height of the pole.

Step 3: Name what we are looking for.

The distance from the base of the pole to either stake is the same as the height of the pole.

Let [latex]x=[/latex] the height of the pole.

Let [latex]x=[/latex] the distance from pole to stake.

Each side is a right triangle. We draw a picture of one of them.

Step 4: Translate into an equation.

We can use the Pythagorean Theorem to solve for [latex]x[/latex].

Write the Pythagorean Theorem.

[latex]{a}^{2}+{b}^{2}={c}^{2}[/latex]

Step 5: Solve the equation. Substitute.

[latex]{x}^{2}+{x}^{2}={10}^{2}[/latex]

Simplify.

[latex]2{x}^{2}=100[/latex]

Divide by [latex]2[/latex] to isolate the variable.

[latex]\frac{2{x}^{2}}{2}=\frac{100}{2}[/latex]

Simplify.

[latex]{x}^{2}=50[/latex]

Use the Square Root Property.

[latex]x=±\sqrt{50}[/latex]

Simplify the radical.

[latex]x=±5\sqrt{2}[/latex]

Rewrite to show two solutions.

[latex]x=5\sqrt{2},\phantom{\rule{0.5em}{0ex}}\overline{)x=-5\sqrt{2}}[/latex]

If we approximate this number to the nearest tenth with a calculator, we find [latex]x\approx 7.1[/latex].

Step 6: Check the answer.

Check on your own in the Pythagorean Theorem.

Step 7: Answer the question.

The pole should be about [latex]7.1[/latex] feet tall.

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for the number of seconds, which is time.

Step 3: Name what we are looking for.

Let [latex]t=[/latex] the number of seconds.

Step 4: Translate into an equation. Use the formula.

[latex]h\;=\;-16t^2\;+\;v_0t[/latex]

Step 5: Solve the equation.

We know the velocity [latex]{v}_{0}[/latex] is [latex]130[/latex] feet per second.

The height is [latex]260[/latex] feet. Substitute the values.

[latex]260\;=\;-16t^2\;+\;130t[/latex]

This is a quadratic equation, rewrite it in standard form.

Solve the equation using the Quadratic Formula.

[latex]\begin{eqnarray}{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{c}}{\color{red}{}}&=&{\color{red}{0}}\\16t^2-130t+260&=&0\end{eqnarray}[/latex]

Identify the values of [latex]a,b,c.[/latex]

[latex]{\color{blue}{a}}{{\color{blue}{\;}}{\color{blue}{=}}{\color{blue}{\;}}{\color{blue}{16}}\;,\;\;{\color{darkred}{b}}{\color{darkred}{=}}{\color{darkred}{\;}}{\color{darkred}{-}}{\color{darkred}{130}}\;,\;{\color{green}{c}}{\color{green}{\;}}{\color{green}{=}}{\color{green}{\;}}}{\color{green}{260}}{}[/latex]

Write the Quadratic Formula.

[latex]t\;=\;\frac{-b\;\pm\;\sqrt{b^2}\;-\;4\;ac}{2a}[/latex]

Then substitute in the values of [latex]a,b,c[/latex].

[latex]{t\;=\;}\frac{-{\color{darkred}{(-130)}}\;\pm\;\sqrt{{{\color{darkred}{(-130)}}}^2}\;-\;4\;\times\;{\color{blue}{16}}\;\times\;({\color{green}{260}}{)}}{{2\;\times\;}{\color{blue}{16}}}[/latex]

Simplify.

[latex]\begin{eqnarray*}t&=&\frac{130\;\pm\;\sqrt{16,900}\;-\;16,640}{32}\\t&=&\frac{130\;\pm\;260}{32}\end{eqnarray*}[/latex]

Rewrite to show two solutions.

[latex]t\;=\;\frac{130\;+\;260}{32}\;,\;t\;=\;\frac{130\;-\;260}{32}[/latex]

Approximate the answer with a calculator.

[latex]t\;\approx\;4.6\;seconds,\;t\;\approx\;3.6\;seconds[/latex]

Step 6: Check the answer.

The check is left to you.

Step 7: Answer the question.

The firework will go up and then fall back down. As the firework goes up, it will reach [latex]260[/latex] feet after approximately [latex]3.6[/latex] seconds. It will also pass that height on the way down at [latex]4.6[/latex] seconds.

This is a uniform motion situation. A diagram will help us visualize the situation.

Step 1: We fill in the chart to organize the information. We are looking for the speed of the jet stream.

Let [latex]r=[/latex] the speed of the jet stream.

When the plane flies with the wind, the wind increases its speed and so the rate is [latex]450 + r[/latex].

When the plane flies against the wind, the wind decreases its speed and the rate is [latex]450 − r[/latex].

