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Solving Quadratic Inequalities Representing Motion

Solving Quadratic Inequalities Representing Motion

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Author: Colleen Atakpu
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This lesson covers solving quadratic inequalities representing motion.

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Today we're going to talk about solving quadratic inequalities representing motion. So we're going to look at a real-world situation and see how we can use quadratic inequalities to answer questions about that situation. So for my first example, I've got a graph of an object flying through the air. The equation that matches the graph is y is equal to negative 10x squared plus 50x.

So let's say I wanted to know how long the object-- in this case, the rocket-- will be above 50 feet. So I'm really looking for how long, or between what two time intervals, the height of my rocket is above 50. So if I find 50 on my graph and come over to my parabola, I can see that the time intervals where the rocket will be above 50 is approximately between 1 and 1/2 seconds, and 3 and 1/2 seconds, approximately, just by estimating from my graph. So we can solve a quadratic inequality to number one, verify our estimation and, number two, to find a more accurate answer. So let's see how we would do that.

So here's my equation that represents the path of my rocket. I want to know when the rocket is going to be higher than 50 feet. And I know that 50 is the height of my rocket. So I want to know when this expression, which is the height of my rocket at any time for x, is going to be bigger than 50. So I want this expression to be greater than 50, which is the same as 50 being less than this expression.

So now that I have my inequality, I can go ahead and solve that for x. And to do that, I'm going to write it as an equation set equal to 0. However, I'm going to then solve that equation using completing the square. And the first step in completing the square is to bring the constant term to the other side. So I'm actually going to write this as an equation. But instead of setting it equal to 0, I'm going to leave the constant term on this side.

So I'll have 50 equals negative 10x squared plus 50x. So now I'm going to solve this again, as I said, using completing the square. And because I already have my constant term on this side, I'm going to go ahead and divide by my coefficient in front of my x squared term to cancel it out.

And this will become negative 5. Here, I'll just have x squared. And then I'll have minus 5x.

My next step is to take my coefficient of my x term. And I'm going to divide that by 2, which will give me negative 2.5. Then I'm going to take that value and square it, which is going to give me a positive 6.25.

I'm going to take 6.25 and add it to both sides of my equation. On this side, I'll be left with 1.25. And here, I'll have x squared minus 5x plus 6.25. Now this I recognize as a perfect square trinomial, which means I can write it as a binomial squared.

And that binomial is going to be x minus 2.5. The negative 2.5 came from this value. As I said, a binomial squared, and then I'll have 1.25 on the other side of my equation.

Now I can continue solving for x by taking the square root of both sides. And when I do that, I know that my value here can have either a positive or a negative value when I square it. So that means I'll have a positive or negative value here. And these will cancel. So now I have positive or negative 1.1 approximately is equal to x minus 2.5.

So now that I have this equation, I can separate it into two separate equations using the positive and negative values. So my first equation will be 1.1 equals x minus 2.5. And my second equation is negative 1.1 equals x minus 2.5. These are both approximately equal to.

Solving both of these equations to find my solutions for x, I see that my first solution will be x is approximately equal to 3.6 and that my second equation will be that x is approximately equal to 1.4. So now that I have my solutions for x for my equation, I can find the range of values that satisfy my inequality by using these two solutions to create intervals on a number line and test a value in each interval.

So I'm going to create a number line. Here, I'll have 0, 1.4, and 3.6. So I can use these as boundaries-- these values as boundaries for my interval.

And I'm going to test a value. So for this first interval, I'll use 0 as my test value. Here, I'll use 2. And here, I'll go ahead and use 4.

So my first test value of 0-- I'm going to see if that satisfies the inequality by substituting 0 in for x into the inequality to see if it makes a true statement. So here I'll have 50 is less than negative 10 times 0 squared plus 50 times 0. Simplifying, this will give me 50 should be less than 0 plus 0, which I see is incorrect. So the value of 0 does not satisfy my inequality. So this interval is not part of my solution set.

Next I'll try my test value of 2. So substituting 2 into my inequality for x, I have, simplifying this, 2 squared is 4. 4 times negative 10 will give me negative 40. 50 times 2 will give me 100. And negative 40 plus 100 will give me 60.

So I see that 60 is greater than 50. So that means the value of 2 does satisfy my inequality. So I know that this interval is part of my solution set.

Testing my last value of 4, I'll again substitute 4 in for my x in my inequality to see if it yields a true statement. 4 squared is 16, times negative 10 will give me negative 160. 50 times 4 is 200. Negative 160 plus 200 will give me a positive 40, which is not greater than 50. So I see that 4 does not satisfy my inequality, which means that this interval is not a part of my solution set.

So I found that the interval that is a part of my solution set is here, in between but not including the values of 1.4 and 3.6. So I found that my rocket is going to be greater than 50 feet between the times of 1.4 and 3.6 seconds.

OK. So now let's say for the same situation we want to know how long the rocket will be above 100 feet. So again, we can start by getting an initial estimate by looking at our graph. If I go to 100 feet for my height, I can see that my parabola will never reach 100 feet, or my rocket will never reach 100 feet. So there is no solution for this question. So we can, again, verify that there's no solution by solving the quadratic inequality algebraically to see that the answer to this question is no solution.

So I've taken my equation for the path of the rocket. I'm going to write an inequality. I want to find when the rocket is higher than 100 feet.

So the height I know is y. I'm going to substitute 100 there. And I want to find when the rocket is greater than that. So I have 100 is less than negative 10x squared plus 50x.

I'm going to solve this using the quadratic formula. And I'm going to write this as an equation set equal to 0. So to set it equal to 0, I'm going to subtract 100 from both sides. So I'll have 0-- and I'll write it as an equation-- equals negative 10x squared plus 50x minus 100.

Using the quadratic formula, I need to identify values for a, b, and c. I know a is negative 10, b is 50, and c is negative 100. So substituting into my formula for the quadratic formula, I'll have negative 50 plus or minus the square root of 50 squared minus 4 times negative 10 times negative 100 over 2 times negative 10. And that will be equal to x.

I'm going to start by simplifying underneath my radical. 50 squared will give me 2,500. 4 times negative 10 times negative 100 is going to give me a positive 4,000 all over-- I'm going to simplify this-- negative 20.

So now I see that my discriminant-- the value underneath my radical-- is going to be negative. Because 2,500 minus 4,000 is going to give me a negative number. So I can stop here. Because I know that this is going to give me a non-real solution or an imaginary solution, which matches up with what we originally saw by looking at our graph-- that the parabola will never reach more than 100 feet.

So let's go over our key points from today. A quadratic equation can be used to model the path of an object rising and falling due to gravity. If x represents time and y represents height, the quadratic inequality can be used to answer questions about the object's path, such as the time interval that an object is greater than or less than a certain height.

So I hope that these key points and examples helped you understand a little bit more about solving quadratic inequalities representing motion. Keep using your notes and keep on practicing, and soon you'll be a pro. Thanks for watching.

Notes on "Solving Quadratic Inequalities Representing Motion"

Overview

(00:00 - 00:10) Introduction

(00:11 - 06:35) Quadratic Inequality Example

(06:36 - 09:01) Quadratic Inequality with No Solution Example

(09:02 - 09:38) Summary

Key Formulas

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Key Terms

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