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Study Guides > ALGEBRA / TRIG I

Applications of Quadratic Equations

Learning Outcomes

  • Solve application problems involving quadratic equations
Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. When working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation.  Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge. Any time we solve a quadratic equation, it is important to make sure that the equation is equal to zero so that we can correctly apply the techniques we have learned for solving quadratic equations. For example, [latex]12x^{2}+11x+2=7[/latex] must first be changed to [latex]12x^{2}+11x+-5=0[/latex] by subtracting [latex]7[/latex] from both sides. In our first example, we will apply the Zero Product Principal to a quadratic equation to solve an equation involving the area of a garden.

Example

The area of a rectangular garden is [latex]30[/latex] square feet. If the length is [latex]7[/latex] feet longer than the width, find the dimensions.

Answer: The formula for the area of a rectangle is [latex]\text{Area}=\text{length}\cdot\text{width}[/latex], or [latex]A=l\cdot{w}[/latex].

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,A=l\cdot{w}\\\,\text{width}=w\\\text{length}=w+7\\\,\,\,\,\,\text{area}=30\\\\30=\left(w+7\right)\left(w\right)\end{array}[/latex]

Multiply.

[latex]30=w^{2}+7w[/latex]

Subtract [latex]30[/latex] from both sides to set the equation equal to [latex]0[/latex].

[latex]w^{2}+7w–30=0[/latex]

Find two numbers whose product is [latex]−30[/latex] and whose sum is [latex]7[/latex], and write the middle term as [latex]10w–3w[/latex].

[latex]w^{2}+10w–3w–30=0[/latex]

Factor w out of the first pair and [latex]−3[/latex] out of the second pair.

[latex]w\left(w+10\right)-3\left(w+10\right)=0[/latex]

Factor out [latex]w+10[/latex].

[latex]\left(w–3\right)\left(w+10\right)=0[/latex]

Use the Zero Product Property to solve for w.

[latex]\begin{array}{l}w-3=0\,\,\,\,\,\,\text{or}\,\,\,\,\,\,w+10=0\\\,\,\,\,\,\,\,w=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=-10\end{array}[/latex]

The solution [latex]w=−10[/latex] does not work for this application, as the width cannot be a negative number, we discard the [latex]−10[/latex]. So, the width is [latex]3[/latex] feet.

The width = [latex]3[/latex] feet

Substitute [latex]w=3[/latex] into the expression [latex]w+7[/latex] to find the length: [latex]3+7=10[/latex].

The length is [latex]3+7=10[/latex] feet

Answer

The width of the garden is [latex]3[/latex] feet, and the length is [latex]10[/latex] feet.

In the example in the following video, we present another area application of factoring trinomials. https://youtu.be/PvXsWZp588o

Applying the Quadratic Formula

A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation is not completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough. In our next example, we will determine how long it takes for a ball to hit the ground when falling from a building.  This time, we will solve the quadratic equation using the quadratic formula.

Example

A ball is thrown off a building from [latex]200[/latex] feet above the ground. Its starting velocity (also called initial velocity) is [latex]−10[/latex] feet per second. The negative value means it is heading toward the ground. The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?

Answer: When the ball hits the ground, the height is [latex]0[/latex]. Substitute [latex]0[/latex] for h.

[latex]\begin{array}{c}h=-16t^{2}-10t+200\\0=-16t^{2}-10t+200\\-16t^{2}-10t+200=0\end{array}[/latex]

This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].

[latex] t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]

Simplify. Be very careful with the signs.

[latex] \begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]

Use a calculator to find both roots.

t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].

Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground. The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.

In the next video, we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground. https://youtu.be/RcVeuJhcuL0 The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.

Example

Bob made a quilt that is [latex]4[/latex] ft [latex]\times[/latex] [latex]5[/latex] ft. He has [latex]10[/latex] sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)

Answer: Sketch the problem. Since you do not know the width of the border, you will let the variable [latex]x[/latex] represent the width. In the diagram. The original quilt is indicated by the red rectangle. The border is the area between the red and blue lines. A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x. Since each side of the original [latex]4[/latex] by [latex]5[/latex] quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex]. (Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.) A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x. You are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]

There are [latex]10[/latex] sq ft of fabric for the border, so set the area of border to be [latex]10[/latex].

[latex]10=\left(4+2x\right)\left(5+2x\right)–20[/latex]

Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].

[latex]10=20+8x+10x+4x^{2}–20[/latex]

Simplify.

[latex]10=18x+4x^{2}[/latex]

Subtract [latex]10[/latex] from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.

[latex]\begin{array}{c}0=18x+4x^{2}-10\\\\\text{or}\\\\4x^{2}+18x-10=0\\\\2\left(2x^{2}+9x-5\right)=0\end{array}[/latex]

Factor out the greatest common factor, [latex]2[/latex], so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].

[latex]\begin{array}{r}2\left(2x^{2}+9x-5\right)=0\\\\\frac{2\left(2x^{2}+9x-5\right)}{2}=\frac{0}{2}\\\\2x^{2}+9x-5=0\end{array}[/latex]

Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].

[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]

Simplify.

[latex] x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]

Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.

[latex]\begin{array}{c}x=\frac{-9+11}{4}=\frac{2}{4}=\frac{1}{2}=0.5\\\\\text{or}\\\\x=\frac{-9-11}{4}=\frac{-20}{4}=-5\end{array}[/latex]

Ignore the solution [latex]x=−5[/latex], since the width could not be negative. The width of the border should be [latex]0.5[/latex] ft.

Try It

[ohm_question]54696[/ohm_question]
Our next video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt. https://youtu.be/Zxe-SdwutxA In this last video, we show how to use the quadratic formula to solve an application involving a picture frame. https://youtu.be/AlIoxXQ-V50

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