CHAPTER 03.01: QUADRATIC EQUATIONS: Example of Quadratic Equation
In this segment we're going to take an example of solving a quadratic equation. So we'll take an example, a physical example from physics, and see that how we can use the solution for a quadratic equation to solve certain problems. So let's suppose somebody's at the top of a building, so let's suppose this is a building here, and he or she throws a ball from the top of the building, and what they do is they throw it down at 50 miles per hour. So they're not just leaving the ball from rest, but simply throwing it down with some initial velocity, down from the top of the building to the bottom. The building is given to be 420 feet tall, so it's telling me that it's 420 feet tall. Now what we want to do is we want to find . . . find the time when the ball reaches ground, so the ground is right here, reaches ground? So that's what we want to be able to do, that somebody is throwing a ball at 50 miles per hour from the top of a building which is 420 feet tall, and we want to find out what is the time when it will hit the . . . hit the ground. So we know that the formula which governs this particular physical model is s is equal to u t plus one half g t squared, where s is the distance, u is the initial velocity, which, in our case, is 50 miles per hour, and we know g, which is the acceleration due to gravity. So you can see that, since we know the distance is 420 feet, we know the initial velocity, which is 50 miles per hour, and g, which is acceleration due to gravity, which we will take as 32.2 feet per second squared, so we can very well see that this is going to result in a quadratic equation. So let's go ahead and see that what kind of numbers we need to put in there. So we already know that s is 420 feet, we know u is 50 miles per hour, but since we have our s in feet, we're going to change this, and our acceleration due to gravity will be in feet per second squared, so g is 32.2 feet per second squared, we're going to convert hours into seconds and miles into feet, so I'll have to multiply this by 1 hour divided by 3600 seconds multiplied by 5280 feet divided by 1 mile, that's the conversion for a mile, and that's the conversion for 1 hour. So from here what I will get is that I will get a value for u which turns out to be 73.33 feet per second, so that's the conversion which is going to take place in order to be able to find out what the value of the initial velocity is, in feet per second. So now I'm going to substitute these values, so s is equal to u t plus one half g t squared, that's the formula which defines this particular model to find out the time it takes for the ball to reach ground, s is 420, u is 73.33, this is time, one half, g is 32.2, and then t squared, that's what I get there. So if I simplify this to write down it in the typical quadratic equation formula, I'll get 16.1 t squared plus 73.33 t minus 420 equal to 0, because this is 16.1, so 16.1 t squared plus 73.3 t, then 420 is taken to the other side, and equal to 0, so that turns out to be my quadratic formula. Now I'm going to use my quadratic equation formula solution to be able to find out the time when it hits the ground, so let's go ahead and see when does it hit the ground? So we know that we have 16.1 t squared plus 73.33 t minus 420 equal to 0. So I know that for a typical quadratic equation a x squared plus b x plus c equal to 0, x is -b plus minus square root of b squared minus 4 a c, divided by 2 a, that's the formula which we can derive, or which we can use. So in this case, the time will be equal to -b, which is -73.33, because b is 73.33 right there, plus minus b squared, which is 73.33 squared, minus 4 a is 16.1, and then c is -420, and then divided by 2 times a, which in this case is 16.1. So a is 16.1 . . . a is 16.1, b is 73.3, and c is -420, and using those numbers for the formula, we are able to get time equal to this, so we get two roots of this equation, 3.315 and -7.870. So what you have to do now is to, from the physics of the problem, you have to choose whether this is the right answer or whether this is the right answer. Since we are talking about time, t greater than or equal to 0, the acceptable root is 3.315 seconds, that is the acceptable root. So you get two roots of the equation, so from the physics of the problem, you've got to decide which one is the acceptable one, because you cannot have two roots of this equation which are acceptable, so only this one is valid because for time t greater than 0, now you have -7.870. So what I would like you to do as homework and see, does this root have a physical significance? It is not a . . . it is not an acceptable root, because you can only accept roots which are time greater than 0, so that makes this to be the acceptable root here, 3.315. But what I would like you do to as homework is to see if this -7.870, does this have some kind of physical significance in the problem itself? And that's the end of this segment. |