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, 12x2+11x+2=7 must first be changed to 12x2+11x+−5=0 by subtracting 7 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 30 square feet. If the length is 7 feet longer than the width, find the dimensions.
Answer: The formula for the area of a rectangle is Area=length⋅width, or A=l⋅w.
A=l⋅wwidth=wlength=w+7area=3030=(w+7)(w)
Multiply.
30=w2+7w
Subtract 30 from both sides to set the equation equal to 0.
w2+7w–30=0
Find two numbers whose product is −30 and whose sum is 7, and write the middle term as 10w–3w.
w2+10w–3w–30=0
Factor w out of the first pair and −3 out of the second pair.
w(w+10)−3(w+10)=0
Factor out w+10.
(w–3)(w+10)=0
Use the Zero Product Property to solve for w.
w−3=0orw+10=0w=3w=−10
The solution w=−10 does not work for this application, as the width cannot be a negative number, we discard the −10. So, the width is 3 feet.
The width = 3 feet
Substitute w=3 into the expression w+7 to find the length: 3+7=10.
The length is 3+7=10 feet
Answer
The width of the garden is 3 feet, and the length is 10 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 200 feet above the ground. Its starting velocity (also called initial velocity) is −10 feet per second. The negative value means it is heading toward the ground.
The equation h=−16t2−10t+200 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 0. Substitute 0 for h.
h=−16t2−10t+2000=−16t2−10t+200−16t2−10t+200=0
This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, x=2a−b±b2−4ac. In this case, the variable is t rather than x. a=−16,b=−10, and c=200.
t=2(−16)−(−10)±(−10)2−4(−16)(200)
Simplify. Be very careful with the signs.
t=−3210±100+12800=−3210±12900
Use a calculator to find both roots.
t is approximately −3.86 or 3.24.
Consider the roots logically. One solution, −3.86, cannot be the time because it is a negative number. The other solution, 3.24 seconds, must be when the ball hits the ground.
The ball hits the ground approximately 3.24 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 4 ft ×5 ft. He has 10 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 x 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.
Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be 5+2x, and the width will be 4+2x.
(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.)
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=(4+2x)(5+2x)–(4)(5)
There are 10 sq ft of fabric for the border, so set the area of border to be 10.
10=(4+2x)(5+2x)–20
Multiply (4+2x)(5+2x).
10=20+8x+10x+4x2–20
Simplify.
10=18x+4x2
Subtract 10 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.
0=18x+4x2−10or4x2+18x−10=02(2x2+9x−5)=0
Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, 2x2+9x–5=0.
2(2x2+9x−5)=022(2x2+9x−5)=202x2+9x−5=0
Use the Quadratic Formula. In this case, a=2,b=9, and c=−5.
x=2a−b±b2−4acx=2(2)−9±92−4(2)(−5)
Simplify.
x=4−9±121=4−9±11
Find the solutions, making sure that the ± is evaluated for both values.
x=4−9+11=42=21=0.5orx=4−9−11=4−20=−5
Ignore the solution x=−5, since the width could not be negative.
The width of the border should be 0.5 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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