Step 2: Write in the rates. Write in the distances.

Since [latex]D={r}\times{t}[/latex], we solve for [latex]t[/latex] and get [latex]t=\frac{D}{r}[/latex].

Step 3: We divide the distance by the rate in each row, and place the expression in the time column.

Step 4: We know the times add to [latex]9[/latex] and so we write our equation.

[latex]\frac{2000}{450-r}+\frac{2000}{450+r}=9[/latex]

Step 5: We multiply both sides by the LCD.

[latex]\left(450-r\right)\left(450+r\right)\left(\frac{2000}{450-r}+\frac{2000}{450+r}\right)=9\left(450-r\right)\left(450+r\right)\phantom{\rule{5.2em}{0ex}}[/latex]

Step 6: Simplify.

[latex]2000\left(450+r\right)+2000\left(450-r\right)=9\left(450-r\right)\left(450+r\right)[/latex]

Step 7: Factor the [latex]2,000[/latex].

[latex]2000\left(450+r+450-r\right)=9\left({450}^{2}-{r}^{2}\right)[/latex]

Step 8: Solve.

[latex]2000\left(900\right)=9\left({450}^{2}-{r}^{2}\right)[/latex]

Step 9: Divide by [latex]9[/latex].

[latex]2000\left(100\right)={450}^{2}-{r}^{2}[/latex]

Step 10: Simplify.

[latex]\begin{eqnarray*}200000&=&202500-{r}^{2}\\-2500&=&-{r}^{2}\\50&=&r \text{The speed of the jet stream.}\end{eqnarray*}[/latex]

Step 11: Check:

Is [latex]50[/latex] mph a reasonable speed for the jet stream? Yes.

If the plane is travelling [latex]450[/latex] mph and the wind is [latex]50[/latex] mph

Tailwind:

[latex]\begin{eqnarray*}450+50&=&500\;\text{mph}\\\frac{2000}{500}&=&4\;\text{ hours}\end{eqnarray*}[/latex]

Headwind:

[latex]\begin{eqnarray*}450-50&=&\;\text{400 mph}\\\frac{2000}{400}&=&\;\text{5 hours}\end{eqnarray*}[/latex]

The times add to [latex]9[/latex] hours, so it checks. The speed of the jet stream was [latex]50[/latex] mph.

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Step 1: State variables.

Let [latex]x=[/latex] the number of hours for Press #2 to complete the job.

Step 2: Enter the hours per job for Press #1, Press #2, and when they work together.

The part completed by Press #1 plus the part completed by Press #2 equals the amount completed together.

Step 3: Translate to an equation.

[latex]\begin{array}{c}\underbrace{\text{Work completed by Press #1 + Press #2 = Together}}\\\frac1{x+12}+\frac1x=\frac18\end{array}[/latex]

Step 4: Solve.

[latex]\frac1{x\;+\;12}\;+\;\frac1x\;=\;\frac18[/latex]

Step 5: Multiply by the LCD, [latex]8x\left(x+12\right)[/latex].

[latex]{\color{darkred}{8}}{{\color{darkred}{x}}{\color{darkred}{(}}{\color{darkred}{x}}{\color{darkred}{\;}}{\color{darkred}{+}}{\color{darkred}{\;}}{\color{darkred}{12}}{\color{darkred}{)}}\left(\frac1{x\;+\;12}+\;\frac1x\right)\;=\;\left(\frac18\right){\color{darkred}{8}}{\color{darkred}{x}}{\color{darkred}{(}}{\color{darkred}{x}}{\color{darkred}{\;}}{\color{darkred}{+}}{\color{darkred}{\;}}{\color{darkred}{12}}}{\color{darkred}{)}}{}[/latex]

Step 6: Simplify.

[latex]\begin{eqnarray*}8x+8(x+12)&=&x(x+12)\\8x+8x+96&=&x^2+12x\\0&=&x^2-4x-96\end{eqnarray*}[/latex]

Step 7: Solve.

[latex]\begin{eqnarray*}0&=&(x-12)(x+96)\\\\x-12&=&0\;\\x&=&12\;\text{hours}\\\\x+8&=&0\\x&=&-8\;\text{hours}\end{eqnarray*}[/latex]

Since the idea of negative hours does not make sense, we use the value [latex]x=12[/latex].

[latex]\begin{eqnarray*}{\color{red}{12}}+12&=&24\;\text{hours}\\{\color{red}{12}}&=&12\;\text{hours}\end{eqnarray*}[/latex]

Step 8: Write our sentence answer.

Press #1 would take [latex]24[/latex] hours and Press #2 would take [latex]12[/latex] hours to do the job alone.