Please respond to both POST1 and POST2 in at least 200 words each. I have included the original post as a reference when answering the post.

Original Post:

For this discussion, complete the following tasks:

1. Use an outside source to search for a quadratic equation that models something from your daily life.
2. Solve the equation in two ways.
3. Discuss which method you liked better and why.
4. In your responses to peers, compare and contrast your preferences for how to solve quadratic equations.

POST1:

The quadratic equation is used to describe the relationship betweenobjects that are thrown and how long it takes to hit the ground.

In spirit of playoffs, I decided to determine how long a punt takesto hit the ground. As I discovered in combing through ways of findingrelevant information, the science of a punt is pretty fascinating. Forinstance, using a right triangle you can determine the necessary anglefor a kick to account for distance. That is a function by itself. Forthis discussion I will just focus on the quadratic equation to determinehow long the average punt takes to hit the ground. This would berelevant to know because members of a special team should be able tocover this distance before the ball hits the return man.

The formula:

y=-4.9t^2+V0(t)+H0 where -4.9t^2 is the constant of gravity, V0 isthe initial velocity in Meters/Second and H0 is the starting height.

This formula is created due to the effect of the football travelingin a parabola. When the ball is first kicked it travels upward. Startingfrom the initial launch, the effect of gravity begins to counteract theinitial velocity, eventually causing the football to fall. The actualfunction output is height.

Y=-4.9t^2+39.2t+2.45

Solving by factoring: I don’t like this method. It usually doesn’t work to solve complex equations.

Solving by completing the square:

y=-4.9t^2+39.2t+2.45

-4.9t^2+39.2t+2.45=0

-2t^2+16t+1=0

-2t^2+16t+1=0 put into the form t^2+2at+a^2+b

-2(t^2-8t+16-.5-16)=0

-2(t-4)^2-.5-16=0

-2(t-4)^2-16.5=0

(t-4)^2=-16.52/2

t= +/-(√16.52+4)/2

t=8.062

t=-0.062

In this case, total time is equal to 8.062 seconds when the ball lands (ignoring -0.062 because it isn’t relevant)

Solving using the quadratic equation:

y=-4.9t2+39.2t+2.45 we set theequation equal to 0 and divide by negative like terms and to give uswhole and positive term value 0=2t^2-16t-1

Using the quadratic formula we plug in the values of a, b and c

t=(-(-16)^2 +/-√(-16)2-42(-1)22)/2*2

Solving this equation we get the same answers: t=-0.062 and t=8.062

I like completing the square and quadratic formula because they aresimilar and allow to manipulate numbers that don’t necessarily factoreasily

Source:

Quadratic Applications: Projectile Motion (n.d). Retrieved from https://www.brainfuse.com/jsp/alc/resource.jsp?s=gre&c=36714&cc=108826

POST2:

Over the weekend, I went on a canoe trip. I spent the morningtraveling upstream and the afternoon traveling downstream. Overall, Ispent 4 hours traveling 10 miles. I know that the stream I was canoeingflows at 3 MPH. Now I need to find out at what speed I traveled.

(distance upstream/speed – current) + (distance downstream/speed + current) = 4 hours

(10 / x -3) + (10 / x+3) = 4 hours

10 (x+3) + 10 (x-3) = 4 (x+3)(x-3)

10x + 30 + 10x – 30 = 4(X^2 – 9)

20X = 4x^2 – 9

F(x) = 4x^2 – 20x – 9

Quadratic Formula:

X = (-b +/- √(b^2-4ac)) / 2a

X = (-(-20)+/- √(-(-20)-4(4)(-20))) / ((2)(4))

X = (20 +/- √(400 + 144))) / 8

X = (20 +/- √(544))/8

X = 5.41

X = -0.41

Therefore, since I cannot travel at a negative speed, I traveled at 5.41 MPH.

Completing the square:

4x^2 – 20x – 9 = 0

(4/4)x^2 – (20/4)x – (9/4) = (0/4)

X^2 -5x – (9/4) = 0

X^2 – 5x = (9/4)

X^2 – 5x + (25/4) = (9/4) + (25/4)

(X – (5/2))^2 = (34/4)

X – (5/2) = +/-√(17/2)

X = (5/2) +/- √(17/2)

X = 5.41

X = -0.41

Overall, I preferred to use the quadratic formula. I found that itwas easier for me to plug numbers into a formula versus what I saw as amassive amount of work by completing the square.